QUESTION 3
A computer repair service found that a random sample of 45 repair costs had a mean cost of \( \$ 659 \). Assume that the po the \( 95 \% \) confidence interval for the population mean repair cost \( \mu \) of all computers.
A. \( (612.5,705.5) \)
B. \( (627.5,690.5) \)
C. \( (622.5,695.5) \)
D. \( (623.5,694.5) \)
QUESTION 4
Let \( x \) be a random variable that represents the length of an Atlantic croaker fish. If \( x \) is normally distributed with a mean of of 2 inches, find the length of Croaker fish at the bottom of the top \( 15 \% \).
A. 7.927 inches
B. 8.234 inches
C. 11.546 inches
D. 12.073 inches
QUESTION 5
Let \( x \) be a random variable that represents the annual salary of an elementary school teacher. The mean annual salary is rep standard deviation is \( \$ 1800 \). If a random sample of 50 elementary school teachers is selected, what is the probability that th \( \$ 50,000 \) ?
A. 0.0102
B. 0.0312
C. 0.373
D. 0.312
QUESTION 6
Let \( x \) be a random variable representing the monthly cost of joining a health club. We may assume that \( x \) has a normal distributi standard deviation is \( \$ 570 \) A fitness manazine advertises that the mean monthlv rost of ioinino a health club is \( \$ 35 \) You work Click Save and Submit to save and submit. Click Save All Answers to save all answers.
Answer
QUESTION 6: \( \mu = 350, \sigma = 570 \), Calculate \( z \) score for your monthly cost and compare with the advertised mean.
Step 1 ：QUESTION 3: n = 45, \( \bar{x} = 659 \), \( \alpha = 0.05 \), Calculate \( \frac{s}{\sqrt{n}} \) for the confidence interval formula.
Step 2 ：QUESTION 4: \( \mu = 12.5, \sigma = 2 \), Calculate \( z \) score for the bottom of the top \( 15\% \) and use \( x = \mu + (z \times \sigma) \) to find the length.
Step 3 ：QUESTION 5: \( \mu = 50,000, \sigma = 1800, n = 50 \), Calculate \( z \) score for \( \bar{x} < 50,000 \) and use the standard normal table to find the probability.
Step 4 ：QUESTION 6: \( \mu = 350, \sigma = 570 \), Calculate \( z \) score for your monthly cost and compare with the advertised mean.
