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About 6% of employed adults in the United States held multiple jobs. A random sample of 69 employed adults is chosen. Use the
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Answer #1

Solution:-

a) It is not appropriate ti use the normal curve, since np = 4.14 is less than 10.

b) The probability that less than 6.6% of the individuals in the sample hold multiple jobs is 0.6791.

p = 0.066

By applying normal distribution:-

z= \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

z= \frac{0.066-0.06}{\sqrt{\frac{0.06(1-0.06)}{339}}}

z = 0.4652

P(z < 0.4652) = 0.6791

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