Question

5.28  Total sleep time of college students. In Example 5.4, the total sleep time per night among college students was approximately Normally distributed with mean μ = 6.78 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.

1. (a) What is the standard deviation for the average time?

2. (b) Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean.

3. (c) What is the probability that your average will be below 6.9 hours?

Answer 1

Solution :

Given that,

mean = = 6.78 hours

standard deviation = = 1.24 hours.

n = 120

= = 6.78 hours.

a) = / n = 1.24 / 120 = 0.1132 hours.

Using Empirical rule,

b) P( - 2 < <   + 2 ) = 95%

= P( 6.78 - 2 * 0.1132 < < 6.78 + 2 * 0.1132 ) = 95%

= P( 6.78 - 0.2264 < < 6.78 + 0.2264 ) = 95%

=P( 6.5536 < < 7.0064 ) = 95%

c) P( < 6.9) = P(( - ) / < (6.9 - 6.78) / 0.1132)

= P(z < 1.06)

Using z table

= 0.8554