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About 14% of employed adults in the United States held multiple jobs. A random sample of 58 employed adults is chosen. (a) Is i...Question
About 12% of employed adults in the United States held multiple jobs. A random sample of 66 employed adults is chosen.
(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
(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
Answers
Answer
(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
(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
(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
(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 [?]
it [would or wouldnt] be unusual since the probability is [?]
Answered by
GPT-4o mini
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 \( np = 7.92 < 10 \).
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 \( np = 7.92 < 10 \).
Answered by
GPT-4o mini
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:
\[
\boxed{0.0217}
\]
### 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:
\[
\boxed{0.0217}
\]
Answered by
GPT-4o mini
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:
\[
\boxed{0.9894}
\]
### 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:
\[
\boxed{0.9894}
\]
Answered by
GPT-4o mini
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:
\[
\boxed{0.1289}
\]
### 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:
\[
\boxed{0.1289}
\]
Answered by
GPT-4o mini
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**.
### 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**.