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 lasting for less than 8.3 minutes based on a sample of 25 sessions (do not round). That is, P( X-bar ≤8.3)=

• A. 0.9332
• B. 0.8413
• C. 0.0668
• D. 0.1587

Answer 1

We know the population parameters.

The population distribution is normally distributed with a mean of 8, and a standard deviation of 1.

A sample of size 25 is taken. Thus, according to the Central Limit Theorem, we know that the sampling distribution of the sample means has mean = to population mean and a standard deviation = to population standard deviation divided by the square root of the sample size.

Thus, sampling distribution mean= 8

Sampling distribution standard deviation = 1/sqrt(25)

= 1/5

=0.2.0

Thus, required z score = (value-mean) / standard_deviation

= (8.3-8) / 0.2

= 0.3/ 0.2

= 1.5

The area under the curve for a z score of 1.5 can be found from a z distribution table.

The required probability is 0.9332.

Thus, the correct answer is A. 0.9332.