Respuesta :
Answer:
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
