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 9.7 hours and a random sample of 45 adults is taken.

a. What is the probability that the sample average is more than 38 hours?

b.  Suppose the population standard deviation is unknown. If 71% 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

Solution :

Given that ,

a) = 36.07

= / n = 9.7 / 45 = 1.45

P( > 38) = 1 - P( < 38)

= 1 - P[( - ) / < (38 - 36.07) /1.45 ]

= 1 - P(z < 1.33)   

= 1 - 0.9082

= 0.0918

b) Using standard normal table,

P(Z > z) = 71%

= 1 - P(Z < z) = 0.71

= P(Z < z) = 1 - 0.71

= P(Z < z ) = 0.29

= P(Z < -0.55 ) = 0.29

z = -0.55

Using z-score formula,

x = z * +

35 = -0.55 * + 36.07

= 35 - 36.07 / -0.55

= 1.95