Question

According to the South Dakota Department of Health, the number of hours of TV viewing per week is higher among adult women than adult men. A recent study showed women spent an average of 33 hours per week watching TV, and men, 24 hours per week. Assume that the distribution of hours watched follows the normal distribution for both groups, and that the standard deviation among the women is 5.0 hours and is 5.9 hours for the men.

What percent of the women watch TV less than 36 hours per week? (Round your z-score computation and final answer to 2 decimal places.)

What percent of the men watch TV more than 20 hours per week? (Round your z-score computation and final answer to 2 decimal places.)

How many hours of TV do the four percent of women who watch the most TV per week watch? Find the comparable value for the men. (Round your answers to 3 decimal places.)

Answer 1

Solution:

1) Let X be a random variable which represents the time spent on watching television per week by women.

Given that, X ~ N(33, 5.0²)

μ = 33 hours and σ = 5.0 hours

We have to find P(X < 36).

We know that, if X ~ (μ, σ^{​​​​​​2}) then

Using "pnorm" function of R we get, P(Z < 0.6) = 0.7257

P(X < 36) = 0.7257 = 72.57%

Hence, 72.57% of women watch TV less than 36 hours per week.

2) Let Y be a random variable which represents the time spent on watching television per week by men.

Given that, Y ~ N(24, 5.9²)

μ = 24 hours and σ = 5.9 hours

We have to find P(Y > 20 hours).

We know that, if Y ~ (μ, σ^{​​​​​​2}) then

Using "pnorm" function of R we get, P(Z > -0.678) = 0.7511

P(Y > 20) = 0.7511 = 75.11%

Hence, 75.11% of men watch TV more than 20 hours per week.

3) The 96th percentile value of X will be the required hours of TV watched by the 4% of women who watch the most TV per week.

Let the 96th percentile be c.

Hence, P(X < c) = 0.96

We have, X~ N(33, 5.0²)

μ = 33 hours and σ = 5.0 hours

We know that, if X ~ (μ, σ^{​​​​​​2}) then

...........................(1)

Using "qnorm" function of R we get, P(Z < 1.7507) = 0.96

Comparing, P(Z < 1.7507) = 0.96 and (1) we get,

The 96th percentile of X is 41.753.

Hence, four percent of women who watch the most TV per week, watch 41.753 hours of TV.

4) To obtain the comparable value we need to obtain How many hours of TV do the four percent of men who watch the most TV per week watch.

The 96th percentile value of Y will be the required hours of TV watched by the 4% of men who watch the most TV per week.

Let the 96th percentile be k.

Hence, P(Y < k) = 0.96

We have, Y~ N(24, 5.9²)

μ = 24 hours and σ = 5.9 hours

We know that, if Y ~ (μ, σ^{​​​​​​2}) then

...........................(2)

Using "qnorm" function of R we get, P(Z < 1.7507) = 0.96

Comparing, P(Z < 1.7507) = 0.96 and (2) we get,

The 96th percentile of Y is 34.329.

Hence, four percent of men who watch the most TV per week watch 34.329 hours of TV. This value is the comparable value.

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