Question

According to Nielsen Media Research, the average number of hours of TV viewing by adults (18 and over) per week in the United States is 36.07 hours. Suppose the standard deviation is 8.7 hours and a random sample of 51 adults is taken. Appendix A Statistical Tables a. What is the probability that the sample average is more than 36 hours? b. What is the probability that the sample average is less than 36.6 hours? c. What is the probability that the sample average is less than 29 hours? If the sample average actually is less than 40 hours, what would it mean in terms of the Nielsen Media Research figures? d. Suppose the population standard deviation is unknown. If 66% of all sample means are greater than 35 hours and the population mean is still 36.07 hours, what is the value of the population standard deviation?

a. What is the probability that the sample average is more than 36 hours?
b. What is the probability that the sample average is less than 36.6 hours?
c. What is the probability that the sample average is less than 29 hours? If the sample average actually is less than 40 hours, what would it mean in terms of the Nielsen Media Research figures?
d. Suppose the population standard deviation is unknown. If 66% of all sample means are greater than 35 hours and the population mean is still 36.07 hours, what is the value of the population standard deviation?

Answer 1

given that

n=51

let X number of hours of TV viewing by adults

so sample average have

a)

we have to find

now

=1-P(Z<-0.06)

=1-0.476=0.524

b)

=0.668

c)

<0.00001

if the sample average is less than 40 then there is chance that sample average will match with population average.

d)

let SD is required standard deviation

given that

now

from Z table

P(Z>-0.4125)=0.66 so on comparing