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shepuryov [24]
3 years ago
15

A gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than

work for someone else. A random sample of 600 18-29 year-olds is obtained today. What is the probability that no more than 70% would prefer to start their own business?
Mathematics
1 answer:
Neko [114]3 years ago
3 0

Answer:

The probability that no more than 70% would prefer to start their own business is 0.1423.

Step-by-step explanation:

We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.

Let \hat p = <u><em>sample proportion of people who prefer to start their own business</em></u>

The z-score probability distribution for the sample proportion is given by;

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

where, p = population proportion who would prefer to start their own business = 72%

            n = sample of 18-29 year-olds = 600

Now, the probability that no more than 70% would prefer to start their own business is given by = P( \hat p \leq 70%)

       P( \hat p \leq 70%) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } \leq \frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } } ) = P(Z \leq -1.07) = 1 - P(Z < 1.07)

                                                                       = 1 - 0.8577 = <u>0.1423</u>

The above probability is calculated by looking at the value of x = 1.07 in the z table which has an area of 0.8577.

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