Question

Time spent using e-mail per session is normally distributed,
with m = 9 minutes and s = 2 minutes. If you select a random
sample of 25 sessions,
a. what is the probability that the sample mean is between 8.8 and
9.2 minutes?
b. what is the probability that the sample mean is between 8.5 and
9 minutes?

c. If you select a random sample of 100 sessions, what is the prob-
ability that the sample mean is between 8.8 and 9.2 minutes?

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

Answer 1

Solution :

Given that,

mean = = 9

standard deviation = = 2

n = 25

= = 9

= / n = 2 / 25 = 0.4

a) P(8.8 < < 9.2)  

= P[(8.8 - 9) / 0.4 < ( - ) / < (9.2 - 9) / 0.4)]

= P(-0.50 < Z < 0.50)

= P(Z < 0.50) - P(Z < -0.50)

Using z table,  

= 0.6915 - 0.3085  

= 0.3830

b) P(8.5 < < 9)  

= P[(8.5 - 9) / 0.4 < ( - ) / < (9 - 9) / 0.4)]

= P(-1.25 < Z < 0)

= P(Z < 0) - P(Z < -1.25)

Using z table,  

= 0.5 - 0.1056

= 0.3944

c) n = 100

= = 9

= / n = 2 / 100 = 0.2

P(8.8 < < 9.2)  

= P[(8.8 - 9) / 0.2 < ( - ) / < (9.2 - 9) / 0.2)]

= P(-1.00 < Z < 1.00)

= P(Z < 1.00) - P(Z < -1.00)

Using z table,  

= 0.8413 - 0.1587

= 0.6826

A probability of part (b) is higher than probability part (a), because higher sample size,then smaller standard deviation