Q: A professor collects 200 papers and estimates that 30% will be missing names. What is the expected number of papers without names?

$$\mu_x = np = 200 \times 0.30 = 60$$

Q: Given $$n = 200$$ and $$p = 0.30$$, what is the standard deviation of the number of papers missing names?

$$\sigma_x = \sqrt{np(1-p)} = \sqrt{200 \times 0.30 \times 0.70} = 6.48$$

Q: If the mean number of papers missing names is 60 and the standard deviation is 6.48, which interval would be considered 'bell-shaped' for this distribution?

$$\mu \pm 2\sigma = 60 \pm 2 \times 6.48 = [47.04, 72.96]$$

Q: What is the probability that at least 75 papers will be missing names out of 200, if $$p = 0.30$$?

$$P(X \geq 75) = 1 - \text{binomcdf}(200, 0.30, 74) = 0.0138$$

Q: Suppose the scores on a statistics exam are normally distributed with mean $$\mu = 72.6$$ and $$\sigma = 10.6. $$What is the score for the 90th percentile?

$$x = 72.6 + 1.28 \times 10.6 = 86.188$$

Q: What is the top 5% of all wait times if the mean is 6.4 minutes and the standard deviation is 1.5 minutes?

$$x = 6.4 + 1.645 \times 1.5 = 8.87$$ minutes

Question 1

A professor collects 200 papers and estimates that 30% will be missing names. What is the expected number of papers without names?

Accepted Answer

$$\mu_x = np = 200 	imes 0.30 = 60$$

Question 2

Given $$n = 200$$ and $$p = 0.30$$, what is the standard deviation of the number of papers missing names?

Accepted Answer

$$\sigma_x = \sqrt{np(1-p)} = \sqrt{200 	imes 0.30 	imes 0.70} = 6.48$$

Question 3

If the mean number of papers missing names is 60 and the standard deviation is 6.48, which interval would be considered 'bell-shaped' for this distribution?

Accepted Answer

$$\mu \pm 2\sigma = 60 \pm 2 	imes 6.48 = [47.04, 72.96]$$

Question 4

What is the probability that at least 75 papers will be missing names out of 200, if $$p = 0.30$$?

Accepted Answer

$$P(X \geq 75) = 1 - 	ext{binomcdf}(200, 0.30, 74) = 0.0138$$

Question 5

A student scores 55 on a statistics exam with a mean of 72.6 and a standard deviation of 10.6. Would this score be considered 'unusual'?

Accepted Answer

Yes, because $$P(X \u003c 55) = 0.0485 \u003c 0.05$$, so it is unusual.

Question 6

Suppose the scores on a statistics exam are normally distributed with mean $$\mu = 72.6$$ and $$\sigma = 10.6. $$What is the score for the 90th percentile?

Accepted Answer

$$x = 72.6 + 1.28 	imes 10.6 = 86.188$$

Question 7

What is the probability that a student got a score of at least 70 on the exam ($$\mu = 72.6$$, $$\sigma = 10.6$$)?

Accepted Answer

$$P(X \geq 70) = 1 - P(Z \u003c -0.25) = 1 - 0.4013 = 0.5987$$

Question 8

A person waits between 5 and 8 minutes. If the mean wait time is 6.4 minutes and the standard deviation is 1.5 minutes, what is the probability that a person waits between 5 and 8 minutes?

Accepted Answer

$$P(5 \u003c X \u003c 8) = P(-0.93 \u003c Z \u003c 1.07) = 0.8577 - 0.1762 = 0.6815$$

Question 9

What is the top 5% of all wait times if the mean is 6.4 minutes and the standard deviation is 1.5 minutes?

Accepted Answer

$$x = 6.4 + 1.645 	imes 1.5 = 8.87$$ minutes

Question 10

A normal probability plot for a data set appears as a straight line. What does this indicate about the data?

Accepted Answer

The data is approximately normally distributed.

Question 11

What is the mean and standard deviation of the sampling distribution of the sample mean $$\bar{x}$$ for a population with mean $$\mu$$ and standard deviation $$\sigma$$, sample size $$n$$?

Accepted Answer

$$\mu_{\bar{x}} = \mu$$, $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$

Question 12

If the total energy needed during pregnancy is normally distributed with $$\mu = 2600$$ kcal/day and $$\sigma = 50$$ kcal/day, what is the probability that a randomly selected pregnant woman has an energy need of at least 2650 kcal/day?

Accepted Answer

$$P(X \geq 2650) = P(Z \geq 1) = 1 - 0.8415 = 0.1585$$

Question 13

Suppose a random sample of 20 pregnant women is taken. What is the standard deviation of the sampling distribution of the sample mean?

Accepted Answer

$$\sigma_{\bar{x}} = \frac{50}{\sqrt{20}} = 11.18$$

Question 14

What is the probability that the mean energy need of a sample of 20 pregnant women exceeds 2625 kcal/day?

Accepted Answer

$$P(\bar{x} \u003e 2625) = P(Z \u003e 2.236) = 1 - 0.9875 = 0.0125$$

Question 15

If $$P(\bar{x} \u003e 2625) = 0.0125$$ for a sample of 20 pregnant women, would this result be considered unusual?

Accepted Answer

Yes, because $$0.0125 \u003c 0.05.$$

Question 16

A binomial experiment has $$n = 12$$ trials and $$p = 0.75$$ probability of success. What is the probability that at least 4 are winners?

Accepted Answer

$$P(X \geq 4) = 1 - 	ext{binomcdf}(12, 0.75, 3) = 0.9999$$

Question 17

In a binomial experiment with $$n = 12$$ and $$p = 0.75$$, what is the probability that at most 7 are winners?

Accepted Answer

$$P(X \leq 7) = 	ext{binomcdf}(12, 0.75, 7) = 0.1574$$

Probability Distributions, Normal Approximation, and Sampling Distributions: Study Notes

Study Guide - Practice Questions

Study Guide - Flashcards