Question

9) Electricity bills in a certain city have mean $104.88. Assume the bills are normally distributed with standard deviation $12.40. A sample of 69 bills was selected for an audit. Find the 38 percentile for the sample mean. Round to two decimal places.

10) According to one survey, the mean serum cholesterol level for US adults was 197.4 with a standard deviation 51.9. A simple random sample of 96 adults is chosen. Find the 46 percentile for the sample mean. Round to one decimal place.

Answer 1

9)

Z score for 38 percentile = -0.31

(x - 104.88) / (12.4 / sqrt(69)) = -0.31

or, x = 104.88 - 0.31 * (12.4 / sqrt(69))

or, x = 104.42

10) Z for 46 percentile = -0.1

(x - 197.4) / (51.9 / sqrt(96)) = -0.1

or, x = 197.4 - 0.1 * (51.9 / sqrt(96))

or, x = 196.9

Answer 2

9. To find the 38th percentile for the sample mean, we need to find the Z-score corresponding to that percentile and then use it to find the sample mean using the formula:

Z = (x̄ - μ) / (σ / √n)

where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

First, we find the Z-score corresponding to the 38th percentile using a standard normal distribution table or calculator:

Z = invNorm(0.38) ≈ -0.26

Next, we plug in the given values and solve for x̄:

-0.26 = (x̄ - 104.88) / (12.40 / √69)

-0.26 * (12.40 / √69) = x̄ - 104.88

x̄ = -0.26 * (12.40 / √69) + 104.88

x̄ ≈ 103.48

Therefore, the 38th percentile for the sample mean is approximately $103.48.

10. To find the 46th percentile for the sample mean, we follow a similar process as in question 9:

Z = invNorm(0.46) ≈ -0.12

x̄ = -0.12 * (51.9 / √96) + 197.4

x̄ ≈ 194.9

Therefore, the 46th percentile for the sample mean is approximately 194.9 mg/dL.