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just olya [345]
3 years ago
12

Stress at work: In a poll conducted by the General Social Survey, 80% of respondents said that their jobs were sometimes or alwa

ys 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
Mathematics
2 answers:
FromTheMoon [43]3 years ago
6 0

Answer:

(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.

Step-by-step 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{\frac{\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 \frac{140}{190} )

     P( \hat p \leq \frac{140}{190} ) = P( \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } \leq \frac{\frac{140}{190} -0.80}{\sqrt{\frac{\frac{140}{190}(1-\frac{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 > \frac{155}{190} )

     P( \hat p > \frac{155}{190} ) = P( \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } > \frac{\frac{155}{190} -0.80}{\sqrt{\frac{\frac{155}{190}(1-\frac{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( \frac{145}{190}  \leq \hat p \leq \frac{158}{190} )

    P( \frac{145}{190}  \leq \hat p \leq \frac{158}{190} )  = P( \hat p \leq \frac{158}{190} ) - P( \hat p < \frac{145}{190} )

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

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

                                                                  = 1 - 0.88298 = 0.11702                                                  

Therefore,  P( \frac{145}{190}  \leq \hat p \leq \frac{158}{190} ) = 0.87698 - 0.11702 = 0.75996

Svet_ta [14]3 years ago
5 0

Answer:

a) P=0.019

b) P=0.263

c) P=0.794

Step-by-step 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=\sqrt{np(1-p)}=\sqrt{190*0.8*0.2}=\sqrt{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

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

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