Q: Given a normal distribution with $$\mu=40$$ and $$\sigma=2$$, between what two X values (symmetrically distributed around the mean) are 80% of the values?

$$40 \pm 1.28 \times 2 = (37.44, 42.56)$$

Q: A survey shows 65% of people prefer working from home. In a sample of 81 workers, what is the probability that between 62% and 68% of the sample prefer working from home?

Standard error: $$SE = \sqrt{\frac{0.65 \times 0.35}{81}} \approx 0.052$$; $$z_1$$ = \frac{0.62-0.65}{0.052} \approx -0.577$$, z_2$ = \frac{0.68-0.65}{0.052} \approx 0.577$$; $$P = P(-0.577 \u003c Z \u003c 0.577) \approx 0.438$$

Q: The diameter of a brand of tennis balls is normally distributed with mean 2.64 inches and standard deviation 0.05 inch. For a sample of 16 balls, between what two values of $$\bar{X} ($$symmetrically around the mean) are 60% of the sample means expected to fall?

$$2.64 \pm 0.524 \times \frac{0.05}{4} = (2.63345, 2.64655)$$

Question 1

A supermarket study found that the time spent shopping for a specific basket of items is uniformly distributed between 20 and 40 minutes. What is the probability that a randomly selected shopper spends less than 35 minutes in the store?

Accepted Answer

$$P(X \u003c 35) = \frac{35-20}{40-20} = 0.75$$

Question 2

The time between arrivals at a traffic intersection follows an exponential distribution with a mean of 30 seconds. What is the probability that the time between arrivals is 32 seconds or less?

Accepted Answer

$$P(X \leq 32) = 1 - e$$^{-32/30}$$ \approx 0.655$$

Question 3

Given a normal distribution with $$\mu=40$$ and $$\sigma=2$$, between what two X values (symmetrically distributed around the mean) are 80% of the values?

Accepted Answer

$$40 \pm 1.28 	imes 2 = (37.44, 42.56)$$

Question 4

A uniform distribution describes the time spent shopping between 10 and 40 minutes. What are the mean and standard deviation of this distribution?

Accepted Answer

Mean: $$\mu = \frac{10+40}{2} = 25$$; Std Dev: $$\sigma = \sqrt{\frac{(40-10)^2$}{12}} \approx 8.66$$

Question 5

A random sample of 50 people includes 42 employed individuals. If the population proportion is 0.85, what is the sample proportion and the standard error of the proportion?

Accepted Answer

Sample proportion: $$\hat{p} = \frac{42}{50} = 0.84$$; SE: $$\sqrt{\frac{0.85 	imes 0.15}{50}} \approx 0.0506$$

Question 6

A population has a median salary of $$60,000 and a mean salary of 65$$,000. If a sample of 9 is selected, is the data definitely normally distributed?

Accepted Answer

No, because the mean and median are not equal, indicating skewness.

Question 7

Given a standardized normal distribution, what is the probability that $$Z is $$between $$-0.75$$ and $$0.5$$?

Accepted Answer

$$P(-0.75 \u003c Z \u003c 0.5) = P(Z \u003c 0.5) - P(Z \u003c -0.75) \approx 0.6915 - 0.2266 = 0.4649$$

Question 8

A survey shows 65% of people prefer working from home. In a sample of 81 workers, what is the probability that between 62% and 68% of the sample prefer working from home?

Accepted Answer

Standard error: $$SE = \sqrt{\frac{0.65 	imes 0.35}{81}} \approx 0.052$$; $$z_1$$ = \frac{0.62-0.65}{0.052} \approx -0.577$$, z_2$ = \frac{0.68-0.65}{0.052} \approx 0.577$$; $$P = P(-0.577 \u003c Z \u003c 0.577) \approx 0.438$$

Question 9

The diameter of a brand of tennis balls is normally distributed with mean 2.64 inches and standard deviation 0.05 inch. For a sample of 16 balls, between what two values of $$\bar{X} ($$symmetrically around the mean) are 60% of the sample means expected to fall?

Accepted Answer

$$2.64 \pm 0.524 	imes \frac{0.05}{4} = (2.63345, 2.64655)$$

Question 10

Given a normal distribution with $$\mu=80$$ and $$\sigma=4$$, what is the probability that $$X \u003e 83$$?

Accepted Answer

$$P(X \u003e 83) = P(Z \u003e \frac{83-80}{4}) = P(Z \u003e 0.75) = 1 - 0.7734 = 0.2266$$

Question 11

For a normal distribution with $$\mu=80$$ and $$\sigma=4$$, five percent of the values are less than what $$X$$ value?

Accepted Answer

$$X = 80 + (-1.645) 	imes 4 = 73.42$$

Question 12

The weight of a product is normally distributed with mean 80 grams and standard deviation 0.04 gram. What is the probability that an individual product weighs less than 80.5 grams?

Accepted Answer

$$P(X \u003c 80.5) = P(Z \u003c \frac{80.5-80}{0.04}) = P(Z \u003c 12.5) \approx 1$$

Question 13

If a sample of 16 products is selected from a population with mean 80 grams and standard deviation 0.04 gram, what is the probability that the sample mean weight is less than 80.5 grams?

Accepted Answer

$$SE = \frac{0.04}{\sqrt{16}} = 0.01$$; $$Z = \frac{80.5-80}{0.01} = 50$$; $$P(Z \u003c 50) \approx 1$$

Question 14

A company wants to estimate the proportion of customers who are satisfied with their service. If the true proportion is 0.7 and a sample of 100 customers is taken, what is the standard error of the sample proportion?

Accepted Answer

$$SE = \sqrt{\frac{0.7 	imes 0.3}{100}} = 0.0458$$

Question 15

A business analyst claims that the average time to complete a transaction is 5 minutes with a standard deviation of 1 minute. If a random sample of 36 transactions is taken, what is the probability that the sample mean is greater than 5.2 minutes?

Accepted Answer

$$SE = \frac{1}{\sqrt{36}} = 0.1667$$; $$Z = \frac{5.2-5}{0.1667} = 1.2$$; $$P(Z \u003e 1.2) = 1 - 0.8849 = 0.1151$$

Question 16

A company finds that the time to resolve a customer complaint is exponentially distributed with a mean of 10 minutes. What is the probability that a complaint is resolved in less than 8 minutes?

Accepted Answer

$$P(X \u003c 8) = 1 - e$$^{-8/10}$$ = 1 - e$$^{-0.8}$$ \approx 0.5507$$

Question 17

A normal distribution has $$\mu=100$$ and $$\sigma=15. $$What value of $$X$$ corresponds to the 90th percentile?

Accepted Answer

$$X = 100 + 1.2816 	imes 15 = 119.22$$

Question 18

A sample of 25 light bulbs has a mean life of 1200 hours and a standard deviation of 100 hours. What is the 95% confidence interval for the mean life of the bulbs?

Accepted Answer

$$CI = 1200 \pm 1.96 	imes \frac{100}{5} = 1200 \pm 39.2 = (1160.8, 1239.2)$$

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