Questions: Working two jobs: About 5% of employed adults in the United States held multiple jobs. A random sample of 68 employed adults is chosen. Use the TI- 84 Plus calculator as needed. Part 1 of 5 (a) Is it appropriate to use the normal approximation to find the probability that less than 6.5% of the individuals in the sample hold multiple jobs? If so, find the probability. If not, explain why not. It is not appropriate to use the normal curve, since np= 3.4 10. Part 2 of 5 (b) A new sample of 336 employed adults is chosen. Find the probability that less than 6.5% of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places as needed. The probability that less than 6.5% of the individuals in this sample hold multiple jobs is 0.8964 Part 3 of 5 (c) Find the probability that more than 6.5% of the individuals in the sample of 336 hold multiple jobs. Round the answer to at least four decimal places as needed. The probability that more than 6.5% of individuals in the sample of 336 hold multiple jobs is .

Solution Steps

(a) To determine if the normal approximation is appropriate, we need to check if both \( n \times p \) and \( n \times (1-p) \) are greater than or equal to 10. If they are, we can use the normal approximation. Here, \( n = 68 \) and \( p = 0.05 \).

(b) For the new sample of 336, we can use the normal approximation to find the probability that less than 6.5% of the individuals hold multiple jobs. We calculate the mean and standard deviation of the binomial distribution and then use the normal distribution to find the probability.

(c) To find the probability that more than 6.5% of the individuals in the sample of 336 hold multiple jobs, we use the complement rule with the probability found in part (b).

To determine if the normal approximation is appropriate for the sample of \( n = 68 \) employed adults with \( p = 0.05 \), we calculate: \[ n p = 68 \times 0.05 = 3.4 \] \[ n (1 - p) = 68 \times 0.95 = 64.6 \] Since \( n p < 10 \), it is not appropriate to use the normal approximation. Thus, we conclude: \[ \text{It is not appropriate to use the normal curve.} \]

For the new sample of \( n = 336 \), we want to find the probability that less than \( 6.5\% \) of the individuals hold multiple jobs. We calculate the mean and standard deviation: \[ \text{Mean} = n p = 336 \times 0.05 = 16.8 \] \[ \text{Standard Deviation} = \sqrt{n p (1 - p)} = \sqrt{336 \times 0.05 \times 0.95} \approx 3.995 \] Next, we find the z-score for \( p = 0.065 \): \[ z = \frac{(0.065 \times 336) - 16.8}{3.995} \approx 1.262 \] Using the z-score, we find the probability: \[ P(X < 0.065) \approx 0.8964 \]

To find the probability that more than \( 6.5\% \) of the individuals in the sample of \( 336 \) hold multiple jobs, we use the complement: \[ P(X > 0.065) = 1 - P(X < 0.065) \approx 1 - 0.8964 = 0.1036 \]

Final Answer