The amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts (a) through (d). (a) According to a certain survey, adults spend 2.35 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 60 adults is obtained, describe the sampling distribution of x, the mean amount of time spent watching television on a weekday. x ▼ is approximately normal has a uniform distribution with μx=nothing and σx=nothing. (Round to six decimal places as needed.) (b) Determine the probability that a random sample of 60 adults results in a mean time watching television on a weekday of between 2 and 3 hours. The probability is nothing. (Round to four decimal places as needed.) (c) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 2.00 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.00 hours or less from a population whose mean is presumed to be 2.35 hours. The likelihood is nothing. (Round to four decimal places as needed.)
The amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts (a) through (d). (a) According to a certain survey, adults spend 2.35 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 60 adults is obtained, describe the sampling distribution of x, the mean amount of time spent watching television on a weekday. x ▼ is approximately normal has a uniform distribution with μx=nothing and σx=nothing. (Round to six decimal places as needed.) (b) Determine the probability that a random sample of 60 adults results in a mean time watching television on a weekday of between 2 and 3 hours. The probability is nothing. (Round to four decimal places as needed.) (c) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 2.00 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.00 hours or less from a population whose mean is presumed to be 2.35 hours. The likelihood is nothing. (Round to four decimal places as needed.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts (a) through (d).
(a) According to a certain survey, adults spend
mean amount of time spent watching television on a weekday.
2.35
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
60
adults is obtained, describe the sampling distribution of
x,
the x
▼
is approximately normal
has a uniform distribution
μx=nothing
and
σx=nothing.
(Round to six decimal places as needed.)
(b) Determine the probability that a random sample of
60
adults results in a mean time watching television on a weekday of between 2 and 3 hours.The probability is
nothing.
(Round to four decimal places as needed.)(c) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of
55
individuals who consider themselves to be avid Internet users results in a mean time of
2.00
hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of
2.00
hours or less from a population whose mean is presumed to be
2.35
hours.The likelihood is
nothing.
(Round to four decimal places as needed.)Expert Solution
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