Question

According to a certain​ survey, adults spend 2.25 hours per day watching television on a weekday. Assume that the standard deviation for​ "time spent watching television on a​ weekday" is 1.93 hours. If a random sample of 50 adults is​ obtained, describe the sampling distribution of x overbarx​, the mean amount of time spent watching television on a weekday. x overbarx is approximately normal with mu Subscript x overbarμxequals=2.25 and sigma Subscript x overbarσxequals=0.272943

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

​(c) Determine the probability that a random sample of 50 adults results in a mean time watching television on a weekday of between 2 and 3 hours. The probability is 0.8172 ​(Round to four decimal places as​ needed.)

​(d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 45 individuals who consider themselves to be avid Internet users results in a mean time of 1.86 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.86 hours or less from a population whose mean is presumed to be 2.25 hours. The likelihood is 0.0876.

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

Interpret this probability. Select the correct choice below and fill in the answer box within your choice.

​(Round to the nearest integer as​ needed.)

A.If 1000 different random samples of size nequals=45 individuals from a population whose mean is assumed to be 2.25 hours is​ obtained, we would expect a sample mean of exactly 1.86 in about ____ of the samples.

B.If 1000 different random samples of size n equals=45 individuals from a population whose mean is assumed to be 2.25 hours is​ obtained, we would expect a sample mean of 1.86 or less in about ______ of the samples.

C.If 1000 different random samples of size nequals=45 individuals from a population whose mean is assumed to be 2.25 hours is​ obtained, we would expect a sample mean of 1.86 or more in about ________ of the samples.

Answer 1

c)

µ = 2.25, σ = 1.93, n = 50
P(2 < X̅ < 3) =
= P( (2-2.25)/(1.93/√50) < (X-µ)/(σ/√n) < (3-2.25)/(1.93/√50) )
= P(-0.9159 < z < 2.7478)
= P(z < 2.7478) - P(z < -0.9159)
Using excel function:
= NORM.S.DIST(2.7478, 1) - NORM.S.DIST(-0.9159, 1)
= 0.8172

d)

µ = 2.25, σ = 1.93, n = 45
P(X̅ < 1.86) =
= P( (X̅-μ)/(σ/√n) < (1.86-2.25)/(1.93/√45) )
= P(z < -1.3555)
Using excel function:
= NORM.S.DIST(-1.3555, 1)
= 0.0876

B.If 1000 different random samples of size n = 45 individuals from a population whose mean is assumed to be 2.25 hours is​ obtained, we would expect a sample mean of 1.86 or less in about 88 of the samples.