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 52 adults is taken.

a. What is the probability that the sample average is more than 35 hours?

b. What is the probability that the sample average is less than 38.8 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 75% of all sample means are greater than 48 hours and the population mean is still 36.07 hours, what is the value of the population standard deviation?

Answer 1

Population mean, µ = 36.07

Population standard deviation, σ = 9.7

Sample size, n = 52

a) probability that the sample average is more than 35 hours, P(X̅ > 35) =

= P( (X̅-μ)/(σ/√n) > (35-36.07)/(9.7/√52) )

= P(z > -0.7955)

= 1 - P(z < -0.7955)

Using excel function:

= 1 - NORM.S.DIST(-0.7955, 1)

= 0.7868

b) probability that the sample average is less than 38.8 hours, P(X̅ < 38.8) =

= P( (X̅-μ)/(σ/√n) < (38.8-36.07)/(9.7/√52) )

= P(z < 2.0295)

Using excel function:

= NORM.S.DIST(2.0295, 1)

= 0.9788

c) probability that the sample average is less than 29 hours, P(X̅ < 29) =

= P( (X̅-μ)/(σ/√n) < (29-36.07)/(9.7/√52) )

= P(z < -5.2559)

Using excel function:

= NORM.S.DIST(-5.2559, 1)

= 0

probability that the sample average is less than 40 hours, P(X̅ < 40) =

= P( (X̅-μ)/(σ/√n) < (40-36.07)/(9.7/√52) )

= P(z < 2.9216)

Using excel function:

= NORM.S.DIST(2.9216, 1)

= 0.9983

d) Z score at p = 0.75 using excel = NORM.S.INV(0.75) = 0.6745

X = µ + z*σ

48 = 36.07 + (0.6745)*σ

σ = (48-36.07)/0.6745 = 17.69