Q: If 200 shoppers enter the supermarket, and the probability that a shopper spends between 26 and 56 minutes in the store is 0.7333, how many shoppers would you expect to spend that amount of time in the store?

$$200 \times 0.7333 = 146.66 \approx 147$$ shoppers

Q: If there are 40 players on a roster, how many would you expect to have a batting average of 0.260 or greater, given the probability is 0.1894?

$$40 \times 0.1894 = 7.576 \approx 8$$ players

Q: Using technology, you find that the probability a state has between 300 and 350 active physicians per 100,000 residents is 0.393. If you randomly select 10 states, how many would you expect to fall in this range?

$$10 \times 0.393 = 3.93 \approx 4$$ states

Question 1

Suppose the time (in minutes) U.S. workers spend commuting to work is normally distributed with a mean of 26.8 and a standard deviation of 17.1. What is the z-score for a worker who commutes for 15 minutes?

Accepted Answer

$$z = \frac{15 - 26.8}{17.1} = -0.69$$

Question 2

A stretch of highway has vehicle speeds that are normally distributed with a mean of 67 mph and a standard deviation of 3.5 mph. What is the probability that a randomly selected vehicle is violating the speed limit of 70 mph or higher?

Accepted Answer

First, calculate $$z = \frac{70 - 67}{3.5} = 0.86. $$The probability is $$P(z \u003e 0.86) \approx 0.1949.$$

Question 3

The time spent in a supermarket is normally distributed with a mean of 47 minutes and a standard deviation of 12 minutes. What is the probability that a shopper spends between 26 and 56 minutes in the store?

Accepted Answer

Calculate $$z_1$$ = \frac{26 - 47}{12} = -1.75$$, z_2$ = \frac{56 - 47}{12} = 0.75$$. $$P(-1.75 \u003c z \u003c 0.75) = 0.7734 - 0.0401 = 0.7333.$$

Question 4

If 200 shoppers enter the supermarket, and the probability that a shopper spends between 26 and 56 minutes in the store is 0.7333, how many shoppers would you expect to spend that amount of time in the store?

Accepted Answer

$$200 	imes 0.7333 = 146.66 \approx 147$$ shoppers

Question 5

A survey indicates that the number of active physicians per state is normally distributed with a mean of 286 and a standard deviation of 57 (per 100,000 residents). What is the probability that a randomly selected state has fewer than 300 active physicians per 100,000 residents?

Accepted Answer

Calculate $$z = \frac{300 - 286}{57} = 0.25$$. $$P(z \u003c 0.25) \approx 0.5987$$

Question 6

The batting averages of Major League Baseball players are normally distributed with a mean of 0.245 and a standard deviation of 0.017. What percent of players have a batting average of 0.260 or greater?

Accepted Answer

Calculate $$z = \frac{0.260 - 0.245}{0.017} = 0.882$$. $$P(z \u003e 0.882) \approx 0.1894$$, or 18.94%.

Question 7

If there are 40 players on a roster, how many would you expect to have a batting average of 0.260 or greater, given the probability is 0.1894?

Accepted Answer

$$40 	imes 0.1894 = 7.576 \approx 8$$ players

Question 8

A shopper spends more than 35 minutes in a supermarket where the mean time is 47 minutes and the standard deviation is 12 minutes. What is the probability that a randomly selected shopper spends more than 35 minutes in the store?

Accepted Answer

Calculate $$z = \frac{35 - 47}{12} = -1.00$$. $$P(z \u003e -1.00) = 0.8413$$

Question 9

Using technology, you find that the probability a state has between 300 and 350 active physicians per 100,000 residents is 0.393. If you randomly select 10 states, how many would you expect to fall in this range?

Accepted Answer

$$10 	imes 0.393 = 3.93 \approx 4$$ states

Question 10

Which of the following statements about the standard normal distribution is TRUE?

Accepted Answer

The area under the curve to the left of $$z = 0 is 0.5.$$

Question 11

A variable $$x is $$normally distributed with mean $$\mu$$ and standard deviation $$\sigma. $$Which formula correctly converts $$x to a $$standard normal variable $$z$$?

Accepted Answer

$$z = \frac{x - \mu}{\sigma}$$

Question 12

Suppose the average time spent in a store is 47 minutes with a standard deviation of 12 minutes. What is the probability that a shopper spends between 29 and 50 minutes in the store?

Accepted Answer

Calculate $$z_1$$ = \frac{29 - 47}{12} = -1.5$$, z_2$ = \frac{50 - 47}{12} = 0.25$$. $$P(-1.5 \u003c z \u003c 0.25) = 0.5987 - 0.0668 = 0.5319.$$

Question 13

Which of the following best describes the relationship between the area under a normal curve and probability?

Accepted Answer

The area under the normal curve for an interval represents the probability that a value falls within that interval.

Question 14

If a random variable $$x is $$normally distributed with mean $$\mu = 100$$ and standard deviation $$\sigma = 15$$, what is the probability that $$x is $$greater than 130?

Accepted Answer

Calculate $$z = \frac{130 - 100}{15} = 2.0$$. $$P(z \u003e 2.0) = 0.0228$$

Question 15

Which of the following is NOT a required input when using technology to find normal probabilities?

Accepted Answer

The sample size

Question 16

A company claims that the average time to resolve a customer complaint is 24 hours with a standard deviation of 6 hours. If the times are normally distributed, what is the probability that a randomly selected complaint is resolved in less than 18 hours?

Accepted Answer

Calculate $$z = \frac{18 - 24}{6} = -1.0$$. $$P(z \u003c -1.0) = 0.1587$$

Normal Probability Distributions: Finding Probabilities

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