Chapter 4: Discrete Random Variables
Chapter 4 Practice
Section 4.1
Use the following information to answer the next six exercises: A baker is deciding how many batches of muffins to make to sell in his bakery. He wants to make enough to sell every one and no fewer. Through observation, the baker has established a probability distribution.
| x | P(x) |
|---|---|
| 1 | 0.15 |
| 2 | 0.35 |
| 3 | 0.40 |
| 4 | 0.10 |
Define the random variable X.
Solution
Let X = the number of batches that the baker will sell. –>
What is the probability the baker will sell more than one batch? P(x > 1) = _______
Solution
0.35 + 0.40 + 0.10 = 0.85
What is the probability the baker will sell exactly one
batch? P(x = 1) = _______
Solution
0.15
–>
On average, how many batches should the baker make?
Solution
1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45
Use the following information to answer the next four exercises: Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random.
Define the random variable X.
Solution
Let X = the number of days Ellen attends practice per week.
–>
Construct a probability distribution table for the data.
Solution
| x | P(x) |
|---|---|
| 0 | 0.03 |
| 1 | 0.04 |
| 2 | 0.08 |
| 3 | 0.85 |
We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic?
Solution
Each probability is between zero and one, inclusive.
–>
Use the following information to answer the next five exercises: Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35% of the time, four events 25% of the time, three events 20% of the time, two events 10% of the time, one event 5% of the time, and no events 5% of the time.
Define the random variable X.
Solution
Let X = the number of events Javier volunteers for each month.
What values does x take on?
Solution
0, 1, 2, 3, 4, 5
–>
Construct a PDF table.
Solution
| x | P(x) |
|---|---|
| 0 | 0.05 |
| 1 | 0.05 |
| 2 | 0.10 |
| 3 | 0.20 |
| 4 | 0.25 |
| 5 | 0.35 |
Find the probability that Javier volunteers for less than three events each month. P(x < 3) = _______
Solution
0.05 + 0.05 + 0.10 = 0.20
–>
Find the probability that Javier volunteers for at least one event each month. P(x > 0) = _______
Solution
1 – 0.05 = 0.95
Section 4.2
Complete the expected value table.
| x | P(x) | x*P(x) |
|---|---|---|
| 0 | 0.2 | |
| 1 | 0.2 | |
| 2 | 0.4 | |
| 3 | 0.2 |
Solution
–>
Find the expected value from the expected value table.
| x | P(x) | x*P(x) |
|---|---|---|
| 2 | 0.1 | 2(0.1) = 0.2 |
| 4 | 0.3 | 4(0.3) = 1.2 |
| 6 | 0.4 | 6(0.4) = 2.4 |
| 8 | 0.2 | 8(0.2) = 1.6 |
Solution
0.2 + 1.2 + 2.4 + 1.6 = 5.4
Find the standard deviation.
| x | P(x) | x*P(x) | (x – μ)2P(x) |
|---|---|---|---|
| 2 | 0.1 | 2(0.1) = 0.2 | (2–5.4)2(0.1) = 1.156 |
| 4 | 0.3 | 4(0.3) = 1.2 | (4–5.4)2(0.3) = 0.588 |
| 6 | 0.4 | 6(0.4) = 2.4 | (6–5.4)2(0.4) = 0.144 |
| 8 | 0.2 | 8(0.2) = 1.6 | (8–5.4)2(0.2) = 1.352 |
Solution
σ=
1.156+0.588+0.144+1.352
=
3.24
=1.8
–>
Identify the mistake in the probability distribution table.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.15 | 0.15 |
| 2 | 0.25 | 0.50 |
| 3 | 0.30 | 0.90 |
| 4 | 0.20 | 0.80 |
| 5 | 0.15 | 0.75 |
Solution
The values of P(x) do not sum to one.
Identify the mistake in the probability distribution table.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.15 | 0.15 |
| 2 | 0.25 | 0.40 |
| 3 | 0.25 | 0.65 |
| 4 | 0.20 | 0.85 |
| 5 | 0.15 | 1 |
Solution
The values of xP(x) are not correct.
–>
Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. He has the following probability distribution.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.35 | |
| 2 | 0.20 | |
| 3 | 0.15 | |
| 4 | ||
| 5 | 0.10 | |
| 6 | 0.05 |
Define the random variable X.
Solution
Let X = the number of years a physics major will spend doing post-graduate research.
Define P(x), or the probability of x.
Solution
Let P(x) = the probability that a physics major will do post-graduate research for x years.
–>
Find the probability that a physics major will do post-graduate research for four years. P(x = 4) = _______
Solution
1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15
Find the probability that a physics major will do post-graduate research for at most three years. P(x ≤ 3) = _______
Solution
0.35 + 0.20 + 0.15 = 0.70
–>
On average, how many years would you expect a physics major to spend doing post-graduate research?
Solution
1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years
Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year’s class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution.
- Let X = the number of years a student will study ballet with the teacher.
- Let P(x) = the probability that a student will study ballet x years.
Complete [link] using the data provided.
| x | P(x) | x*P(x) |
|---|---|---|
| 1 | 0.10 | |
| 2 | 0.05 | |
| 3 | 0.10 | |
| 4 | ||
| 5 | 0.30 | |
| 6 | 0.20 | |
| 7 | 0.10 |
In words, define the random variable X.
Solution
X is the number of years a student studies ballet with the teacher.
P(x = 4) = _______
Solution
1 – 0.10 – 0.05 – 0.10 – 0.30 – 0.20 – 0.10 = 0.15 –>
P(x < 4) = _______
Solution
0.10 + 0.05 + 0.10 = 0.25
On average, how many years would you expect a child to study ballet with this teacher?
Solution
1(0.10) + 2(0.05) + 3(0.10) + 4(0.15) + 5(0.30) + 6(0.20) + 7(0.10) = 4.5 years –>
What does the column “P(x)” sum to and why?
Solution
The sum of the probabilities sum to one because it is a probability distribution.
What does the column “x*P(x)” sum to and why?
Solution
The sum of xP(x) = 4.5; it is the mean of the distribution.
–>
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win 💲30. If it is not a face card, you pay 💲2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?
Solution
[latex]-2\left(\frac{40}{52}\right)+30\left(\frac{12}{52}\right)=-1.54+6.92=5.38[/latex]
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win 💲30. If it is not a face card, you pay 💲2. There are 12 face cards in a deck of 52 cards. Should you play the game?
Solution
Yes, because there is a positive expected value, and the more you play, the more likely you are to get closer to the expected value. –>
Section 4.3
Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status.
In words, define the random variable X.
Solution
X = the number that reply “yes”
X ~ _____(_____,_____)
Solution
B(8,0.713)
–>
What values does the random variable X take on?
Solution
0, 1, 2, 3, 4, 5, 6, 7, 8
Construct the probability distribution function (PDF).
| x | P(x) |
|---|---|
Solution
–>
On average (μ), how many would you expect to answer yes?
Solution
5.7
What is the standard deviation (σ)?
Solution
1.2795
–>
What is the probability that at most five of the freshmen reply “yes”?
Solution
0.4151
What is the probability that at least two of the freshmen reply “yes”?
Solution
0.9990 –>
Section 4.4
Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select a freshman from the study until you find one who replies “yes.” You are interested in the number of freshmen you must ask.
In words, define the random variable X.
Solution
X = the number of freshmen selected from the study until one replied “yes” that same-sex couples should have the right to legal marital status.
X ~ _____(_____,_____)
Solution
G(0.713) –>
What values does the random variable X take on?
Solution
1,2,…
Construct the probability distribution function (PDF). Stop at x = 6.
| x | P(x) |
|---|---|
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
| 6 |
Solution
–>
On average (μ), how many freshmen would you expect to have to ask until you found one who replies “yes?”
Solution
1.4
What is the probability that you will need to ask fewer than three freshmen?
Solution
0.9176 –>
Section 4.5
Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample.
In words, define the random variable X.
Solution
X = the number of business majors in the sample.
X ~ _____(_____,_____)
Solution
H(16, 7, 9) –>
What values does X take on?
Solution
2, 3, 4, 5, 6, 7, 8, 9
Find the standard deviation.
Solution
–>
On average (μ), how many would you expect to be business majors?
Solution
6.26
Section 4.6
Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day.
Assume the event occurs independently on any given day. Define the random variable X.
Solution
X ~ P(120)
–>
What values does X take on?
Solution
0, 1, 2, 3, 4, …
What is the probability of getting 150 customers in one day?
Solution
0.0010
–>
What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.
Solution
0.0485
What is the probability that the store will have more than 12 customers in the first hour?
Solution
0.2084
–>
What is the probability that the store will have fewer than 12 customers in the first two hours?
Solution
0.0214
Which type of distribution can the Poisson model be used to approximate? When would you do this?
Solution
The Poisson distribution can approximate a binomial distribution, which you would do if the probability of success is small and the number of trials is large.
–>
Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age.
Assume the event occurs independently on any given day. In words, define the random variable X.
Solution
X = the number of U.S. teens who die from motor vehicle injuries per day.
X ~ _____(_____,_____)
Solution
P(8)
–>
What values does X take on?
Solution
0, 1, 2, 3, 4, …
For the given values of the random variable X, fill in the corresponding probabilities.
Solution
–>
Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.
Solution
No
Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.
Solution
No
–>
the long-term average of many trials of a statistical experiment
