Chapter 6: The Normal Distribution and The Central Limit Theorem

6.6 Using Technology for the Normal Distribution

Learning Objectives

By the end of this section, the student should be able to:

  • use technology to solve problems involving the Normal Distribution

This is another section to show technology tools that can be used to help with the statistics concepts taught in this chapter. This section will cover tools to help solve problems involving the Normal Distribution.

Graphing Calculator

Similar to all the other technology sections, the main tool introduced will be the TI-83, 83+, 84, or 84+ Graphing Calculator. The graphing calculator can help to find probabilities for a normal distribution and find percentiles for the normal distribution.

In 6.2 Using the Normal Distribution, we discussed the shaded area under the normal curve as the area to the left of [latex]x[/latex] - represented by the probability [latex]P(X \lt x)[/latex] called the cumulative distribution function. In that section we looked up the value in a table; however, it was mentioned that it can be calculated either by a calculator or a computer as well, and that technology has made the tables virtually obsolete. Remember that to use the table, it required a standardizing process that involved z-scores. Well, again, as mentioned in Section 6.2, "There are many technologies such as calculators and various statistical software that let us skip the entire standardizing process and instantaneously provide us with a probability." This page is here to show you that process.

Probabilities can be quickly calculated using technology. The technology that will be used for the following examples are the TI-83+ and TI-84 calculators. The best news about using the TI graphing calculator is that there is no need to standardize. You don't have to go from [latex]X \sim N(\mu, \sigma) \rightarrow Z \sim N(0, 1)[/latex], you just use [latex]X \sim N(\mu, \sigma)[/latex]. Although the calculator is a quick method, drawing the image of the shaded region can help you to see if your answer is valid.

Using the TI-83+ and TI-84+ Calculators for Normal Probabilities

Access [DISTR] by pressing 2nd key, vars key. (Note: DISTR stands for "Distributions")

Press 2: normalcdf.

The syntax for the instructions are as follows: [latex]\text{normalcdf(lower value, upper value, mean, standard deviation)}[/latex]

Another way to think of it is normalcdf(left bound, right bound, μ, σ) which corresponds to P(left bound < X < right bound)

In some instances:

  • the upper number of the area might be [latex]1\text{E}99[/latex] (which equals [latex]10^{99}[/latex]). You get [latex]1\text{E}99 = 10^{99})[/latex] by pressing [latex]1[/latex], the [latex]\text{EE}[/latex] key (a 2nd key) and then [latex]99[/latex]. Or, you can enter [latex]10^{99}[/latex] instead. The number [latex]10^{99}[/latex] is way out in the right tail of the normal curve.
  • the lower number of the area might be [latex]-1 \text{E} 99[/latex] (which equals  [latex]-10^{99}[/latex]). The number [latex]-10^{99}[/latex] is way out in the left tail of the normal curve.

Note

The graphing calculator has the following options to be used as well:

  • normalpdf(x,μ,σ) yields a probability density function value (only useful to plot the normal curve, in which case "x" is the variable)
  • normalcdf(left bound, right bound) corresponds to P(left bound < Z < right bound) – standard normal

Let's see this in action with some examples.

Example

The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

  1. Find the probability that a randomly selected student scored more than 65 on the exam.
  2. Find the probability that a randomly selected student scored less than 85.

Solution

  1. Draw the graph, then find [latex]P(x > 65)[/latex]. [latex]\text{normalcdf}(65, 1 \text{E}99, 63, 5) = 0.3446[/latex].
  2. [latex]\text{normalcdf}(-1 \text{E} 99, 85, 63, 5) = 1[/latex] (rounds to one). The probability that one student scores less than 85 is approximately one (or 100%).
A normal distribution with 63 at the peak. The value 65 has a shaded region to the right that represents the probability 0.3446.
Figure 1. P(x > 65) = 0.3446

Example

There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.

  1. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
  2. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.

Solution

  1. [latex]\text{normalcdf}(23, 64.7, 36.9, 13.9) = 0.8186[/latex]
  2. [latex]\text{normalcdf}(-10^{99}, 50.8, 36.9, 13.9) = 0.8413[/latex]

Your Turn!

A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.

Solution

Let [latex]X =[/latex] the amount of time (in hours) a household personal computer is used for entertainment. [latex]X \sim N(2, 0.5)[/latex] where [latex]\mu = 2[/latex] and [latex]\sigma = 0.5[/latex]. Find [latex]P(1.8 \lt x \lt 2.75)[/latex].

The probability for which you are looking is the area between [latex]x = 1.8[/latex] and [latex]x = 2.75[/latex].

A normal curve with the peak at 2. Values of 1.8 and 2.75 have a shaded region between them.
Figure 2. P(1.8< x < 2.75)

[latex]\text{normalcdf}(1.8, 2.75, 2, 0.5) = 0.5886[/latex]

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.

Using the TI-83+ and TI-84+ Calculators for Finding Percentiles (Z-scores)

Access [DISTR] by pressing 2nd key,vars key. (Note: DISTR stands for "Distributions")

Press 3: invNorm

The syntax for the instructions are as follows: for the [latex]k^{\text{th}}[/latex] percentile: [latex]\text{invNorm(area to the left, mean, standard deviation)}[/latex]

Another way to think of it is invNorm(p,μ,σ) yields the critical value, k: P(X < k) = p

Notes:

  • Percentiles are values that are left area, but be careful to read if you are given a right area in the problem. Remember, [latex]\text{Right Area} = 1 - \text{Left Area}[/latex].
  • invNorm(p) yields the critical value, k: P(Z < k) = p for the standard normal

Example

The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

  1. Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).
  2. Find the 70th percentile (that is, find the score [latex]k[/latex] such that 70% of scores are below k and 30% of the scores are above [latex]k[/latex]).

Solution

  1. Draw a graph and shade the area that corresponds to the 90th percentile. Let [latex]k = \text{the }90^{\text{th}}\text{ percentile}[/latex]. The variable [latex]k[/latex] is located on the x-axis. [latex]P(x \lt k)[/latex] is the area to the left of [latex]k[/latex]. The 90th percentile [latex]k[/latex] separates the exam scores into those that are the same or lower than [latex]k[/latex] and those that are the same or higher. Ninety percent of the test scores are the same or lower than [latex]k[/latex], and ten percent are the same or higher. For this problem, [latex]\text{invNorm}(0.90, 63, 5) = 69.4[/latex]. The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above.
  2. [latex]\text{invNorm}(0.70, 63, 5) = 65.6[/latex]. The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.
Normal distribution curve with peak at 63. A point, k, is to the right of 63. The area under the curve to the left of k is shaded representing P(x < k) = 0.90
Figure 3. The shaded region is the 90th percentile.

Example

Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean [latex]\mu = 81 \text{ points}[/latex] and standard deviation [latex]\sigma = 15 \text{ points}[/latex].

  1. Calculate the first and third quartile scores for this exam.
  2. The middle 50% of the exam scores are between what two values?

Solution

  1. [latex]Q_1 = 25^{\text{th}} \text{percentile} = \text{invNorm}(0.25, 81, 15) = 70.9[/latex]. [latex]Q_3 = 75^{\text{th}} \text{percentile} = \text{invNorm}(0.75, 81, 15) = 91.1[/latex].
  2. In part a, The middle 50% of the scores are between 70.9 and 91.1.

Example

There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).

  1. Calculate the interquartile range (IQR).
  2. Forty percent of the ages that range from 13 to 55+ are at least what age?

Solution

  1. [latex]IQR = Q_3 – Q_1[/latex]. Calculate [latex]Q_3 = 75^{\text{th}} \text{percentile}[/latex] and [latex]Q_1 = 25^{\text{th}} \text{percentile}[/latex]. [latex]\text{invNorm}(0.75, 36.9, 13.9) = Q_3 = 46.2754[/latex]. [latex]\text{invNorm}(0.25, 36.9, 13.9) = Q_1 = 27.5246[/latex]. [latex]IQR = Q_3 – Q_1 = 18.7508[/latex].
  2. Find [latex]k[/latex] where [latex]P(x > k) = 0.40[/latex] ("At least" translates to "greater than or equal to."). [latex]0.40 = \text{Area to the right}[/latex]. [latex]\text{Area to the left} = 1 – 0.40 = 0.60[/latex]. The area to the left of [latex]k = 0.60[/latex]. [latex]\text{invNorm}(0.60, 36.9, 13.9) = 40.4215[/latex]. [latex]k = 40.42[/latex]. Forty percent of the ages that range from 13 to 55+ are at least 40.42 years.

Your Turn!

A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

Solution

To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, [latex]k[/latex], where [latex]P(x \lt k) = 0.25[/latex].

A normal distribution curve. The area under the left tail of the curve is shaded representing the probability that x is less than k is 0.25.
Figure 4. The shaded area is the 25th percentile.

[latex]\text{invNorm}(0.25, 2, 0.5) = 1.66[/latex]

The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

Your Turn!

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. The middle 40% of mandarin oranges from this farm are between [latex]\underline{\hspace{2cm}}[/latex] and [latex]\underline{\hspace{2cm}}[/latex].

Solution

The middle area = 0.40, so each tail has an area of 0.30.

[latex]1 - 0.40 = 0.60[/latex]

The tails of the graph of the normal distribution each have an area of 0.30.

Find [latex]k_1[/latex], the 30th percentile and [latex]k_2[/latex], the 70th percentile ([latex]0.40 + 0.30 = 0.70[/latex]).

[latex]k_1 = \text{invNorm(0.30, 5.85, 0.24)} = 5.72 \text{ cm}[/latex]

[latex]k_2 = \text{invNorm(0.70, 5.85, 0.24)} = 5.98 \text{ cm}[/latex]

Helpful Videos for the Graphing Calculator

Below are links to helpful videos for using the graphing calculator for the concepts covered on this page:

 

Additional Technology Tools

In addition to the graphing calculator, there are some additional technology tools that can be used for the concepts covered on this page. Below are links to helpful videos for those tools:

 

Section Practice

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 [latex]n = 100[/latex] cars. Let [latex]X[/latex] represent the number of defective cars in the sample.

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:

  1. There is about a 68% chance that the number of heads will be somewhere between [latex]\underline{\hspace{2cm}}[/latex] and [latex]\underline{\hspace{2cm}}[/latex].
  2. There is about a [latex]\underline{\hspace{2cm}}[/latex] chance that the number of heads will be somewhere between 12 and 28.
  3. There is about a [latex]\underline{\hspace{2cm}}[/latex] chance that the number of heads will be somewhere between eight and 32.
Solution
  1. 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].
  2. There is about a 95% chance that the number of heads will be somewhere between 12 and 28. For this problem: normalcdf(12,28,20,4) = 0.9545 = 95.45%
  3. There is about a 99.73% chance that the number of heads will be somewhere between eight and 32. For this problem: normalcdf(8,32,20,4) = 0.9973 = 99.73%.

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

  1. somewhere between 34 and 54 prizes.
  2. somewhere between 54 and 64 prizes.
  3. more than 64 prizes.
Solution
  • n = 190; p = [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{(190)(0.2)(0.8)} = 5.5136[/latex]
  1. For this problem: [latex]P(34 \lt x \lt 54) = \text{normalcdf}(34,54,48,5.5136) = 0.7641[/latex]
  2. For this problem: [latex]P(54 \lt x \lt 64) = \text{normalcdf}(54,64,48,5.5136) = 0.0018[/latex]
  3. 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.

  1. 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.
  2. Find the 95th percentile, and express it in a sentence.
Solution

[latex]X =[/latex] 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[/latex]; normalcdf(30,1EE99,28,5) = 0.3446

invNorm(0.95,0.28,0.05) = 0.3622.95% 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]

Media Attributions

  • fig-ch06_05_01-1

License

Icon for the Creative Commons Attribution-ShareAlike 4.0 International License

Introductory Statistics Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book