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 34 hours per week watching TV, and men, 29 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 4.5 hours and is 5.1 hours for the men. a. What percent of the women watch TV less than 40 hours per week? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)b. What percent of the men watch TV more than 25 hours per week? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)c. How many hours of TV do the one 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.)

a) P(x<40) = 0.90824

Therefore, the percent of the women watch TV less than 40 hours per week is 0.90824 × 100 = 90.8240%

b)P(x>25) = 1 - P(z = -0.78) = 0.7823

Therefore, percent of the men watch TV more than 25 hours per week?is 0.7823 × 100 = 78.230%

c)The number of hours that the one percent of WOMEN who watch the most TV per week watch is for 44.485hours

While, for the MEN, the number of hours that the one percent of men who watch the most TV per week watch is for 40.883 hours

Step-by-step explanation:

To solve this question, we would be using z score formula:

z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

a. What percent of the women watch TV less than 40 hours per week? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

z = (x-μ)/σ,

where x is the raw score = 40 hours

μ is the population mean = 34 hours

σ is the population standard deviation = 4.5

z = (40 - 34)/4.5

z = 1.33333

Approximately to 2 decimal places = z score = 1.33

Using the normal distribution z score table

Probabilty value from Z-Table:

P(z = 1.33) = P(x<40) = 0.90824

Therefore, the percent of the women watch TV less than 40 hours per week is 0.90824 × 100 = 90.8240%

b. What percent of the men watch TV more than 25 hours per week? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

z = (x-μ)/σ,

where x is the raw score = 25 hours

μ is the population mean = 29 hours

σ is the population standard deviation = 5.1

z = (25 - 29)/5.1

z = -0.78431

Approximately to 2 decimal places

z score = -0.78

Using the z score normal distribution table:

Probability value from Z-Table:

P(z = -0.78) = P(x<Z) = 0.2177

P(x>25) = 1 - P(z = -0.78) = 0.7823

Therefore, percent of the men watch TV more than 25 hours per week?is 0.7823 × 100 = 78.230%

c. How many hours of TV do the one 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.)

First, we find what the z score is.

We were asked in the question to find how many hours 1% of the women watch TV the most.

We have to find the confidence interval

100 - 1% = 99%

The z score for the confidence interval of 99% or 0.99(in decimal form) = 2.33

z score = 2.33

Since we know the z score now, we proceed to find x = raw score.

z = (x-μ)/σ,

where x is the raw score = unknown

μ is the population mean = 34 hours

σ is the population standard deviation = 4.5

2.33= (x - 34)/4.5

Cross Multiply

2.33 × 4.5 = x - 34

10.485 = x - 34

x = 10.485 + 34

x = 44.485 hours.

Therefore, the number of hours that the one percent of women who watch the most TV per week watch is for 44.485hours

In the question, we were also asked to find the comparable value for men.

Hence, for one percent of the men.

We determine what the z score is.

We were asked in the question to find how many hours 1% of the men watch TV the most.

We have to find the confidence interval

100 - 1% = 99%

The z score for the confidence interval of 99% or 0.99(in decimal form) = 2.33

We already have our z score as 2.33

z = (x-μ)/σ,

where x is the raw score = unknown

μ is the population mean = 29 hours

σ is the population standard deviation = 5.1

2.33= (x - 29)/5.1

Cross Multiply

2.33 × 5.1 = x - 29

11.883 = x - 29

x = 11.883 + 29

x = 40.883 hours.

Therefore, the number of hours that the one percent of men who watch the most TV per week watch is for 40.883 hours

i believe the answer is d.

hope it you! : d

cant open file sorry!

step-by-step explanation: