Question

The mean time to prepare a tax return is 90 minutes and the standard deviation of this distribution is 14 minutes. Suppose 49 returns from this year are selected and analyzed regarding the preparation time. What is the probability that the mean time for the sample of 49 returns is between 86 minutes and 92 minutes?

Answer 1

P ( 86<X<92 )=P ( 86−90< X−μ<92−90 )=P ((86−90)/2<(X−μ)/σ<(92−90)/2)
Since Z=(x−μ)/σ , (86−90)/2=−2 and (92−90)/2=1 we have:
P ( 86<X<92 )=P ( −2<Z<1 )
Use the standard normal table to conclude that:
P ( −2<Z<1 )=0.8185