About $13 \%$ of employed aduits in the United States held multiple jobs. A random sample of 74 employed adults is chosen. Use the TI- 84 Plus calculator as needed.
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Part 1 of 5
(a) Is it appropriate to use the normal approximation to find the probability that less than $7.1 \%$ of the individuals in the sample hold multiple jobs? If so, find the probability. If not, explain why not.
It is not $\quad \mathbf{\nabla}$ appropriate to use the normal curve, since $n p=9.62$
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Part: $1 / 5$
Part 2 of 5
(b) A new sample of 225 employed adults is chosen. Find the probability that less than $7.1 \%$ of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places.
The probability that less than $7.1 \%$ of the individuals in this sample hold multiple jobs is $\square$.
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Answer
Final Answer: The probability that less than 7.1% of the individuals in this sample hold multiple jobs is \(\boxed{0.0042}\)
Step 1 ：The problem is asking for the probability that less than 7.1% of a sample of 225 employed adults hold multiple jobs, given that 13% of employed adults in the United States hold multiple jobs. This is a binomial problem, but since the sample size is large, we can use the normal approximation to the binomial distribution.
Step 2 ：The mean of the distribution is np, where n is the sample size and p is the probability of success (in this case, holding multiple jobs). So, \(mean = np = 225 * 0.13 = 29.25\)
Step 3 ：The standard deviation is \(\sqrt{np(1-p)}\). So, \(std\_dev = \sqrt{225 * 0.13 * (1 - 0.13)} = 5.044551516240071\)
Step 4 ：We want to find the probability that the proportion is less than 7.1%, or 0.071. To do this, we can standardize the value we're interested in (0.071) by subtracting the mean and dividing by the standard deviation to get a z-score. So, \(z = \frac{0.071 * 225 - mean}{std\_dev} = -2.6315520730164828\)
Step 5 ：Then we can use a z-table or a function like scipy's norm.cdf to find the probability that a standard normal random variable is less than this z-score. So, \(probability = 0.004249792038589428\)
Step 6 ：Final Answer: The probability that less than 7.1% of the individuals in this sample hold multiple jobs is \(\boxed{0.0042}\)
