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Brrunno [24]
3 years ago
5

Applicants for temporary office work at Carter Temporary Help Agency who have successfully completed an administrative assistant

course are then placed in suitable positions by Nancy Dwyer and Darla Newberg. Employers who hire temporary help through the agency return a card indicating satisfaction or dissatisfaction with the work performance of those hired. From past experience it is known that 80% of the employees placed by Nancy are rated as satisfactory, and 65% of those placed by Darla are rated as satisfactory. Darla places 55% of the temporary office help at the agency, and Nancy places the remaining 45%. If a Carter office worker is rated unsatisfactory, what is the probability that he or she was placed by Darla? (Round your answer to three decimal places.)
Mathematics
1 answer:
dezoksy [38]3 years ago
5 0

Answer:

0.681

Step-by-step explanation:

Let's define the following events:

S: a Carter office worker is rated satisfactory

U : a Carter office worker is rated unsatisfactory

ND: a Carter office worker is placed by Nancy Dwyer

DN: a Carter office worker is placed by Darla Newberg

We have from the original text that

P(S | ND) = 0.8, this implies that P(U | ND) = 0.2.

P(S | DN) = 0.65, this implies that P(U | DN) = 0.35. Besides

P(DN) =  0.55 and P(ND) = 0.45, then we are looking for

P(DN | U), using the Bayes' formula we have

P(DN | U) = \frac{P(U | DN)P(DN)}{P(U | DN)P(DN) + P(U | ND)P(ND)} = \frac{(0.35)(0.55)}{(0.35)(0.55)+(0.2)(0.45)}=0.681

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According to the South Dakota Department of Health, the number of hours of TV viewing per week is higher among adult women than
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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

8 0
2 years ago
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