Question

Stress at work: In a poll conducted by the General Social Survey, 80% of respondents said that their jobs were sometimes or always stressful. One hundred and ninety workers are chosen at random. Use the TI-84 Plus calculator as needed. Round your answer to at least four decimal places. (a) Approximate the probability that 140 or fewer workers find their jobs stressful. (b) Approximate the probability that more than 155 workers find their jobs stressful. (c) Approximate the probability that the number of workers who find their jobs stressful is between 145 and 158 inclusive. Part 1 of 3

Answer 1

(a) Probability that 140 or fewer workers find their jobs stressful is 0.02385

(b) Probability that more than 155 workers find their jobs stressful is 0.28774

(c) Probability that the number of workers who find their jobs stressful is between 145 and 158 inclusive is 0.75996.

Explanation:

We are given that in a poll conducted by the General Social Survey, 80% of respondents said that their jobs were sometimes or always stressful.

Let p = % of respondents said that their jobs were sometimes or always stressful = 80%

The z score probability distribution for proportion is given by;

Z =
`\frac{\hat p -p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = % of respondents said that their jobs were stressful in a sample of one hundred and ninety workers

(a) The probability that 140 or fewer workers find their jobs stressful is given by = P(
`\hat p`
`\leq`
`(140)/(190)` )

P(
`\hat p`
`\leq`
`(140)/(190)` ) = P(
`\frac{\hat p -p}{\sqrt{(\hat p(1-\hat p))/(n) } }`
`\leq`
`\frac{(140)/(190) -0.80}{\sqrt{((140)/(190)(1-(140)/(190)))/(190) } }` ) = P(Z
`\leq` -1.98) = 1 - P(Z < 1.98)

= 1 - 0.97615 = 0.02385

(b) The probability that more than 155 workers find their jobs stressful is given by = P(
`\hat p` >
`(155)/(190)` )

P(
`\hat p` >
`(155)/(190)` ) = P(
`\frac{\hat p -p}{\sqrt{(\hat p(1-\hat p))/(n) } }` >
`\frac{(155)/(190) -0.80}{\sqrt{((155)/(190)(1-(155)/(190)))/(190) } }` ) = P(Z > 0.56) = 1 - P(Z
`\leq` 0.56)

= 1 - 0.71226 = 0.28774

(c) The probability that the number of workers who find their jobs stressful is between 145 and 158 inclusive is given by = P(
`(145)/(190)`
`\leq`
`\hat p`
`\leq`
`(158)/(190)` )

P(
`(145)/(190)`
`\leq`
`\hat p`
`\leq`
`(158)/(190)` ) = P(
`\hat p`
`\leq`
`(158)/(190)` ) - P(
`\hat p` <
`(145)/(190)` )

P(
`\hat p`
`\leq`
`(158)/(190)` ) = P(
`\frac{\hat p -p}{\sqrt{(\hat p(1-\hat p))/(n) } }`
`\leq`
`\frac{(158)/(190) -0.80}{\sqrt{((158)/(190)(1-(158)/(190)))/(190) } }` ) = P(Z
`\leq` 1.16) = 0.87698

P(
`\hat p` <
`(145)/(190)` ) = P(
`\frac{\hat p -p}{\sqrt{(\hat p(1-\hat p))/(n) } }` <
`\frac{(145)/(190) -0.80}{\sqrt{((145)/(190)(1-(145)/(190)))/(190) } }` ) = P(Z < -1.19) = 1 - P(Z
`\leq` 1.19)

= 1 - 0.88298 = 0.11702

Therefore, P(
`(145)/(190)`
`\leq`
`\hat p`
`\leq`
`(158)/(190)` ) = 0.87698 - 0.11702 = 0.75996

Answer 2

a) P=0.019

b) P=0.263

c) P=0.794

Explanation:

We assume that the poll gives the population's proportion of respondents said that their jobs were sometimes or always stressful (p=0.8).

Then, a sample of size n=190 is taken.

The sample mean is:

`\mu=np=190*0.8=152`

The sample standard deviation is:

`\sigma=√(np(1-p))=√(190*0.8*0.2)=√(30.4)=5.514`

The probability that 140 or fewer workers find their jobs stressful is:

`z=(X-\mu)/\sigma=(140.5-152)/5.514=-11.5/5.514=-2.08 \\\\P(X\leq140)=P(X<140.5)=P(z<-2.08)=0.019`

Note: a correction for continuity is applied.

The probability that more than 155 workers find their jobs stressful is

`z=(X-\mu)/\sigma=(155.5-152)/5.514=3.5/5.514= 0.635 \\\\P(X>155)=P(X>155.5)=P(z>0.635)=0.263`

The probability that the number of workers who find their jobs stressful is between 145 and 158 inclusive is:

`z_1=(X_1-\mu)/\sigma=(158.5-152)/5.514=6.5/5.514= 1.179\\\\ z_2=(X_2-\mu)/\sigma=(144.5-152)/5.514=-7.5/5.514= -1.360 \\\\ P(145\leq X\leq 158)=P(144.5< X< 158.5)\\\\P(145\leq X\leq 158)=P(X<158.5)-P(X<144.5)\\\\P(145\leq X\leq 158)=P(z<1.179)-P(z<-1.360)\\\\P(145\leq X\leq 158)=0.881-0.087=0.794`