Question

The random variable X (population) is the time spent (in minutes) using e-mail per session and it is known that X~N(8,1^2). An average session X-bar is based on a sample of 25 sessions. Determine the probability of an average session of e-mail lasting for more than 8.5 minutes.

A. 0.6554

B. 0.6915

C. 0.3085

D. 0.0062

Answer 1

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We have been given the following:

Xbar = 8.5, Mu = 8, Stdev = 1 ( from the normal distribution notation). Also, sample size, n = 25

We will use normal distribution test statistic(Z) and not student' distribution statistic (t) because population standard deviation is known to us. if it wasn't then we should have used the t-statistic.

using the above parameters we normalize our distribution :

P(X>8.5) = ?

= P(Z > (8.5-8)/(1/sqrt(25))

= P(Z > 2.5)

= .00621 or .0062

Option D. is correct