* Box one: Choose between “ less or greater” * Box Two: Chose between “increases
ID: 2946799 • Letter: #
Question
* Box one: Choose between “ less or greater” * Box Two: Chose between “increases or decreases” * Box three: Choose between “decrease or increase”According to a social media blog, time spent on a certain social networking website has a mean of 16 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 4 minutes. Complete parts (a) through (d) below a. If you t a random sample of 36 sessions, what is the probability that the sample mean is between 15 5 and 16 5 minutes? (Round to three decimal places as needed) b. If you select a random sample of 36 sessions, what is the probability that the sample mean is between 15 and 16 minutes? (Round to three decimal places as needed) c. If you select a random sample of 144 sessions, what is the probabiity that the sample mean is between 15 5 and 16.5 minutes? (Round to three decimal places as needed) d. Explain the difference in the resuilts 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 distibution) in (c)is than in (a) As the standard error that includes the population mean will always values become more concentrated around the mean. Therefore, the probability that the sample mean will fall in a region ywhen the sample size increases
Explanation / Answer
mean=16
sd=4
n=36
P(15.5<X<16.5)
P(15.5-16/4/sqrt(36)<Z<16.5-16/4/sqrt(36)
P(-0.75<Z<0.75)
=P(Z<0.75)-P(Z<-0.75)
=0.7734-(1-0.7734)
=0.5468
answer:0.5468
Solutionb:
P(15.5<X<16.5)
P(15-16/4/sqrt(36)<Z<16-16/4/sqrt(36)
P(-1.5<Z<0)
=0.4332
ANSWER:0.4332
Solutionc:
n=144
P(15.5<X<16.5)
P(15.5-16/4/sqrt(144)<Z<16.5-16/4/sqrt(144)
P(-1.5<Z<1.5)
=0.8664
ANSWER:0.8664
Solutiond:
less
decreases
increases
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