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 8.8 hours and a random sample of 41 adults is taken. Appendix A Statistical Tables 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 30 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 71% 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

a) 0.782

b) 0.976

c) 0.00

d) 21.57 hours

Explanation:

We are given the following information in the question:

Mean, μ = 36.07 hours

Standard Deviation, σ = 8.8 hours

Sample size, n = 41

We are given that the distribution of the average number of hours of TV viewing by adults is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/((\sigma)/(√(n)))`

a) P( the sample average is more than 35 hours)

P(x > 35)

`P( x > 35) = P( z > \displaystyle(35 - 36.07)/((8.8)/(√(41)))) = P(z > -0.779)`

`= 1 - P(z \leq -0.779)`

Calculation the value from standard normal z table, we have,

`P(x > 35) = 1 - 0.218 = 0.782 = 78.2\%`

b) P(sample average is less than 38.8 hours)

`P(x < 38.8) =P( z < \displaystyle(38.8 - 36.07)/((8.8)/(√(41))))\\\\P( z < 1.98)`

Calculating the value from the standard normal table we have,

`P( x < 38.8) = 0.976 = 97.6%`

c)P(sample average is less than 30 hours)

`P(x < 30) =P( z < \displaystyle(30 - 36.07)/((8.8)/(√(41))))\\\\P( z < -4.41)`

Calculating the value from the standard normal table we have,

`P( x < 38.8) = 0`

Thus, the sample cannot have a mean or average less than 30 hours.

d) The population standard deviation is unknown.

P( the sample average is more than 48 hours)

P(x > 48)

`P( x > 48) = P( z > \displaystyle(48 - 36.07)/(\sigma)) = 0.71`

`1 - P(z \leq \displaystyle(48 - 36.07)/(\sigma) ) = 0.71`

`P(z \leq \displaystyle(48 - 36.07)/(\sigma) ) = 1-0.71= 0.29`

Calculation the value from standard normal z table, we have,

`P(z < 0.553) = 0.29`

`\displaystyle(48 - 36.07)/(\sigma) = 0.553`

Solving, we get,

`\sigma = 21.57\text{ hours}`

e) The sample average actually is less than 40 hours. This means that 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 less than 40 hours.