Title: Probability and Probability Distributions
Course: Question Banks

Q: Which blood type is the most unusual?
- Blood Type O (this is the most usual type; it has the highest probability.)
- Blood Type A (Type A is not unusual. There is a 41% chance that someone has Type A blood.)
- Blood Type B (There is a 10% chance that someone has Type B blood. This is more unusual than O or A. But it is not the most unusual.)
- Blood Type AB (There is a 4% chance that someone has Type AB blood. This is the most unusual because it has the smallest probability.)
A: Blood Type AB

Q: The 2014 U.S. Census Bureau Survey of Income and Program Participation includes information about living arrangements, partially summarized in the table above (Source: https://www.census.gov/content/dam/Census/library/publications/2017/demo/p70br-148.pdf)

Suppose we collect a random sample of 2000 people residing in the U.S. Based on the probability distribution, which result would be surprising?
- 230 live alone. (Probability is a long-run relative frequency. For a random sample of 2000 people, the relative frequency is 230/ 2000 = 0.115. This is close to the probability 0.116 (11.6%) given in the table. So this is not surprising.)
- 750 live with a parent / child. (Probability is a long-run relative frequency. For a random sample of 2000 students, the relative frequency is 750/ 2000= 0.375. This is close to the probability 0.374 (37.4%) given in the table. So this is not surprising.)
- 400 live with their spouse (Probability is a long-run relative frequency. For a random sample of 2000 people, the relative frequency is 400/ 2000 = 0.2. This is not close to the probability 0.392 (39.2%) given in the table. Therefore, this is a surprising result based on the probabilities given in the table.)
A: 400 live with their spouse

Q: This data represents the responses of 200 community college students. The variable is “the number of credit cards” the students had.

What is the probability that a student has at least one credit card?
- 0.61 (This is the probability that a student has no credit cards. “At least” one credit card means 1 or more than 1. The complement of “at least one credit card” is “no credit cards.”)
- 0.20 (This is the probability that a student has exactly one credit card. “At least” one credit card means 1 or more than 1. The complement of “at least one credit card” is “no credit cards.”)
- 0.80 (Did you calculate 1 – P(1 credit card)? “At least” one credit card means 1 or more than 1. The complement is “no credit cards.”)
- 0.39 (P(at least one credit card) = 1 – P(0 credit cards) = 1 – 0.61 = 0.39)
A: 0.39

Q: This data represents the responses of 200 community college students. The variable is “the number of credit cards” the students had.

What is the probability that a student has at most 4 credit cards?
- 0.01 (This is the probability that a student has exactly 4 credit cards. “At most 4” credit cards means 4 or fewer than 4. For this probability distribution, the complement of “at most 4 credit cards” is “5 credit cards.”)
- 0.04 (This is the probability that a student as 4 or 5 credit cards. “At most 4” credit cards means 4 or fewer than 4. For this probability distribution, the complement of “at most 4 credit cards” is “5 credit cards.”)
- 0.96 (Did you calculate 1 – P(4 or 5 credit cards)? “At most 4” credit cards means 4 or fewer than 4. For this probability distribution, the complement of “at most 4 credit cards” is “5 credit cards.”)
- 0.97 (P(at most 4 credit cards) = 1 – P(5 credit cards) = 1 – 0.03 = 0.97)
A: 0.97

Q: According to results from the Gallup’s 2016 annual Lifestyle Poll,

“Pet ownership among the American public breaks down this way: 27% own a dog but not a cat, 12% own a cat but not a dog, 17% own both, 3% own pets other than cats or dogs.”

Which of the following is the probability that a randomly selected person does not have any pets?
- 0.59 (This is the probability that a randomly chosen person has a pet.)
- 0.67 (Looks like you used the probability of a union formula. Note that there is no overlap between the categories based on the wording!)
- 0.41 (P(No pets) = 1 - P(has a pet) = 1 - (0.27 + 0.12 + 0.17 + 0.03) = .41)
A: 0.41

Q: Suppose the above table summarizes some results of an observational study of hockey and football athletes.

Which pair of statistics is useful in determining if being a football player independent of incurring brain trauma?
- 61/ 76 and 61/ 82 (These are the conditional probabilities P(Football | brain trauma) and P(brain trauma | Football). Recall how to test for independence.)
- 61/76 and 15/ 76 (These are the conditional probabilities that compare brain trauma risk between the two groups: P(Football | brain trauma) and P( Hockey | brain trauma) .)
- 61/ 82 and 82/ 151 (These are the probabilities: P(brain trauma | Football) and P(trauma). If these probabilities are approximately equal, we have brain trauma and football are independent.)
A: 61/ 82 and 82/ 151

Q: A student poll asked the question “Do you prefer cats, dogs, or neither?”. The results of the poll are as follows:

Which of the following pairs of statistics is NOT useful in determining if a preference for cats is independent of gender?
- 47/ 99 and 99/ 238 (This is P(overweight) and P(overweight | male). This is a useful comparison to investigate independence.)
- 52/ 99 and 133/ 238 (This is P(male) and P(male | overweight). This is a useful comparison to investigate independence.)
- 52/ 99 and 47/ 99 (This is P(woman | cats) and P(male | cats). These statistics only look at the people with a cat preference (note how these probabilities sum to 1!). Recall, if P(A) is approximately equal to P(A | B), then A and B are independent events.)
A: 52/ 99 and 47/ 99

Q: The above table summarizes some results of a small clinical trial of the medication N^ω-nitro-L-arginine methyl ester (TRT), used to treat cardiogenic shock.

Which pair of statistics is useful in determining if TRT is independent of survival?
- 11/15 and 11/16 (These are the conditional probabilities P(Survived | TRT) and P( TRT | Survived). Recall how to test for independence.)
- 11/15 and 5/ 15 (These are the conditional probabilities that compare survival rates between the two groups TRT and NoTRT.)
- 11/ 16 and 16/ 30 (These are the probabilities: P(TRT | Survived) and P(Survived). If these probabilities are approximately equal, we have TRT and survivall are independent.)
A: 11/ 16 and 16/ 30

Q: Choose a college student at random. Decide whether the following is a discrete or continuous random variable:

The number of courses the student takes this semester
- Continuous (Number of courses is a discrete random variable, since it can take any value from a list of distinct values: 1,2,3,4, etc.)
- Discrete (Number of courses is indeed a discrete random variable, since it can take any value from a list of distinct values: 1, 2, 3, 4, etc.)
A: Discrete

Q: Choose a college student at random. Decide whether the following is a discrete or continuous random variable:

The student's height
- Discrete (Student's height is a continuous random variable, since it can take any value in an interval. It's true that when rounded to the nearest inch (or centimeter) it looks like it is discrete, but the exact height of a person is continuous. You cannot list all possible heights.)
- Continuous (Student's height is indeed continuous, since it can take any value in an interval. It's true that when rounded to the nearest inch (or centimeter) it looks like it is discrete, but the exact height of a person is continuous. You cannot list all possible heights.)
A: Continuous

Q: Choose a college student at random. Decide whether the following is a discrete or continuous random variable:

The student's (exact) body temperature
- Discrete (Student's exact body temperature is a continuous random variable, since it can take any value in an interval. It's true that when rounded to the nearest degree Fahrenheit (or Celsius) it looks like it is discrete, but the exact body temperature of a person is continuous. You cannot list all possible temperatures.)
- Continuous (Student's exact body temperature is indeed a continuous random variable since it can take any value in an interval. It's true that when rounded to the nearest degree Fahrenheit (or Celsius) it looks like it is discrete, but the exact body temperature of a person is continuous. You cannot list all possible temperatures.)
A: Continuous

Q: Choose a college student at random. Decide whether the following is a discrete or continuous random variable:

The number of siblings the student has
- Continuous (Number of siblings is a discrete random variable, since it can take any value from a list of distinct values: 0,1,2,3,4, etc.)
- Discrete (Student's number of siblings is indeed a discrete random variable, since it can take any value from a list of distinct values: 0,1,2,3,4, etc.)
A: Discrete

Q: Choose a college student at random. Decide whether the following is a discrete or continuous random variable:

The (exact) time the student spends doing school work during a week
- Discrete (The exact time a student spends doing school work during a week is a continuous random variable, since it can take any value in an interval. It's true that when rounded to the nearest hour or minute it looks like it is discrete, but the exact time is continuous. You cannot list all possible exact times.)
- Continuous (The exact time a student spends doing school work during a week is indeed a continuous random variable since it can take any value in an interval. It's true that when rounded to the nearest hour or minute it looks like it is discrete, but the exact time is continuous. You cannot list all possible exact times.)
A: Continuous

Q: Choose a college student at random. Decide whether the following is a discrete or continuous random variable:

The number of alcoholic beverages that the student drinks in a typical week
- Continuous (The number of alcoholic beverages the student drinks in a typical week is a discrete random variable, since it can take any value from a list of distinct values: 0,1,2,3,4, etc.)
- Discrete (The number of alcoholic beverages the student drinks in a typical week is indeed a discrete random variable, since it can take any value from a list of distinct values: 0,1,2,3,4, etc.)
A: Discrete

Q: A couple plans to have no more than two children, and they will keep having children until they have a girl. So, if their first child is a girl, they will stop and have only one child. However, if their first child is a boy, they will try again and have a second child.

As it turns out, the probability of having a boy is slightly greater than having a girl. Here is the probability distribution for the number of boys the couple could have.

Boys | 0 boys | 1 boy | 2 boys
Probability | 0.49 | 0.25 | 0.26

What is the expected number of boys the couple will have? (Recall: the expected value is the mean).
- 1.5 (You may have averaged the number of boys. But the expected number of boys is the weighted average using the probabilities as weights.)
- 0.257 (You may have calculated the weighted mean and then divided by 3. But you should not divide by 3. To calculate the weighted mean, we multiply each probability by its corresponding number of boys, and then add these products together.)
- 0.77 (The expected number of boys is the weighted mean which is 0.490(0) + 0.25(1) + 0.26(2))
A: 0.77

Q: Consider the probability distribution of X, where X is the number of times a college graduate changed majors.

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8
P(X) | 0.135 | 0.271 | 0.271 | 0.180 | 0.090 | 0.036 | 0.012 | 0.003 | 0.002

What is the probability that a randomly selected college graduate has changed majors more than two times?
- 0.594 (This is the probability that a student changed majors at least twice (2 or more) not more than two times (must be greater than 2 times).)
- 0.271 (This is the probability that a student changed majors exactly twice, P(X=2). P(change majors at least two times) = 1 - P(change majors 0, 1, or 2 times).)
- 0.406 (This is the probability that a student did not change majors twice, 1 – P(X=2). However, the probability we are asking for is P(change majors more than two times) = 1 - P(change majors 0, 1, or 2 times).)
- 0.323 (P(change majors more than two times) = 1 - P(change majors 0, 1, or 2 times) = 1 - (0.135 + 0.271 + 0.271) = 1 - 0.677 = 0.323.)
A: 0.323

Q: The shaded area in the probability histogram above represents the probability of a male’s shoe size being:
- At least 12 (“At least 12” means all shoes sizes 12 or bigger will be counted. However, the shoe size 15 is not shaded in the diagram (but is captured by the statement “at least 12”).)
- More than 12 (The statement “more than 12” does not include 12 (but 12 is shaded), while it does include 15 (but 15 is not shaded).)
- Between (not including) 12 and 14 (The sizes 12 and 14 are shaded in the diagram, but the statement does not include them.)
- Less than 14 (The statement “less than 14” does not include 14 (but 14 is shaded), while it does include 11 (but 11 is not shaded).)
- Between (and including) 12 and 14 (The shoe sizes 12, 12.5, 13, 13.5, 14 are all included, but all other sizes are excluded.)
A: Between (and including) 12 and 14

Q: Which notation represents the shaded area?
- P(X ≥ 12) (P(X ≥ 12) means X is at least 12, that is, X is 12 or greater than 12. The shaded area corresponds to the interval where X is 12 or greater and also less than or equal to 14. Review the Spotlight on Inequality Notation.)
- P(X > 12) (P(X > 12) means X is greater than 12. The shaded area corresponds to the interval where X is greater than or equal to 12, and also less than or equal to 14. So X is between (and including) 12 and 14. Review the Spotlight on Inequality Notation.)
- P(12 < X < 14) (P(12 < X < 14) means X is between 12 and 14 (but 12 and 14 are not included). Review the Spotlight on Inequality Notation.)
- P(X < 14) (P(X < 14) means X is less than 14. The shaded area corresponds to the interval where X is greater than or equal to 12, and also less than or equal to 14. So X is between (and including) 12 and 14. Review the Spotlight on Inequality Notation.)
- P(12 ≤ X ≤ 14) (P(12 ≤ X ≤ 14) means X is between 12 and 14 (including 12 and 14).)
A: P(12 ≤ X ≤ 14)

Q: Suppose we have a large group of people and record their body temperatures. We then make probability histograms with the data in the following two ways:

Method 1: With temperatures measured to the nearest degree Fahrenheit (°F)

Method 2: With temperatures measured to the nearest tenth degree Fahrenheit (0.1°F)

Which of the resulting probability histograms will fit a curve most closely?
- Method 1 (Measuring temperature to the nearest degree will result in a probability histogram with wider rectangles than measuring to the nearest tenth degree.)
- Method 2 (The probability histogram constructed using temperature measured to the nearest tenth degree will fit a curve better because it will have narrower rectangles.)
A: Method 2

Q: Which of the following probabilities is represented by the shaded region under the probability density curve?
- P(X > 2) (Only values to the right of 3 are shaded, but X > 2 refers to some values to the left of 3 (like 2.5, for example).)
- P(X < 2) (Values to the right of 3 are shaded, but X < 2 refers to values to the left of 2.)
- P(2 < X < 3) (Only values to the right of 3 are shaded But P(2 < X < 3) refers to values between 2 and 3.)
- P(X < 3) (The shaded values are all those to the right of 3, but X < 3 refers to all values to the left of 3.)
- P(X > 3) (X > 3 refers to all values to the right of 3, which is exactly the interval that is shaded on the picture.)
A: P(X > 3)

Q: If X is the continuous random variable, is it true that P(X > 3) = P(X ≥ 3)? Why?
- No, because the interval “X > 3” excludes 3, while the interval “X ≥ 3” includes 3. (It is true that “X > 3” and “X ≥ 3” refer to different intervals (one includes 3 and the other doesn’t), but the corresponding probabilities are actually equal. It might be a good idea to take another look at Comment 1 above.)
- Yes, because X = 3 cannot happen. (It is true that P(X > 3) = P(X ≥ 3), but the reasoning here is incorrect. If X is a continuous random variable, it can take on any value in a range, and 3 is in the range of possible values for this variable.)
- Yes, because the difference between the two shaded regions is that one includes the boundary and the other doesn’t, but the boundary line has no area. (There is no area over a single point, so the two areas under the probability density curve that represent P(X > 3) and P(X ≥ 3) are equal. In other words, P(X = 3) = 0.)
A: Yes, because the difference between the two shaded regions is that one includes the boundary and the other doesn’t, but the boundary line has no area.

Q: Which of the following statements most precisely describes the probability that a randomly chosen 11th grader scored 90 or more on this test?

The scores follow a normal distribution with μ = 71 and σ = 9.
- about 0.16 (0.16 is the probability that randomly chosen student scored 80 or above (since 80 is one standard deviation above the mean). So we can make a statement that is much more precise than this one.)
- about 0.05 (0.05 is the probability that a randomly chosen student scored more than 2 standard deviations away from the mean (it’s the area of both tails in the second statement of the Empirical Rule). 90 is more than 2 standard deviation above the mean, so we can make a more precise statement.)
- about 0.025 (90 is a little more than 2 standard deviations above the mean (71 + 2*9 = 71 + 18 = 89). So the area is a little less than 2.5% of the total area under the normal curve.)
A: about 0.025

Q: Scores on the final exam in Professor Meyer’s statistics class follow a normal distribution, with a mean of 82 and a standard deviation of 5.

Ron, who took Professor Meyer's class, scored 88 on the final. Ron's z-score is:
- 88 (88 is Ron's raw score. Recall that the z-score of a normal value is (value - mean) / standard deviation. Please try again.)
- 6 (It seems like you started well by calculating the difference (value - mean) = 88 - 82. Recall however that z-score is calculated: (value - mean) / standard deviation.)
- -1.2 (It seems like your mistake is that you calculated: (mean - value) / standard deviation. You switched "mean" and "value." Note that z-scores are calculated: (value - mean) / standard deviation. Please try again.)
- 1.2 (Indeed, Ron's z-score = (88 - 82) / 5 = 1.2.)
A: 1.2

Q: Scores on the final exam in Professor Meyer’s statistics class follow a normal distribution, with a mean of 82 and a standard deviation of 5.

Heather took Professor Meyer's class, and the z-score of her final is 1.2. This means that Heather's final score is:
- 1.2 points above the mean. (A z-score tells us how many standard deviations above or below the mean the value is. Positive z-scores indicate that the value is above the mean, and negative values indicate that the value is below the mean. Try again.)
- 1.2 points below the mean. (A z-score tells us how many standard deviations above or below the mean the value is. Positive z-scores indicate that the value is above the mean, and negative values indicate that the value is below the mean. Try again.)
- 1.2 standard deviations below the mean. (Recall that a positive z-score indicates that the value is above the mean, and a negative z-score indicates that the value is below the mean.)
- 1.2 standard deviations above the mean. (Indeed, a z-score of 1.2 indicates that Ron's score of 88 is 1.2 standard deviations above the mean score.)
A: 1.2 standard deviations above the mean.

Q: Scores on the final exam in Professor Meyer’s statistics class follow a normal distribution, with a mean of 82 and a standard deviation of 5.

Dan took Professor Meyer's class, and the z-score of his final is -2.4. This means that Dan's score is:
- slightly lower than average. (While it is true that Dan's score is below average, his z-score indicates that his score is 2.4 standard deviations below the mean, which is much below average. In particular, from the standard deviation rule, we know that since Dan's score is less than 2 standard deviations below the mean, his score is in the lowest 2.5% of scores.)
- slightly above average. (A negative z-score indicates that the value is below the mean. Try again.)
- much higher than average. (A negative z-score indicates that the value is below the mean. Try again.)
- much lower than average. (Indeed, Dan's z-score indicates that his score is 2.4 standard deviations below the mean, which is much below average. In particular, from the standard deviation rule, we know that, since Dan's score is less than 2 standard deviations below the mean, his score is in the lowest 2.5% of scores.)
A: much lower than average.

Q: Scores on the final exam in Professor Meyer’s statistics class follow a normal distribution, with a mean of 82 and a standard deviation of 5.

Dan took Professor Meyer's class, and the z-score of his final is -2.4.

Dan's actual score on the final is:
- 94 (Dan's z-score of -2.4 indicates that his score is 2.4 standard deviations below the mean. It seems like you added 2.4 standard deviations to the mean. Try again.)
- 84.4 (Dan's z-score of -2.4 indicates that his score is 2.4 standard deviations below the mean. It seems like you added (rather than subtracted) 2.4 points (rather than standard deviations) to the mean. Try again.)
- 79.6 (Dan's z-score of -2.4 indicates that his score is 2.4 standard deviations below the mean. It seems like you subtracted 2.4 points from the mean rather than 2.4 standard deviations. Try again.)
- 70 (Indeed, Dan's z-score of -2.4 indicates that his score is 2.4 standard deviations below the mean. Since the mean is 82 and each standard deviation is 5, we get that Dan's score = 82 - 2.4(5) = 70.)
A: 70

Q: Scores on the final exam in Professor Meyer’s statistics class follow a normal distribution, with a mean of 82 and a standard deviation of 5.

Ron, who took Professor Meyer's class, scored 88 on the final.

Julie was assigned to take her statistics class with Professor Fisher, whose final scores follow a normal distribution with mean 75 and standard deviation of 6. Her score on the final was 84.

Which of the two scores, Ron's or Julie's, is more impressive?
- Ron's score is more impressive. (While it is true that Ron's raw score (88) is higher than Julie's (84), it makes more sense to compare their z-scores. In other words, it would make more sense to compare how they scored relative to other students in their respective classes. Try again.)
- Both scores are equally impressive. (In order to decide which score is more impressive, compare the z-scores of both Ron's and Julie's final scores. Bear in mind that each of the raw scores comes from a different normal distribution.)
- Julie's score is more impressive. (Indeed, Ron's z-score is (88 - 82) / 5 = 1.2, and Julie's z-score is (84 - 75) / 6 = 1.5. Julie's score is more impressive, since she scored 1.5 standard deviations above the mean, while Ron scored only 1.2 standard deviations above the mean.)
A: Julie's score is more impressive.

Q: The National Collegiate Athletic Association, NCAA, requires a combined SAT score of 820 or better for a student to be eligible to participate in college athletics. Suppose that SAT scores are normally distributed with a mean of 1026 and a standard deviation of 209.

What is the probability that a randomly selected student who takes the SAT has a qualifying SAT score?
- 0.99 (Did you calculate a z-score of 0.99? If so, you reversed X and µ in the calculation the z-score. The z-score is −0.99. Now find the probability, using the simulation to determine the area.)
- 0.1611 (You either reversed X and µ in the calculation the z-score and found P(z > 0.99) = 0.1611 or you have the correct z-score (z = −0.99), but you found P(z < −0.99) instead of P(z > −0.99).)
- 0.8389 (The z-score is −0.99. The P(X > 820) = P(z > −0.99) = 0.8389.)
A: 0.8389

Q: Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies above the 95th percentile have a condition called macrosomia (also known as “big baby syndrome”).

What birth weight marks the 95th percentile?
- 3985 grams (Did you use 0.95 as the Z-score? You need to use the simulation to find the Z-score for a probability of 0.95.)
- 2571 grams (Did you mark of the lowest 5% instead of the highest 5%? The 95 percentile means that 95% of the Z-scores are below our Z-score.)
- 4361 grams (x = μ + 1.64 · σ = 3466 + 1.64(546) = 3466 + 895.44 =4361.44, which is about 4361 grams.)
A: 4361 grams