Question

in a poll about work, 62% of respondents said that their jobs were sometimes or always stressful. twelve workers are chosen at random. part 1 of 2 (a) what is the mean number who find their jobs stressful in a sample of 12 workers? round the answer to two decimal places. the mean number who find their jobs stressful is . part 2 of 2 (b) what is the standard deviation of the number who find their jobs stressful in a sample of 12 workers? round the answer to four decimal places. the standard deviation of the number who find their jobs stressful is .

Answer 1

The mean and standard deviation of number who find their jobs stressful are 7.44 and 1.6814 respectively.

Given,

number of workers chosen , n=12

percentage of respondents =62%

from given data the calculation are done according to binomial distribution,

then,

probability of respondents ,$p=\frac{62}{100}=0.62$

a)calculating mean,

$\mu=n*p\\\\\mu=12*0.62\\\\\mu=7.44$

b)calculating standard deviation,

$\sigma=\sqrt{n*p*(1-p)}\\\\\sigma=\sqrt{12*0.62*(1-0.62)}\\\\\sigma=\sqrt{7.44(0.38)}\\\\\sigma=\sqrt{2.8272}\\\\\sigma=1.6814$

Thus, the mean and standard deviation of number who find their jobs stressful are 7.44 and 1.6814 respectively.

To learn more about binomial distribution refer here

brainly.com/question/15303817

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