Question

Suppose we somehow know that for the entire population of all students, the mean time spent exercising per week is 90 minutes, with a standard deviation of 40 minutes.

a) Would you expect time spent exercising per week to be normally distributed? Explain your reasoning. Consider your exercise habits as well as those of people you know.
b) We plan to collect random samples of 50 students and compute the sample mean time spent exercising, ?̅, for each sample. Would you expect the sampling distribution of ?̅, to be normally distributed? Explain briefly.
c) In approximately what proportion of samples would ?̅ be between 80 and 100 minutes?
d) Now suppose that the sample size is increased to 100 students per sample. In what proportion of
samples would ?̅ be between 80 and 100 minutes?

Answer 1

Solution :

Given that,

mean = = 90

standard deviation = = 40

a ) n = 50

?̅ = 90

?̅ ₌ / n = 40 50 = 5.6568

P (80< ?̅ < 100 )

P ( 80 - 90 / 5.6568) < ( ?̅ -  _?̅ / _?̅ ) < ( 100 - 90 / 5.6568)

P ( - 10 / 5.6568 < z < 10 / 5.6568 )

P (-1.77 < z < 1.77 )

P ( z < 1.77 ) - P ( z < -1.77)

Using z table

= 0.9616 - 0.0384

= 0.9232

Probability = 0.9232

b ) n = 100

?̅ = 90

?̅ ₌ / n = 40 100 = 4

P (80< ?̅ < 100 )

P ( 80 - 90 / 4) < ( ?̅ -  _?̅ / _?̅ ) < ( 100 - 90 /4)

P ( - 10 / 4 < z < 10 / 4 )

P (-2.5 < z < 2.5 )

P ( z < 2.5 ) - P ( z < -2.5)

Using z table

= 0.9938 - 0.0062

= 0.9876

Probability = 0.9876