Question

Time spent using​ e-mail per session is normally​ distributed, with

mu equals μ=9

minutes and

sigma equals σ=2

minutes. Assume that the time spent per session is normally distributed. Complete parts​ (a) through​ (d).

a. If you select a random sample of

25

​sessions, what is the probability that the sample mean is between

8.8

and

9.2

​minutes?

​(Round to three decimal places as​ needed.)

b. If you select a random sample of

25

​sessions, what is the probability that the sample mean is between

8.5

and

9

​minutes?

​(Round to three decimal places as​ needed.)

c. If you select a random sample of

100

​sessions, what is the probability that the sample mean is between

8.8

and

9.2

​minutes?

​(Round to three decimal places as​ needed.)

d. Explain the difference in the results of​ (a) and​ (c). Choose the correct answer below.

The sample size in​ (c) is greater than the sample size in​ (a), so the standard error of the mean​ (or the standard deviation of the sampling​ distribution) in​ (c) is

▼

greater

less

than in​ (a). As the standard error

▼

decreases

increases

​, values become

▼

more

less

concentrated around the mean.​ Therefore, the probability of a region that includes the mean will always

▼

increase

decrease

when the sample size increases.

Answer 1

Solution :

Given that,

mean = $\mu$ = 9 minutes

standard deviation = $\sigma$ = 2 minutes

n = 25

$\mu$_{$\bar x$} =  $\mu$ = 9 minutes

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 2 / $\sqrt$ 25 = 0.4

a) P( 8.8 < $\bar x$ < 9.2)  

= P[(8.8 - 9) / 0.4 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (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 < $\bar x$ < 9)  

= P[(8.5 - 9) / 0.4 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (9 - 9) /0.4 )]

= P(-1.25 < Z < 0.00)

= P(Z < 0.00) - P(Z < -1.25)

Using z table,  

= 0.5 - 0.1056  

= 0.3944

n = 100

$\mu$_{$\bar x$} =  $\mu$ = 9 minutes

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 2 / $\sqrt$ 100 = 0.2

c) P( 8.8 < $\bar x$ < 9.2)  

= P[(8.8 - 9) / 0.2 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (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

d) The sample size in​ (c) is greater than the sample size in​ (a), so the standard error of the mean​ (or the standard deviation of the sampling​ distribution) in​ (c) is

less than in​ (a). As the standard error decreases values become more concentrated around the mean.​ Therefore, the probability of a region that includes the mean will always increase when the sample size increases.