Chapter 6: The Normal Distribution and The Central Limit Theorem
Chapter 6 Homework
Section 6.1 Homework
Use the following information to answer the next two exercises:
The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
1. What is the median recovery time?
a. 2.7
b. 5.3
c. 7.4
d. 2.1
Solution
b. 5.3
2. What is the z-score for a patient who takes ten days to recover?
a. 1.5
b. 0.2
c. 2.2
d. 7.3
Solution
c. 2.2
The length of time to find it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true?
- The data cannot follow the uniform distribution.
- The data cannot follow the exponential distribution.
- The data cannot follow the normal distribution.
- 1 only
- 2 only
- 3 only
- 1, 2, and 3
Solution
2 only
The heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with mean, [latex]\mu= 79 \text{ inches}[/latex] and a standard deviation, [latex]\sigma= 3.89 \text{ inches}[/latex]. For each of the following heights, calculate the z-score and interpret it using complete sentences.
- 77 inches
- 85 inches
- If an NBA player reported his height had a z-score of 3.5, would you believe him? Explain your answer.
Solution
- Use the z-score formula. [latex]z = –0.5141[/latex]. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average.
- Use the z-score formula. [latex]z = 1.5424[/latex]. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average.
- [latex]\text{Height} = 79 + 3.5(3.89) = 90.67 \text{ inches}[/latex], which is over 7.7 feet tall. There are very few NBA players this tall so the answer is no, not likely.
The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean [latex]\mu = 125[/latex] and standard deviation [latex]\sigma = 14[/latex]. Systolic blood pressure for males follows a normal distribution.
- Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters.
- If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him?
Solution
- [latex]z \approx -1.8[/latex] and [latex]z \approx 1.8[/latex]
- I would tell him that 2.5 standard deviations below the mean would give him a blood pressure reading of 90, which is below the range of 100 to 150.
Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean [latex]\mu = 125[/latex] and standard deviation [latex]\sigma = 14[/latex]. If X = a systolic blood pressure score then [latex]X \sim N(125, 14)[/latex].
- Which answer(s) is/are correct?
- Kyle’s systolic blood pressure is 175.
- Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age.
- Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age.
- Kyle’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.
- Calculate Kyle’s blood pressure.
Solution
- Kyle’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.
- Kyle’s blood pressure is equal to [latex]125 + (1.75)(14) = 149.5[/latex].
Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean [latex]\mu = 10.2 \text{ kg}[/latex] and standard deviation [latex]\sigma= 0.8 \text{ kg}[/latex]. Weights are normally distributed. [latex]X \sim N(10.2, 0.8)[/latex].
Calculate the z-scores that correspond to the following weights and interpret them:
- 11 kg
- 7.9 kg
- 12.2 kg
Solution
- [latex]z = 1[/latex]; A child who weighs 11 kg is one standard deviation above the mean of 10.2 kg.
- [latex]z = -2.875[/latex]; A child who weighs 7.9 kg is 2.875 standard deviations below the mean of 10.2 kg.
- [latex]z = 2.5[/latex]; A child who weighs 12.2 kg is 2.5 standard deviation above the mean of 10.2 kg.
In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean [latex]\mu = 520[/latex] and standard deviation [latex]\sigma = 115[/latex].
- Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence.
- What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score?
- For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took?
Solution
Let X = an SAT math score and Y = an ACT math score.
- [latex]X = 720[/latex]; [latex]\frac{\text{720 – 520}}{\text{15}} = 1.74[/latex]. The exam score of 720 is 1.74 standard deviations above the mean of 520.
- [latex]z = 1.5[/latex];The math SAT score is [latex]520 + 1.5(115) \approx 692.5[/latex]. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.
- [latex]\frac{X - \mu }{\sigma} = \frac{700 – 514}{117} =\approx 1.59[/latex], the z-score for the SAT. [latex]\frac{Y - \mu }{\sigma } = \frac{30 - 21}{5.3} \approx 1.70[/latex], the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z-score).
Section 6.2 Homework
Use the following information to answer the next two exercises:
The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the probability of spending more than two days in recovery?
a. 0.0580
b. 0.8447
c. 0.0553
d. 0.9420
Solution
d. 0.9420
The 90th percentile for recovery times is?
a. 8.89
b. 7.07
c. 7.99
d. 4.32
Solution
c. 7.99
Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space?
a. Yes
b. No
c. Unable to determine
Solution
a. Yes
Find the probability that it takes at least eight minutes to find a parking space.
a. 0.0001
b. 0.9270
c. 0.1862
d. 0.0668
Solution
d. 0.0668
Seventy percent of the time, it takes more than how many minutes to find a parking space?
a. 1.24
b. 2.41
c. 3.95
d. 6.05
Solution
c. 3.95
According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement.
- Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically.
- The middle 40% of heights fall between what two values? Sketch the graph, and write the probability statement.
Solution
- [latex]X \sim N(66, 2.5)[/latex]
- 0.5404
- No, the probability that an Asian male is over 72 inches tall is 0.0082
- The middle 40% of heights scores falls between 64.7 and 67.3 inches.
IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual.
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement.
- MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement.
- The middle 50% of IQs fall between what two values? Sketch the graph and write the probability statement.
Solution
- [latex]X \sim N(100, 15)[/latex]
- The probability that a person has an IQ greater than 120 is 0.0918.
- A person has to have an IQ over 130 to qualify for MENSA.
- The middle 50% of IQ scores falls between 89.95 and 110.05.
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories.
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the percent of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined.
- Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement.
Solution
- [latex]X \sim N(36, 10)[/latex]
- The probability that a person consumes more than 40% of their calories as fat is 0.3446.
- Approximately 25% of people consume less than 29.26% of their calories as fat.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet.
- If X = distance in feet for a fly ball, then [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability.
- Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.
Solution
- [latex]X \sim N(250, 50)[/latex]
- The probability that a fly ball travels less than 220 feet is 0.2743.
- Eighty percent of the fly balls will travel less than 292 feet.
In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.
- In words, define the random variable X.
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.
- What percent of the children spend over ten hours per day unsupervised?
- Seventy percent of the children spend at least how long per day unsupervised?
Solution
- X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
- [latex]X \sim N(3, 1.5)[/latex]
- The probability that the child spends less than one hour a day unsupervised is 0.0918.
- The probability that a child spends over ten hours a day unsupervised is less than 0.0001.
- 2.21 hours
In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district.
- State the approximate distribution of X.
- Is 1,956.8 a population mean or a sample mean? How do you know?
- Find the probability that a randomly selected district had fewer than 1,600 votes for President Clinton. Sketch the graph and write the probability statement.
- Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton.
- Find the third quartile for votes for President Clinton.
Solution
- [latex]X \sim N(1956.8, 572.3)[/latex]
- This is a population mean, because all election districts are included.
- The probability that a district had less than 1,600 votes for President Clinton is 0.2676.
- 0.3798
- Seventy-five percent of the districts had fewer than 2,340 votes for President Clinton.
Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of 7 days.
- In words, define the random variable X.
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement.
- Sixty percent of all trials of this type are completed within how many days?
Solution
- X = the distribution of the number of days a particular type of criminal trial will take
- [latex]X \sim N(21, 7)[/latex]
- The probability that a randomly selected trial will last more than 24 days is 0.3336.
- 22.77
Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps.
- In words, define the random variable X.
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the percent of her laps that are completed in less than 130 seconds.
- The fastest 3% of her laps are under [latex]\text{_____}[/latex].
- The middle 80% of her laps are from [latex]\text{_______}[/latex] seconds to [latex]\text{_______}[/latex] seconds.
Solution
- X = the distribution of race times that Terry Vogel produces
- [latex]X \sim N(129.71, 2.28)[/latex]
- Terri completes 55.17% of her laps in less than 130 seconds.
- Terri completes 55.17% of her laps in less than 130 seconds.
- 124.4 and 135.02
Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. The table below displays the ordered real data (in minutes):
| 0.50 | 4.25 | 5 | 6 | 7.25 |
| 1.75 | 4.25 | 5.25 | 6 | 7.25 |
| 2 | 4.25 | 5.25 | 6.25 | 7.25 |
| 2.25 | 4.25 | 5.5 | 6.25 | 7.75 |
| 2.25 | 4.5 | 5.5 | 6.5 | 8 |
| 2.5 | 4.75 | 5.5 | 6.5 | 8.25 |
| 2.75 | 4.75 | 5.75 | 6.5 | 9.5 |
| 3.25 | 4.75 | 5.75 | 6.75 | 9.5 |
| 3.75 | 5 | 6 | 6.75 | 9.75 |
| 3.75 | 5 | 6 | 6.75 | 10.75 |
- Calculate the sample mean and the sample standard deviation.
- Construct a histogram.
- Draw a smooth curve through the midpoints of the tops of the bars.
- In words, describe the shape of your histogram and smooth curve.
- Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes.
- Determine the cumulative relative frequency for waiting less than 6.1 minutes.
- Why aren’t the answers to part f and part g exactly the same?
- Why are the answers to part f and part g as close as they are?
- If only ten customers have been surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion.
Solution
- mean = 5.51, s = 2.15
- Check student’s solution.
- Check student’s solution.
- Check student’s solution.
- [latex]X \sim N(5.51, 2.15)[/latex]
- 0.6029
- The cumulative frequency for less than 6.1 minutes is 0.64.
- The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.
- The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.
- The approximation would have been less accurate, because the smaller sample size means that the data does not fit the normal curve as well.
Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false.
- Ricardo’s actual GPA is lower than Anita’s actual GPA.
- Ricardo is not passing because his z-score is zero.
- Anita is in the 70th percentile of students at her college.
Solution
- If the average GPA is less at Anita’s school than it is at Ricardo’s, then Ricardo’s actual score could be higher.
- Passing can be defined differently at different schools. Also, since Ricardo’s z-score is 0, his GPA is actually the average for his school, which is typically a passing GPA.
- Anita’s percentile is higher than the 70th percentile.
The table below shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums. The table does not include horse-racing or motor-racing stadiums.
| 40,000 | 40,000 | 45,050 | 45,500 | 46,249 | 48,134 |
| 49,133 | 50,071 | 50,096 | 50,466 | 50,832 | 51,100 |
| 51,500 | 51,900 | 52,000 | 52,132 | 52,200 | 52,530 |
| 52,692 | 53,864 | 54,000 | 55,000 | 55,000 | 55,000 |
| 55,000 | 55,000 | 55,000 | 55,082 | 57,000 | 58,008 |
| 59,680 | 60,000 | 60,000 | 60,492 | 60,580 | 62,380 |
| 62,872 | 64,035 | 65,000 | 65,050 | 65,647 | 66,000 |
| 66,161 | 67,428 | 68,349 | 68,976 | 69,372 | 70,107 |
| 70,585 | 71,594 | 72,000 | 72,922 | 73,379 | 74,500 |
| 75,025 | 76,212 | 78,000 | 80,000 | 80,000 | 82,300 |
- Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums (the data).
- Construct a histogram.
- Draw a smooth curve through the midpoints of the tops of the bars of the histogram.
- In words, describe the shape of your histogram and smooth curve.
- Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators.
- Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint: Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample.
- Why aren’t the answers to part f and part g exactly the same?
Solution
- mean = 60,136; s = 10,468
- Answers will vary.
- Answers will vary.
- Answers will vary.
- [latex]X \sim N(60136, 10468)[/latex]
- 0.7440
- The cumulative relative frequency is [latex]\frac{43}{60} = 0.717[/latex].
- The answers for part f and part g are not the same, because the normal distribution is only an approximation.
An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention.
What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability.
Solution
For [latex]x = 240[/latex], [latex]z =−3.0769[/latex]
For [latex]x = 306[/latex], [latex]z = 2[/latex]
[latex]P(240 \lt x \lt 306) = P(–3.0769 \lt z \lt 2) = \text{normalcdf}(-3.0769, 2, 0, 1) = 0.9762[/latex]
According to the scenario given, this means that there is a 97.62% chance that he is not the father.
To answer the second part of the question, there is a [latex]1 - 0.9762 = 0.0238 = 2.38 \%[/latex] chance that he is the father.
A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample.
What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.
Solution
- [latex]n = 100; p = 0.1; q = 0.9[/latex]
- [latex]\mu = np = (100)(0.10) = 10[/latex]
- [latex]\sigma = \sqrt{npq} = \sqrt{\text{(100)(0}\text{.1)(0}\text{.9)}} = 3[/latex]
- [latex]z = \pm 1[/latex]: [latex]x_1 = \mu + z \sigma = 10 + 1(3) = 13[/latex] and [latex]x_2 = µ – z \sigma = 10 – 1(3) = 7[/latex]. 68% of the defective cars will fall between 7 and 13.
- [latex]z = \pm 2[/latex]: [latex]x_1 = \mu + z \sigma = 10 + 2(3) = 16[/latex] and [latex]x_2 = \mu – z \sigma = 10 – 2(3) = 4[/latex]. 95 % of the defective cars will fall between four and 16
- [latex]z = \pm 3[/latex]: [latex]x_1 = \mu + z \sigma = 10 + 3(3) = 19[/latex] and [latex]x_2 = \mu – z \sigma = 10 – 3(3) = 1[/latex]. 99.7% of the defective cars will fall between one and 19.
We flip a coin 100 times ([latex]n = 100[/latex]) and note that it only comes up heads 20% ([latex]p = 0.20[/latex]) of the time. The mean and standard deviation for the number of times the coin lands on heads is [latex]\mu = 20[/latex] and [latex]\sigma = 4[/latex] (verify the mean and standard deviation). Solve the following:
- There is about a 68% chance that the number of heads will be somewhere between [latex]\text{___}[/latex] and [latex]\text{___}[/latex].
- There is about a [latex]\text{___}[/latex] chance that the number of heads will be somewhere between 12 and 28.
- There is about a [latex]\text{___}[/latex] chance that the number of heads will be somewhere between eight and 32.
Solution
- There is about a 68% chance that the number of heads will be somewhere between 16 and 24. [latex]z = \pm 1: x_1 = \mu + z \sigma = 20 + 1(4) = 24[/latex] and [latex]x_2 = \mu-z \sigma = 20 – 1(4) = 16[/latex].
- There is about a 95% chance that the number of heads will be somewhere between 12 and 28. For this problem: [latex]\text{normalcdf}(12, 28, 20, 4) = 0.9545 = 95.45 \%[/latex]
- There is about a 99.73% chance that the number of heads will be somewhere between eight and 32. For this problem: [latex]\text{normalcdf}(8, 32, 20, 4) = 0.9973 = 99.73 \%[/latex].
A $1 scratch-off lotto ticket will be a winner one out of five times. Out of a shipment of [latex]n = 190[/latex] lotto tickets, find the probability for the lotto tickets that there are
- somewhere between 34 and 54 prizes.
- somewhere between 54 and 64 prizes.
- more than 64 prizes.
Solution
- [latex]n = 190; p = \frac{1}{5}= 0.2; q = 0.8[/latex]
- [latex]\mu = np = (190)(0.2) = 38[/latex]
- [latex]\sigma = \sqrt{npq} = \sqrt{\text{(190)(0}\text{.2)(0}\text{.8)}} = 5.5136[/latex]
- For this problem: [latex]P(34 \lt x \lt 54) = \text{normalcdf}(34,54,48,5.5136) = 0.7641[/latex]
- For this problem: [latex]P(54 \lt x \lt 64) = \text{normalcdf}(54,64,48,5.5136) = 0.0018[/latex]
- For this problem: [latex]P(x > 64) = \text{normalcdf}(64, 10^{99}, 48, 5.5136) = 0.0000012[/latex] (approximately 0)
Facebook provides a variety of statistics on its website that detail the growth and popularity of the site.
On average, 28 percent of 18- to 34-year-olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of 5 percent.
- Find the probability that the percent of 18- to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
- Find the 95th percentile, and express it in a sentence.
Solution
X = the percent of 18- to 34-year-olds who check Facebook before getting out of bed in the morning.
[latex]X \sim N(28, 5)[/latex]
[latex]P(x \ge 30) = 0.3446; \text{normalcdf}(30, 1 \text{EE} 99, 28, 5) = 0.3446[/latex]
[latex]\text{invNorm}(0.95, 0.28, 0.05) = 0.3622.95 \%[/latex] of the percentage of 18- to 34-year-olds who check Facebook before getting out of bed in the morning is at most 36.22%.
[latex]P(25 \lt x \lt 55) = \text{normalcdf}(25, 55, 28, 5) = 0.7257(0.7257)(400) = 290.28[/latex]
Section 6.3 Homework
Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.
- In words, [latex]Χ = \text{____________}[/latex]
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- In words, [latex]\overline{X} = \text{____________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
- Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
- Explain why there is a difference in part e and part f.
Solution
- Χ = amount of change students carry
- [latex]Χ \sim E(0.88, 0.88)[/latex]
- [latex]\overline{X} = \text{average amount of change carried by a sample of 25 students}[/latex]
- [latex]\overline{X} \sim N(0.88, 0.176)[/latex]
- 0.0819
- 0.1882
- The distributions are different. Part a is exponential and part b is normal.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.
- If [latex]\overline{X}[/latex] = average distance in feet for 49 fly balls, then [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for [latex]\overline{X}[/latex]. Shade the region corresponding to the probability. Find the probability.
- Find the 80th percentile of the distribution of the average of 49 fly balls.
Solution
- [latex]N(250, \frac{50}{7})[/latex]
- 0.0808
- 256.01 feet
According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.
- In words, [latex]Χ = \text{_____________}[/latex]
- In words, [latex]\overline{X} = \text{_____________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.
- Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.
Solution
- length of time for an individual to complete IRS form 1040, in hours.
- mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours.
- [latex]N\left(10.53, \frac{1}{3}\right)[/latex]
- Yes. I would be surprised, because the probability is almost 0.
- No. I would not be totally surprised because the probability is 0.2312
Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let [latex]\overline{X}[/latex] the average of the 49 races.
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.
- Find the 80th percentile for the average of these 49 marathons.
- Find the median of the average running times.
Solution
- [latex]N(145, 2)[/latex]
- 0.6247
- 146.68
- 145 minutes
The length of songs in a collector’s iTunes album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.
- In words, [latex]Χ = \text{_________}[/latex]
- [latex]Χ \sim _____________[/latex]
- In words, [latex]\overline{X} = \text{_____________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the first quartile for the average song length.
- The IQR(interquartile range) for the average song length is from [latex]\text{______________}[/latex].
Solution
- the length of a song, in minutes, in the collection
- [latex]U(2, 3.5)[/latex]
- the average length, in minutes, of the songs from a sample of five albums from the collection
- [latex]N(2.75, 0.0220)[/latex]
- 2.74 minutes
- 0.03 minutes
In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.
- In words, [latex]Χ = \text{_____________}[/latex]
- In words, [latex]\overline{X} = \text{_____________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- The IQR for [latex]\overline{X}[/latex] is from [latex]\text{_______}[/latex] acres to [latex]\text{_______}[/latex] acres.
Solution
- the size of a U.S. farm in 1940
- the average size of a U.S. farm, in acres
- [latex]N(174, 8.9222)[/latex]
- 168.0, 180.0
Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.
- When the sample size is large, the mean of [latex]\overline{X}[/latex] is approximately equal to the mean of Χ.
- When the sample size is large, [latex]\overline{X}[/latex] is approximately normally distributed.
- When the sample size is large, the standard deviation of [latex]\overline{X}[/latex] is approximately the same as the standard deviation of Χ.
Solution
- True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.
- True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.
- The standard deviation of the sampling distribution of the means will decrease making it approximately the same as the standard deviation of X as the sample size increases.
The percentage of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let [latex]\overline{X}[/latex] = average percent of fat calories.
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined.
- Find the first quartile for the average percent of fat calories.
Solution
- [latex]N(36, 2.5)[/latex]
- 1
- 34.31
The distribution of income in some Third World countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.
- In words, [latex]Χ = \text{_____________}[/latex]
- In words, [latex]\overline{X} = \text{_____________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- How is it possible for the standard deviation to be greater than the average?
- Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200?
Solution
- X = the yearly income of someone in a third world country
- the average salary from samples of 1,000 residents of a third world country
- [latex]\overline{X} \sim N \left(2000, \frac{8000}{\sqrt{1000}}\right)[/latex]
- Very wide differences in data values can have averages smaller than standard deviations.
- The distribution of the sample mean will have higher probabilities closer to the population mean.
[latex]P(2000 \lt \overline{X} \lt 2000) = 0.1537[/latex]; [latex]P(2100 \lt \overline{X} \lt 2200) = 0.1317[/latex]
Which of the following is NOT TRUE about the distribution for averages?
- The mean, median, and mode are equal.
- The area under the curve is one.
- The curve never touches the x-axis.
- The curve is skewed to the right.
Solution
The curve is skewed to the right.
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.
The distribution to use for the average cost of gasoline for the 16 gas stations is:
a. [latex]\overline{X} \sim N(4.59, 0.10)[/latex]
b. [latex]\overline{X} \sim N \left(4.59, \frac{0.10}{\sqrt{16}}\right)[/latex]
c. [latex]\overline{X} \sim N \left(4.59, \frac{16}{0.10}\right)[/latex]
d. [latex]\overline{X} \sim N \left(4.59, \frac{\sqrt{16}}{0.10}\right)\right)[/latex]
Solution
b. [latex]\overline{X} \sim N \left(4.59, \frac{0.10}{\sqrt{16}}\right)[/latex]
Section 6.4 Homework
Which of the following is NOT TRUE about the theoretical distribution of sums?
- The mean, median and mode are equal.
- The area under the curve is one.
- The curve never touches the x-axis.
- The curve is skewed to the right.
Solution
The curve is skewed to the right.
Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials.
- In words, [latex]\Sigma{X} = \text{______________}[/latex]
- [latex]\Sigma{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the total length of the nine trials is at least 225 days.
- Ninety percent of the total of nine of these types of trials will last at least how long?
Solution
- the total length of time for nine criminal trials
- [latex]N(189, 21)[/latex]
- 0.0432
- 162.09; ninety percent of the total nine trials of this type will last 162 days or more.
Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children.
- In words, [latex]X = \text{_____________}[/latex]
- The distribution is [latex]\text{_______}[/latex].
- In words, [latex]\Sigma{X} = \text{______________}[/latex]
- [latex]\Sigma{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the total weight of open boxes is less than 250 pounds.
- Find the 35th percentile for the total weight of open boxes of cereal.
Solution
- X = weight of open cereal boxes
- The distribution is uniform.
- [latex]\Sigma{X} = the sum of the weights of 64 randomly chosen cereal boxes[/latex]
- [latex]N(256, 9.24)[/latex]
- 0.2581
- 252.44 pounds
Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district.
- In words, [latex]X = \text{_____________}[/latex]
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- In words, [latex]\Sigma{X} = \text{______________}[/latex]
- [latex]\Sigma{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Find the probability that the teachers earn a total of over $400,000.
- Find the 90th percentile for an individual teacher’s salary.
- Find the 90th percentile for the sum of ten teachers’ salary.
- If we surveyed 70 teachers instead of ten, graphically, how would that change the distribution in part d?
- If each of the 70 teachers received a $3,000 raise, graphically, how would that change the distribution in part b?
Solution
- X = the salary of one elementary school teacher in the district
- [latex]X \sim N(44,000, 6,500)[/latex]
- [latex]\Sigma{X} \text{sum of the salaries of ten elementary school teachers in the sample}[/latex]
- [latex]\Sigma{X} \sim N(44000, 20554.80)[/latex]
- 0.9742
- $52,330.09
- 466,342.04
- Sampling 70 teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.
- If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000.
Section 6.5 Homework
The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds.
- In words, [latex]X = \text{_____________}[/latex]
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- In words, [latex]\overline{X} = \text{______________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Before doing any calculations, which do you think will be higher? Explain why.
- The probability that an individual attention span is less than ten minutes.
- The probability that the average attention span for the 60 children is less than ten minutes?
- Calculate the probabilities in part e.
- Explain why the distribution for [latex]\overline{X}[/latex] is not exponential.
Solution
- X = the attention span of a two-year-old
- [latex]Χ \sim Exp(18)[/latex]
- [latex]\overline{X}= \text{the mean (average) attention span of a two-year-old}[/latex]
- [latex]\overline{X} \sim N(8, 860)[/latex]
- The standard deviation is smaller, so there is more area under the normal curve.
- Exponential: 0.7135
- Normal: 0.7579
- By the central limit theorem, as n gets larger, the means tend to follow a normal distribution.
The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows.
8.625; 30.25; 27.625; 46.75; 32.875; 18.25; 5; 0.125; 2.9375; 6.875; 28.25; 24.25; 21; 1.5; 30.25; 71; 43.5; 49.25; 2.5625; 31; 16.5; 9.5; 18.5; 18; 9; 10.5; 16.625; 1.25; 18; 12.87; 7; 12.875; 2.875; 60.25; 29.25
- In words, [latex]Χ = \text{______________}[/latex]
- [latex]\overline{x} = \text{_____}[/latex]
- [latex]s_x = \text{_____}[/latex]
- [latex]n = \text{_____}[/latex]
- Construct a histogram of the distribution of the averages. Start at [latex]x = -0.0005[/latex]. Use bar widths of ten.
- In words, describe the distribution of stock prices.
- Randomly average five stock prices together. (Use a random number generator.) Continue averaging five pieces together until you have ten averages. List those ten averages.
- Use the ten averages from part e to calculate the following.
- [latex]\overline{x} = \text{_____}[/latex]
- [latex]s_x = \text{_____}[/latex]
- Construct a histogram of the distribution of the averages. Start at [latex]x = -0.0005[/latex]. Use bar widths of ten.
- Does this histogram look like the graph in part c?
- In one or two complete sentences, explain why the graphs either look the same or look different?
- Based upon the theory of the central limit theorem, [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
Solution
- X = the closing stock prices for U.S. semiconductor manufacturers
- $20.71
- $17.31
- $35
- Check student solutions
- Exponential distribution, [latex]X \sim \text{Exp}\left(\frac{1}{20.71}\right)[/latex]
- Answers will vary.
- i. $20.71; ii. $11.14
- Answers will vary.
- Answers will vary.
- Answers will vary.
- [latex]N \left(20.71, \frac{17.31}{\sqrt{5}}\right)[/latex]
Use the following information to answer the next three exercises:
Richard’s Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery.
1. [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
a. [latex]U(0,4)[/latex]
b. [latex]U(10,2)[/latex]
c. [latex]\text{Exp}(2)[/latex]
d. [latex]N(2,1)[/latex]
Solution
a. [latex]U(0,4)[/latex]
2. The average wait time is:
a. one hour.
b. two hours.
c. two and a half hours.
d. four hours.
Solution
b. two hours
3. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is:
a. [latex]\frac{1}{4}[/latex]
b. [latex]\frac{1}{2}[/latex]
c. [latex]\frac{3}{4}[/latex]
d. [latex]\frac{3}{8}[/latex]
Solution
a. [latex]\frac{1}{4}[/latex]
Use the following information to answer the next two exercises:
The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited.
1. The 90th percentile sample average wait time (in minutes) for a sample of 100 riders is:
a. 315.0
b. 40.3
c. 38.5
d. 65.2
Solution
40.3b.
2. Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes?
a. yes
b. no
c. There is not enough information.
Solution
a. yes
Use the following to answer the next two exercises:
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.
What’s the approximate probability that the average price for 16 gas stations is over $4.69?
a. almost zero
b. 0.1587
c. 0.0943
d. unknown
Solution
a. almost zero
Find the probability that the average price for 30 gas stations is less than $4.55.
a. 0.6554
b. 0.3446
c. 0.0142
d. 0.9858
e. 0
Solution
c. 0.0142
Suppose in a local Kindergarten through 12th grade (K – 12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate following using the normal approximation to the binomial distribution.
- Find the probability that less than 100 favor a charter school for grades K through 5.
- Find the probability that 170 or more favor a charter school for grades K through 5.
- Find the probability that no more than 140 favor a charter school for grades K through 5.
- Find the probability that there are fewer than 130 that favor a charter school for grades K through 5.
- Find the probability that exactly 150 favor a charter school for grades K through 5.
If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology.
Solution
- 0
- 0.1123
- 0.0162
- 0.0003
- 0.0268
Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for 96 days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places.
- Find the probability that Janice is the driver for at most 20 days.
- Find the probability that Roberta is the driver for more than 16 days.
- Find the probability that Barbara drives exactly 24 of those 96 days.
Solution
- 0.2047
- 0.9615
- 0.0938
[latex]X \sim N(60, 9)[/latex]. Suppose that you form random samples of 25 from this distribution. Let [latex]\overline{X}[/latex] be the random variable of averages. Let ΣX be the random variable of sums. For parts c through f, sketch the graph, shade the region, label and scale the horizontal axis for [latex]\overline{X}[/latex], and find the probability.
- Sketch the distributions of X and [latex]\overline{X}[/latex] on the same graph.
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- [latex]P(\overline{x} \lt 60) = \text{_____}[/latex]
- Find the 30th percentile for the mean.
- [latex]P(56 \lt \overline{x} \lt 62) = \text{_____}[/latex]
- [latex]P(18 \lt \overline{x} \lt 58) = _____[/latex]
- [latex]\Sigma{x} \sim \text{_____}(\text{_____},\text{_____})[/latex]
- Find the minimum value for the upper quartile for the sum.
- [latex]P(1,400 \lt \Sigma{x} \lt 1,550) = \text{_____}[/latex]
Solution
- Check student’s solution.
- [latex]\overline{X} \sim N \left(60, \frac{9}{\sqrt{25}}\right)[/latex]
- 0.5000
- 59.06
- 0.8536
- 0.1333
- [latex]N(1500, 45)[/latex]
- 1530.35
- 0.6877
Suppose that the length of research papers is uniformly distributed from ten to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers.
- In words, [latex]X = \text{_____________}[/latex]
- [latex]X \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- [latex]\mu_x = \text{_____}[/latex]
- [latex]\sigma_x = \text{_____}[/latex]
- In words, [latex]\overline{X} = \text{______________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- In words, [latex]\Sigma{X} = \text{_____________}[/latex]
- [latex]\Sigma{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Without doing any calculations, do you think that it’s likely that the professor will need to read a total of more than 1,050 pages? Why?
- Calculate the probability that the professor will need to read a total of more than 1,050 pages.
- Why is it so unlikely that the average length of the papers will be less than 12 pages?
Solution
- the length of the research papers
- U(10, 25)
- 17.5
225
12
= 4.3301
[latex]N(17.5, 0.5839)[/latex]
[latex]N(962.5, 32.11)[/latex]
0.0032
Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district.
- Find the 90th percentile for an individual teacher’s salary.
- Find the 90th percentile for the average teacher’s salary.
Solution
- $52,330
- $46,634
The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital.
- In words, [latex]X = \text{_____________}[/latex]
- In words, [latex]\overline{X} = \text{_____________}[/latex]
- [latex]\overline{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- In words, [latex]\Sigma{X} = \text{_____________}[/latex]
- [latex]\Sigma{X} \sim \text{_____}(\text{_____}, \text{_____})[/latex]
- Is it likely that an individual stayed more than five days in the hospital? Why or why not?
- Is it likely that the average stay for the 80 women was more than five days? Why or why not?
- Which is more likely:
a. An individual stayed more than five days.
b. The average stay of 80 women was more than five days. - If we were to sum up the women’s stays, is it likely that, collectively, they spent more than a year in the hospital? Why or why not?
Solution
- the length of a maternity stay in a U.S. hospital, in days
- the average length of a maternity stay in a U.S. hospital, in days
- [latex]N(2.4, 0.980)[/latex]
- [latex]N(192, 8.05)[/latex]
- Not likely, but possible. The mean stay is 2.4 days.
- No, the probability is 0.
- a. An individual stayed more than five days.
- No, the probability is 0.
For each problem, wherever possible, provide graphs and use the calculator.
1. NeverReady batteries has engineered a newer, longer lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean lifetime is 16.7 hours or less? Is the company’s claim reasonable?
Solution
- We have [latex]\mu = 17[/latex], [latex]\sigma = 0.8[/latex], [latex]\overline{x}= 16.7[/latex], and [latex]n = 30[/latex]. To calculate the probability, we use \text{normalcdf(lower, upper, }\mu, [latex]\frac{\sigma }{\sqrt{n}}) = \text{normalcdf}\left(\text{E}-99, 16.7,17,\frac{0.8}{\sqrt{30}}\right) = 0.0200[/latex].
- If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 lifetime hours is only 2%. Therefore, the class was justified to question the claim.
2. Men have an average weight of 172 pounds with a standard deviation of 29 pounds.
- Find the probability that 20 randomly selected men will have a sum weight greater than 3,600 lbs.
- If 20 men have a sum weight greater than 3,500 lbs., then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain.
Solution
To calculate the probability, we use [latex]\text{normalcdf}(3600, \text{E}99, 3440, 129.69) = 0.1087[/latex]
While the probability is not exceptionally large, [latex]P(\overline{X} > 3600) = 0.1087[/latex], it could be wise to be concerned. After all, we do have a 1 in 10 chance that a random sample of men could have an average sum weight that exceeds the safety limit.
3. M&M candies large candy bags have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class.
| Red | Orange | Yellow | Brown | Blue | Green |
|---|---|---|---|---|---|
| 0.751 | 0.735 | 0.883 | 0.696 | 0.881 | 0.925 |
| 0.841 | 0.895 | 0.769 | 0.876 | 0.863 | 0.914 |
| 0.856 | 0.865 | 0.859 | 0.855 | 0.775 | 0.881 |
| 0.799 | 0.864 | 0.784 | 0.806 | 0.854 | 0.865 |
| 0.966 | 0.852 | 0.824 | 0.840 | 0.810 | 0.865 |
| 0.859 | 0.866 | 0.858 | 0.868 | 0.858 | 1.015 |
| 0.857 | 0.859 | 0.848 | 0.859 | 0.818 | 0.876 |
| 0.942 | 0.838 | 0.851 | 0.982 | 0.868 | 0.809 |
| 0.873 | 0.863 | 0.803 | 0.865 | ||
| 0.809 | 0.888 | 0.932 | 0.848 | ||
| 0.890 | 0.925 | 0.842 | 0.940 | ||
| 0.878 | 0.793 | 0.832 | 0.833 | ||
| 0.905 | 0.977 | 0.807 | 0.845 | ||
| 0.850 | 0.841 | 0.852 | |||
| 0.830 | 0.932 | 0.778 | |||
| 0.856 | 0.833 | 0.814 | |||
| 0.842 | 0.881 | 0.791 | |||
| 0.778 | 0.818 | 0.810 | |||
| 0.786 | 0.864 | 0.881 | |||
| 0.853 | 0.825 | ||||
| 0.864 | 0.855 | ||||
| 0.873 | 0.942 | ||||
| 0.880 | 0.825 | ||||
| 0.882 | 0.869 | ||||
| 0.931 | 0.912 | ||||
| 0.887 |
The bag contained 465 candies, and the listed weights in the table came from randomly selected candies. Count the weights.
- Find the mean sample weight and the standard deviation of the sample weights of candies in the table.
- Find the sum of the sample weights in the table and the standard deviation of the sum of the weights.
- If 465 M&Ms are randomly selected, find the probability that their weights sum to at least 396.9.
- Is the Mars Company’s M&M labeling accurate?
Solution
- For the sample, we have [latex]n = 100[/latex], [latex]\overline{x} = 0.862[/latex], [latex]s = 0.05[/latex]
- [latex]\Sigma \overline{x} = 85.65[/latex], [latex]\Sigma{s} = 5.18[/latex]
- [latex]\text{normalcdf}(396.9, \text{E}99, (465)(0.8565), (0.05)(\sqrt{465})) \approx 1[/latex]
- Since the probability of a sample of size 465 having at least a mean sum of 396.9 is approximately 1, we can conclude that Mars is correctly labeling their M&M packages.
4. The Screw Right Company claims their [latex]\frac{3}{4}[/latex]-inch screws are within ±0.23 of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded.
| 0.757 | 0.723 | 0.754 | 0.737 | 0.757 | 0.741 | 0.722 | 0.741 | 0.743 | 0.742 |
| 0.740 | 0.758 | 0.724 | 0.739 | 0.736 | 0.735 | 0.760 | 0.750 | 0.759 | 0.754 |
| 0.744 | 0.758 | 0.765 | 0.756 | 0.738 | 0.742 | 0.758 | 0.757 | 0.724 | 0.757 |
| 0.744 | 0.738 | 0.763 | 0.756 | 0.760 | 0.768 | 0.761 | 0.742 | 0.734 | 0.754 |
| 0.758 | 0.735 | 0.740 | 0.743 | 0.737 | 0.737 | 0.725 | 0.761 | 0.758 | 0.756 |
The screws were randomly selected from the local home repair store.
- Find the mean diameter and standard deviation for the sample
- Find the probability that 50 randomly selected screws will be within the stated tolerance levels. Is the company’s diameter claim plausible?
Solution
[latex]\overline{x}= 0.75[/latex] and [latex]s = 0.01[/latex]
We have [latex]\text{normalcdf}(0.52, 0.98, 0.75, 0.11550) \approx 1[/latex] and we can conclude that the company’s diameter claim is justified.
5. Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?
Solution
Use [latex]\text{normalcdf} \left(\text{E} - 99, 1.1, 1, \frac{1}{\sqrt{70}} \right) = 0.7986[/latex]. This means that there is an 80% chance that the service time will be less than 1.1 hours. It could be wise to schedule more time since there is an associated 20% chance that the maintenance time will be greater than 1.1 hours.
6. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points?
Solution
The probability that the sample score is between 85 and 125 points is given by [latex]\text{normalcdf}(85, 125, 105, 2020)= 0.9999[/latex]. Therefore, it is almost a guarantee that a well selected sample of 20 adults will have an average score between 85 and 125.
7. Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine?
Solution
Since we have [latex]\text{normalcdf} \left(5.111, 5.291, 5.201, \frac{0.\text{065}}{\sqrt{280}} \right) \approx 1[/latex], we can conclude that practically all the coins are within the limits, therefore, there should be no rejected coins out of a well selected sample of size 280.