Time spent using e-mail per session is normally distributed, with -14 minutes an

Question

Time spent using e-mail per session is normally distributed, with -14 minutes and ?: 3 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 13 8 and 14.2 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 13 5 and 14 minutes? (Round to three decimal places as needed) c. If you select a random sample of 200 sessions, what is the probability that the sample mean is between 13.8 and 14.2 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 values become concentrated when the sample size sampling distribution) in (c) is Vthan in (a). As the standard error around the mean. Therefore, the probability of a region that includes the mean will always increases

Explanation / Answer

Solution:- Given that ? = 14 minutes and ? = 3 minutes.

a. For n = 25

=> P(13.8 < x < 14.2)
= P((13.8-14)/(3/sqrt(25)) < (x-?)/(?/sqrt(n) < (14.2-14)/(3/sqrt(25))

= P(-0.333 < Z < 0.333)

= 0.259

b. For n = 25
=> P(13.5 < x < 14)
= P((13.5-14)/(3/sqrt(25)) < (x-?)/(?/sqrt(n) < (14-14)/(3/sqrt(25))
= P(-0.833 < Z < 0)
= 0.297

c. For n = 200

=> P(13.8 < x < 14.2)
= P((13.8-14)/(3/sqrt(200)) < (x-?)/(?/sqrt(n) < (14.2-14)/(3/sqrt(200))
= -0.943 < Z < 0.943
= 0.6528

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

Increasing the sample size increases probability that sample mean is within a specific distance from the population mean

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