Question

The time spent completing problem sets per week is normally distributed with a mean of 8 hours and a standard deviation of 2 hours. What is the probability that at a randomly selected week, time spent on doing problem sets:

a) will take more than 9 hours?

b) will take less than 9.5 hours?

c) will take less than 6 hours?

If random samples of 25 weeks are taken, what proportion of the sample means:

d) would be more than 9 hours?

e) would be between 7.5 and 8 hours?

f) Explain the difference in the results of (a) and (d).

g) Which is more likely to occur – a particular week doing problem sets for more than 9 hours, a sample mean above 9 hours in a sample of 36 weeks, or a sample mean above 8.4 hours in a sample of 100 weeks? Explain.

Answer 1

X - Hime spent x- Normal (4=8, 02-22) will take more than ghown. PCX29) = 1- P(x<9) El-PCX-429-8 = P(Z<0.5) =1-0.6915 /P(x79) = 0.3086 by P(x499) = P(x-HL 29.53). = P(Z < 0.45) /P(X4 9 5) = 0.1134] celebrello 9 P(x <6) = P(X=4 <0-8) = P(Z <-)) = o 158 7 P(x< 6) = 0.15877

0 0.4 n = 25 x = sample mean. - NN1 ) (4% ) Их = ox = 5 Mx=8 x=2 -1.0.4 In Normal (8, 0.42) dy p(x > 9) = P(X-487 9-8) - = P(Z Z 2-5) El P(Z <2.5) Sl - 0.9938 = 0.0062 g) = 0.0062 c) P(7.5 <x<8) = P(1.5-8 < X 4 = 28-8) = P(-1.25 <2<0) = P(2<0) - P(Z <-1.25) = 0oo.5-0.105 - 0.3944 S<x<8) = 0.3944