According to a certain survey, adults spend 2 hours per day watching television

Question

According to a certain survey, adults spend 2 hours per day watching television on a weekday. Assume that the standard deviation for "time spent watching television on a weekday" is 1.5 hours. If a random sample of 49 adults is obtained, describe the sampling distribution of x , the mean amount of time spent watching television on a weekday. Determine the probability that a random sample of 49 adults results in a mean time watching television on a weekday of between 2 and 3 hours. One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 36 individuals who consider themselves to be enthusiastic Internet users results in a mean time of 2.5 hours watching television on a weekday. The likelihood is?

Explanation / Answer

(A) sample mean is approximately normal with population mean of _2_____ and standard deviation of _s/vn =_1.5/sqrt(49)_ =0.21428 ~ 0.2143

(B)the probability is

P(2<xbar<3) = P((2-2)/0.2143 <(X-mean)/(s/vn) <(3-2)/0.2143)

=P(0 <Z< 4.66) = 0.5 (from standard normal table)

(C) The likelihood is

P(xbar<2.5) = P((xbar-mean)/(s/vn ) <(2.5- 2)/ 2/sqrt(36))

=P(Z<1.50) =0.9332

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