P Take a Test Gabrielle Bur x CChegg Study I Guided S C https:// /Student/Player
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P Take a Test Gabrielle Bur x CChegg Study I Guided S C https:// /Student/PlayerTe aspx 124658643 www.mathxl.com EE Apps UF University of Florida D UF Registrar UF University of Florida:... UF www.honors.ufl.edu D myUFL C University of Florida... D STARS College Por... Agenda creator I Ev MyMathLab Gabrielle Burch 1/15/16 4:59 PM Test: Chapter 7 Post-Test Time Remaining: 02:59:38 Submit Test This Question: 4 pts 2 of 5 (0 complete This Test: 18 pts possible E Question Help HCH Time spent using e-mail per session is normally distributed, with H 14 minutes and o 3 minutes. Complete parts (a) through (d) but but 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. Y than in (a). As the standard deviation values become Y concentrated around the 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 mean. Therefore, the probability of a region that includes the mean will always Y when the sample size increases. Enter your answer in each of the answer boxes 5:59 PM W 1/15/2016Explanation / Answer
Mean = 14
S.D = 3
a)
S.E for 25 sessions = 3 / sqrt(25)
= 3 / 5
= 0.6
P ( 13.8 < X < 14.2)
= P ( 13.8 - 14 / 0.6 < Z < 14.2 -14 / 0.6)
= P( -0.33333 < Z < 0.3333)
= 0.63056 - 0.36944
= 0.26112
b)
P ( 13.5 < X < 14)
= P ( 13.5 - 14 / 0.6 < Z < 14 -14 / 0.6)
= P( -0.833333 < Z < 0)
= 0.5 - 0.20233
= 0.39767
c)
S.E for 200 sessions = 3 / sqrt(200)
= 3 / 14.14
= 0.42426
P ( 13.8 < X < 14.2)
= P ( 13.8 - 14 / 0.42426 < Z < 14.2 -14 / 0.42426)
= P( -0.47140 < Z < 0.47140)
= 0.68133 - 0.31867
= 0.36266
d)
As Sample size increases, the standard error decreases and the standard deviation also decreases. The value of standard deviation becomes more concentrated around the mean.
Therefore, the probability of a region that it will include the mean will always increase as the sample size increases
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