WOW gyms conducted a study to find the average amount of time women spend in the

Question

WOW gyms conducted a study to find the average amount of time women spend in the gym per week. They surveyed 144 women and they found out that the women population spent an average of 20 hours per week in the gym with a standard deviation of 10 hours.

(a) What is the standard error of the mean?

(b) Compute the probability the sample mean is greater than 20 hours?

(c) Compute the probability the sample mean is less than 18 hours?

(d) Compute the probability the sample mean is between 18 and 22 hours?

Explanation / Answer

Standard error is given by : s / sqrt(n)

= 10 / sqrt ( 144)

= 10 / 12

= 5/6

= 0.8333

b)

Mean = 20

P ( X > 20) = 0.5000 (since the distribution is assumed to be symmetric around the mean )

c)

P (X < 18)

= P ( Z < 18 - 20 / (10 / sqrt(144) )

= P ( Z < -2.4 )

= 0.0082

= 0.82%

d )

P ( 18 < X < 22)

= P ( X < 22) - P (X < 18)

= 0.99180 - 0.00820

= 0.98360

= 98.36%

Hope this helps.

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