Question

18 the mean temperature in gen falls for the month of february is 23 degrees with a standard deviation of 4.2 degrees. what is the z - score for a temperature of 17 degrees? a. 1 b. 1.43 c. - 1.43 d. 11.5 19. the 40 yards sprint times for a soccer team are found to be normally distributed with a mean of 5.2 seconds and a standard deviation on 0.3 seconds. what is the z - score for a player who runs a time of 5.6 seconds? a. - 1.33 b. 0.88 c. 1.02 d. 1.33 20. the mean number of accidents a week at a company is 6.4 with a standard deviation of 1.5. what proportion of weeks would you expect to have less than 5 accidents? a. 0.6915 b. 0.1762 c. 0.8238 d. - 0.93

Explanation:

Step1: Recall the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step2: Solve for problem 18

Given \(\mu = 23\), \(\sigma=4.2\), and \(x = 17\).
Substitute into the formula: \(z=\frac{17 - 23}{4.2}=\frac{-6}{4.2}\approx - 1.43\)

Step3: Solve for problem 19

Given \(\mu = 5.2\), \(\sigma = 0.3\), and \(x = 5.6\)
Substitute into the formula: \(z=\frac{5.6-5.2}{0.3}=\frac{0.4}{0.3}\approx1.33\)

Step4: Solve for problem 20

First, find the z - score. Given \(\mu = 6.4\), \(\sigma=1.5\), and \(x = 5\)
\(z=\frac{5 - 6.4}{1.5}=\frac{-1.4}{1.5}\approx - 0.93\)
Then, look up the z - score in the standard normal table. The area to the left of \(z=-0.93\) is \(0.1762\)

Answer:

1. C. - 1.43
2. D. 1.33
3. B. 0.1762