Question

Use this scenario for questions 1-5

Studies indicate that men are spending more money on personal care products. In a recent study, men are reported to spend on average $588 per year for personal care products. The population distribution of the amounts spent follows a normal distribution with a standard deviation of $45 per year.

1. What is the probability that a randomly selecting man spends $627.15 or more on personal care products?  

Answer with three decimal places.

2.Which dollar amounts define the middle 88% of this distribution of $’s spent on personal-care products for men?

$535.1 and $640.9

$518 and $640.9

$518 and $658 Can't be determined

3.A random sample of 75 men is selected. What is the probability that the mean amount spent for those sampled is $580.36 or less? Answer with three decimal places.

4. A random sample of 75 men is selected and the sample mean of dollars spent on personal care products was $586.75. The 95% confidence interval for the true mean is

Answer 1

Solution:-

Mean = 588, S.D = 45

1) The probability that a randomly selecting man spends $627.15 or more on personal care products is 0.192.

x = 627.15

By applying normal distribution:-

z = 0.87

P(z > 0.87) = 0.192

2) 88% of this distribution of $’s spent on personal-care products for men is 518 and 658.

p-value for the middle 88% = 0.06 and 0.94

z-score for the p-value = + 1.555

By applying normal distribution:-

x₁ = 518.025

x₂ = 657.975

3) The probability that the mean amount spent for those sampled is $580.36 or less is 0.071.

x = 580.36

By applying normal distribution:-

z = - 1.47

P(z < -1.47) = 0.071.

4) The 95% confidence interval for the true mean is C.I = (576.566, 596.934).

C.I = 586.75 + 1.96 × 5.19615

C.I = 586.75 + 10.1845

C.I = (576.566, 596.934)