Question

According to a social media​ blog, time spent on a certain social networking website has a mean of 21 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 3 minutes. Complete parts​ (a) through​ (d) below.

a. If you select a random sample of 16 ​sessions, what is the probability that the sample mean is between 20.5 and 21.5 ​minutes? nothing ​(Round to three decimal places as​ needed.)

b. If you select a random sample of 16 ​sessions, what is the probability that the sample mean is between 20 and 21 ​minutes? nothing ​(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 20.5 and 21.5 ​minutes? nothing ​(Round to three decimal places as​ needed.)

this part fill in blanks where the arrows are

d. Explain the difference in the results of​ (a) and​ (c). 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 ▼ increases, decreases, values become more concentrated around the mean.​ Therefore, the probability that the sample mean will fall in a region that includes the population mean will always ▼ decrease increase when the sample size increases.

Answer 1

(a)

= 21

= 3

n = 16

SE = /

= 3/ = 0.75

To find P(20.5 < < 21.5):

Case 1: For from 20.5 to mid value:

Z = (20.5 - 21)/0.75 = - 0.67

Table of Area Under Standard Normal Curve gives area = 0.2486

Case 2: For from mid value to 21.5:

Z = (21.5 - 21)/0.75 = 0.67

Table of Area Under Standard Normal Curve gives area = 0.2486

So,

P(20.5 < < 21.5):= 0.2486 X 2 = 0.4972

So,

Answer is:

0.497

(b)

= 21

= 3

n = 16

SE = /

= 3/ = 0.75

To find P(20 < < 21):

Z = (20 - 21)/0.75 = - 1.33

Table of Area Under Standard Normal Curve gives area = 0.4082

So,

P(20 < 21):= 0.4082

So,

Answer is:

0.408

(c)

= 21

= 3

n = 100

SE = /

= 3/ = 0.3

To find P(20.5 < < 21.5):

Case 1: For from 20.5 to mid value:

Z = (20.5 - 21)/0.3 = - 1.67

Table of Area Under Standard Normal Curve gives area = 0.4525

Case 2: For from mid value to 21.5:

Z = (21.5 - 21)/0.3 = 1.67

Table of Area Under Standard Normal Curve gives area = 0.4525

So,

P(20.5 < < 21.5):= 0.4525 X 2 = 0.9050

So,

Answer is:

0.905

(d)

Correct option:

The sample size in (c) is greater than 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 that the sample mean will fall in a region that includes the population mean will always decrease when the sample size increases.