About 12% of employed adults in the United States held multiple jobs. A random sample of 66 employed adults is chosen.

Question

(a) Is it appropriate to use the normal approximation to find the probability that less than 8.4% of the individuals in the sample hold multiple jobs? If so, find the probability. If not, explain why not.
It [is or isnt] appropriate since np= ? [> or <] 10

Answer 1

To determine whether it is appropriate to use the normal approximation for the distribution of the sample proportion, we need to check the conditions for the Central Limit Theorem (CLT). Specifically, we look at the sample size \( n \) and the probability \( p \) of the event of interest.

Given:

• The proportion of employed adults holding multiple jobs (\( p \)) is 12%, or \( p = 0.12 \).
• The sample size (\( n \)) is 66.

To check the conditions:

1. Calculate \( np \): \[ np = 66 \times 0.12 = 7.92 \]

2. Calculate \( n(1 - p) \): \[ n(1 - p) = 66 \times (1 - 0.12) = 66 \times 0.88 = 58.08 \]

Now we need to check if both \( np \) and \( n(1 - p) \) are greater than 10:

• \( np = 7.92 \) (which is less than 10)
• \( n(1 - p) = 58.08 \) (which is greater than 10)

Since \( np < 10 \), it is not appropriate to use the normal approximation in this case. The requirement is that both \( np \) and \( n(1 - p) \) should be greater than or equal to 10 for the normal approximation to be applicable.

Therefore, the conclusion is: It isn't appropriate since \( np = 7.92 < 10 \).

Answer 2

(b) A new sample of 333 employed adults is chosen. Find the probability that less than 8.4% of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places.

Answer 3

To solve this problem, we can use the normal approximation to the binomial distribution because the sample size \( n = 333 \) is large enough to satisfy the conditions for the central limit theorem (CLT).

Step 1: Calculate the parameters

Given:

• Proportion of employed adults holding multiple jobs, \( p = 0.12 \)
• Sample size \( n = 333 \)

First, we need to calculate the mean \( \mu \) and the standard deviation \( \sigma \) of the sampling distribution of the sample proportion \( \hat{p} \):

1. Mean \( \mu \): \[ \mu = p = 0.12 \]

2. Standard deviation \( \sigma \): \[ \sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.12 \times 0.88}{333}} \] \[ \sigma = \sqrt{\frac{0.1056}{333}} \approx \sqrt{0.0003171} \approx 0.0178 \]

Step 2: Standardization

Now we want to find the probability that less than 8.4% of the individuals in the sample hold multiple jobs. We need to calculate the z-score for \( \hat{p} = 0.084 \).

\[ z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.084 - 0.12}{0.0178} \] \[ z = \frac{-0.036}{0.0178} \approx -2.023 \]

Step 3: Find the probability

Using the z-score we calculated, we can look up the probability for \( z = -2.023 \) in the standard normal distribution table or use a calculator.

Using the standard normal distribution (either via a z-table or calculator), we find: \[ P(Z < -2.023) \approx 0.0217 \]

Step 4: Rounding

Therefore, the probability that less than 8.4% of the individuals in this sample hold multiple jobs is approximately: \[ \boxed{0.0217} \]

Answer 4

(c) Find the probability that more than 7.9% of the individuals in the sample of 333 hold multiple jobs. Round the answer to at least four decimal places.

Answer 5

To find the probability that more than 7.9% of the individuals in the sample of 333 hold multiple jobs, we will again use the normal approximation to the binomial distribution.

Step 1: Calculate the parameters again

Using the same values as before:

• Proportion of employed adults holding multiple jobs, \( p = 0.12 \)
• Sample size \( n = 333 \)

We already calculated the mean \( \mu \) and the standard deviation \( \sigma \):

1. Mean \( \mu \): \[ \mu = p = 0.12 \]

2. Standard deviation \( \sigma \): \[ \sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.12 \times 0.88}{333}} \approx 0.0178 \]

Step 2: Standardization

Now we want to find the probability that more than 7.9% of the individuals in the sample hold multiple jobs. We convert \( \hat{p} = 0.079 \) to a z-score using the formula:

\[ z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.079 - 0.12}{0.0178} \] \[ z = \frac{-0.041}{0.0178} \approx -2.303 \]

Step 3: Find the probability

We want to find \( P(\hat{p} > 0.079) \), which corresponds to \( P(Z > -2.303) \).

Using the z-table or a calculator, we can find \( P(Z < -2.303) \) first and then subtract from 1:

From the standard normal distribution: \[ P(Z < -2.303) \approx 0.0106 \]

Thus: \[ P(Z > -2.303) = 1 - P(Z < -2.303) \approx 1 - 0.0106 = 0.9894 \]

Step 4: Rounding

Therefore, the probability that more than 7.9% of the individuals in this sample hold multiple jobs is approximately: \[ \boxed{0.9894} \]

Answer 6

(d) Find the probability that the proportion of individuals in the sample of 333 who hold multiple jobs is between .14 and .17. Round the answer to at least four decimal places.

Answer 7

To find the probability that the proportion of individuals in the sample who hold multiple jobs is between 0.14 and 0.17, we will again use the normal approximation.

Step 1: Calculate the parameters

Using the same values as before:

• Proportion of employed adults holding multiple jobs, \( p = 0.12 \)
• Sample size \( n = 333 \)

We already calculated the mean \( \mu \) and the standard deviation \( \sigma \) as:

1. Mean \( \mu \): \[ \mu = p = 0.12 \]

2. Standard deviation \( \sigma \): \[ \sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.12 \times 0.88}{333}} \approx 0.0178 \]

Step 2: Standardization

Now we want to find the probability that the sample proportion \( \hat{p} \) is between 0.14 and 0.17. We will convert both values of \( \hat{p} \) to z-scores.

1. Calculate the z-score for \( \hat{p} = 0.14 \): \[ z_1 = \frac{0.14 - 0.12}{0.0178} = \frac{0.02}{0.0178} \approx 1.1236 \]

2. Calculate the z-score for \( \hat{p} = 0.17 \): \[ z_2 = \frac{0.17 - 0.12}{0.0178} = \frac{0.05}{0.0178} \approx 2.8089 \]

Step 3: Find the probabilities

Now we want to find \( P(0.14 < \hat{p} < 0.17) \), which corresponds to \( P(z_1 < Z < z_2) \).

Using a standard normal distribution table or calculator:

• For \( z_1 \approx 1.1236 \): \[ P(Z < 1.1236) \approx 0.8686 \quad \text{(using standard normal table or calculator)} \]

• For \( z_2 \approx 2.8089 \): \[ P(Z < 2.8089) \approx 0.9975 \]

Step 4: Calculate the probability

Now, we can find the probability that the proportion is between 0.14 and 0.17:

\[ P(0.14 < \hat{p} < 0.17) = P(Z < 2.8089) - P(Z < 1.1236) \] \[ P(0.14 < \hat{p} < 0.17) \approx 0.9975 - 0.8686 = 0.1289 \]

Step 5: Rounding

Therefore, the probability that the proportion of individuals in the sample who hold multiple jobs is between 0.14 and 0.17 is approximately: \[ \boxed{0.1289} \]

Answer 8

(e) Would it be unusual if less than 7% of the individuals in the sample of 333 held multiple jobs?
it [would or wouldnt] be unusual since the probability is [?]

Answer 9

To determine whether it would be unusual for less than 7% of the individuals in the sample of 333 to hold multiple jobs, we can find the probability associated with this event using the normal approximation that we calculated previously.

Step 1: Calculate the z-score for \( \hat{p} = 0.07 \)

Using the mean \( \mu = 0.12 \) and the standard deviation \( \sigma \approx 0.0178 \):

\[ z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.07 - 0.12}{0.0178} = \frac{-0.05}{0.0178} \approx -2.8045 \]

Step 2: Find the probability

Now we want to find the probability that the sample proportion is less than 7%, which corresponds to \( P(\hat{p} < 0.07) \) or \( P(Z < -2.8045) \).

Using a standard normal distribution table or calculator, we find:

\[ P(Z < -2.8045) \approx 0.0025 \]

Step 3: Interpretation

We typically consider probabilities less than 0.05 (or equivalently less than 5%) as unusual events. Given that \( P(Z < -2.8045) \approx 0.0025 \), which is less than 0.05, it would indeed be considered unusual.

Conclusion

It would be unusual since the probability is approximately \( 0.0025 \).

So you would complete the sentence as follows:

It would be unusual since the probability is approximately 0.0025.