Question

In a recent survey, the total sleep time per night among college students was approximately Normally distributed with mean μ = 6.76 hours and standard deviation σ = 1.24 hours. You plan to take an SRS of size n = 120 and compute the average total sleep time.

(a) What is the standard deviation for the average time? (Round your answer to three decimal places.) hours

(b) Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean. (Round your answers to three decimal places.) 95% of the time, we'll expect the sample mean to be between a lower value of and an upper value of hours.

(c) What is the probability that your average will be below 6.9 hours? (Round your answer to four decimal places.)

Answer 1

Answer)

A)

Standard deviation = s.d/√n = 1.24/√120 = 0.113

B)

According to the emperical rule

If the data is normally distributed

Then 68% lies in between mean - s.d and mean + s.d

95% lies in between mean - 2*s.d and mean + 2*s.d

99.7% lies in between mean - 3*s.d and mean + 3*s.d

Lower value = 6.76 - (2*0.113) = 6.534

Upper value = 6.76 + (2*0.113) = 6.986

C)

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/(s.d/√n)

Given mean

P(x<6.9)

Z = (x - mean)/(s.d/√n)

Z = (6.9 - 6.76)/0.113 = 1.24

From z table, P(z<1.24) = 0.8925