About 14% of employed adults in the United States held multiple jobs. A random sample of 58 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.

GPT-4o mini · Answer

To determine if it is appropriate to use the normal approximation for the binomial distribution in this problem, we can use the condition related to the sample size and the probabilities involved. Specifically, we want to check if both $np$ and $n(1-p)$ are greater than or equal to 5, where $n$ is the sample size and $p$ is the probability of success (the proportion of individuals with multiple jobs).

Calculate $np$ and $n(1-p)$:
- Here, $n = 58$ (the sample size) and $p = 0.14$ (the proportion holding multiple jobs).
- First, calculate $np$: \[ np = 58 \times 0.14 = 8.12 \]
- Next, calculate $n(1-p)$: \[ n(1-p) = 58 \times (1 - 0.14) = 58 \times 0.86 = 49.88 \]

Both values are greater than 5:

$np = 8.12 \geq 5$
$n(1-p) = 49.88 \geq 5$

Since both conditions are satisfied, it is appropriate to use the normal approximation.

Finding the probability that less than 8.4% of the sample holds multiple jobs:
- We want to find the probability that the proportion with multiple jobs is less than 0.084, or that the count is less than $0.084 \times 58$. \[ 0.084 \times 58 \approx 4.872 \quad (\text{we round to } 4 \text{ for integer count}) \]
- Thus, we want $P(X < 5)$ where $X$ is the number of individuals holding multiple jobs in our sample of 58.
Normal approximation calculations:
- The mean $\mu$ of the binomial distribution is: \[ \mu = np = 8.12 \]
- The standard deviation $\sigma$ is: \[ \sigma = \sqrt{np(1-p)} = \sqrt{58 \times 0.14 \times 0.86} \approx \sqrt{7.0304} \approx 2.65 \]
Converting to the standard normal distribution:
- We apply the continuity correction because we are approximating a discrete distribution with a continuous one. So, we find $P(X < 5)$ by calculating $P(Z < \frac{4.5 - \mu}{\sigma})$: \[ Z = \frac{4.5 - 8.12}{2.65} \approx \frac{-3.62}{2.65} \approx -1.37 \]
Using the standard normal distribution table:
- We look up the Z-value of -1.37 in the standard normal distribution table: \[ P(Z < -1.37) \approx 0.0853 \]

Thus, the probability that less than 8.4% of the individuals in the sample hold multiple jobs is approximately 0.0853 or 8.53%.

Answer

It [is or is not] appropriate to use the normal curve, since np = [> or <?] 10 .

GPT-4o mini · Answer

It **is** appropriate to use the normal curve since $np = 8.12$ is **not** greater than 10 (it is greater than 5). The conditions for using the normal approximation typically require that both $np \geq 5$ and $n(1-p) \geq 5$. As shown in the previous response: - $np = 8.12 \geq 5$ (satisfied) - $n(1-p) = 49.88 \geq 5$ (satisfied) Both conditions are met, so it is indeed appropriate to use the normal approximation in this case. For a complete statement: It **is** appropriate to use the normal curve since $np = 8.12$ is **not** greater than 10 (but is greater than 5), and $n(1-p) = 49.88$ is also greater than 5.

Answer

(b) A new sample of 352 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.

GPT-4o mini · Answer

To find the probability that less than 8.4% of a new sample of 352 employed adults hold multiple jobs, we can follow a similar process as before, using the normal approximation to the binomial distribution.

Define the parameters:
- Sample size ($n$): $352$
- Proportion of employed adults holding multiple jobs ($p$): $0.14$
Calculate $np$ and $n(1-p)$:
- Calculate $np$: \[ np = 352 \times 0.14 = 49.28 \]
- Calculate $n(1-p)$: \[ n(1-p) = 352 \times (1 - 0.14) = 352 \times 0.86 = 303.52 \] Since $np = 49.28 \geq 5$ and $n(1-p) = 303.52 \geq 5$, it is appropriate to use the normal approximation.
Mean and standard deviation:
- The mean $\mu$: \[ \mu = np = 49.28 \]
- The standard deviation $\sigma$: \[ \sigma = \sqrt{np(1-p)} = \sqrt{352 \times 0.14 \times 0.86} \approx \sqrt{49.28 \times 0.86} \approx \sqrt{42.7888} \approx 6.54 \]
Finding the cut-off value:
- The threshold for less than 8.4%: \[ 0.084 \times 352 \approx 29.568 \quad (\text{we round to } 29 \text{ for integer count}) \] Thus, we want $P(X < 30)$.
Normal approximation:
- Convert to the standard normal distribution: \[ Z = \frac{X - \mu}{\sigma} = \frac{29.5 - 49.28}{6.54} \approx \frac{-19.78}{6.54} \approx -3.02 \]
Using the standard normal distribution table:
- Look up the Z-value of $-3.02$ in the standard normal distribution table. This Z-value corresponds to a cumulative probability.
- Typically, $P(Z < -3.02) \approx 0.0013$.

Thus, the probability that less than 8.4% of the individuals in this new sample hold multiple jobs is approximately 0.0013.

So, rounding to at least four decimal places: \[ \text{Probability} \approx 0.0013. \]

Answer

(c) Find the probability that more than 7.9% of the individuals in the sample of 352 hold multiple jobs. Round the answer to at least four decimal places.

GPT-4o mini · Answer

To find the probability that more than 7.9% of the individuals in a sample of 352 employed adults hold multiple jobs, we will again use the normal approximation to the binomial distribution.

Define the parameters:
- Sample size ($n$): $352$
- Proportion of employed adults with multiple jobs ($p$): $0.14$
Calculate $np$ and $n(1-p)$:
- $np = 352 \times 0.14 = 49.28$
- $n(1-p) = 352 \times (1 - 0.14) = 352 \times 0.86 = 303.52$

Both conditions $np \geq 5$ and $n(1-p) \geq 5$ hold true, so we can proceed with the normal approximation.

Mean and standard deviation:
- Mean ($\mu$): \[ \mu = np = 49.28 \]
- Standard deviation ($\sigma$): \[ \sigma = \sqrt{np(1-p)} = \sqrt{352 \times 0.14 \times 0.86} \approx \sqrt{49.28 \times 0.86} \approx \sqrt{42.7888} \approx 6.54 \]
Finding the threshold for 7.9%:
- The threshold for more than 7.9%: \[ 0.079 \times 352 \approx 27.848 \quad (\text{we round to } 27 \text{ for integer count}) \] Thus, we want $P(X > 27)$.
Using the normal approximation:
- Convert to the standard normal distribution: \[ Z = \frac{X - \mu}{\sigma} = \frac{27.5 - 49.28}{6.54} \approx \frac{-21.78}{6.54} \approx -3.32 \] (Using 27.5 for continuity correction)
Using the standard normal distribution table:
- To find $P(X > 27)$, we need to find $P(Z > -3.32)$: \[ P(Z > -3.32) = 1 - P(Z < -3.32) \]
- From the standard normal distribution table or using a calculator: \[ P(Z < -3.32) \approx 0.0004 \]
- Thus, \[ P(Z > -3.32) \approx 1 - 0.0004 = 0.9996 \]

Thus, the probability that more than 7.9% of the individuals in the sample of size 352 hold multiple jobs is approximately 0.9996 when rounded to at least four decimal places.

Answer

(d) Find the probability that the proportion of individuals in the sample of 352 who hold multiple jobs is between 0.13 and 0.19. Round the answer to at least four decimal places.

GPT-4o mini · Answer

To find the probability that the proportion of individuals in a sample of 352 who hold multiple jobs is between 0.13 and 0.19, we will use the normal approximation based on the number of successes (people holding multiple jobs).

Step 1: Define the parameters

Sample size ($n$): 352
Proportion ($p$) of employed adults holding multiple jobs: 0.14

Step 2: Convert proportions to counts

For $p = 0.13$: \[ X_1 = 0.13 \times 352 \approx 46.56 \quad (\text{round to } 46 \text{ for integer count}) \]
For $p = 0.19$: \[ X_2 = 0.19 \times 352 \approx 66.88 \quad (\text{round to } 66 \text{ for integer count}) \]

Thus, we want to find $P(46 < X < 67)$.

Step 3: Normal approximation

We already know:

Mean ($\mu$): \[ \mu = np = 352 \times 0.14 = 49.28 \]
Standard deviation ($\sigma$): \[ \sigma = \sqrt{np(1-p)} = \sqrt{352 \times 0.14 \times 0.86} \approx \sqrt{42.7888} \approx 6.54 \]

Step 4: Convert counts to Z-scores

For $X = 46$: \[ Z_1 = \frac{46.5 - 49.28}{6.54} \approx \frac{-2.78}{6.54} \approx -0.425 \]
For $X = 66$: \[ Z_2 = \frac{66.5 - 49.28}{6.54} \approx \frac{17.22}{6.54} \approx 2.64 \]

Step 5: Find probabilities for Z-scores

Now we will look up the values in the standard normal distribution table or calculate them using a normal distribution calculator.

Probability for $Z_1 = -0.425$: \[ P(Z < -0.425) \approx 0.3358 \]
Probability for $Z_2 = 2.64$: \[ P(Z < 2.64) \approx 0.9958 \]

Step 6: Calculate the final probability

To find the probability that the proportion of individuals in the sample holding multiple jobs is between 0.13 and 0.19: \[ P(46 < X < 67) = P(Z < 2.64) - P(Z < -0.425) \approx 0.9958 - 0.3358 = 0.6600 \]

Final Answer

Thus, the probability that the proportion of individuals in the sample of 352 who hold multiple jobs is between 0.13 and 0.19 is approximately 0.6600 when rounded to at least four decimal places.

Answer

(e) Would it be unusual if less than 9.5% of the individuals in the sample of 352 held multiple jobs?

GPT-4o mini · Answer

To determine if it would be unusual for less than 9.5% of the individuals in the sample of 352 to hold multiple jobs, we can follow these steps:

Step 1: Define the parameters

Sample size ($n$): 352
Proportion ($p$) of employed adults holding multiple jobs: 0.14

Step 2: Calculate the expected number of individuals holding multiple jobs

For $9.5%$: \[ 0.095 \times 352 \approx 33.44 \] We round it to 33 for integer count.

Step 3: Verify the normal approximation eligibility

Mean ($\mu$): \[ \mu = np = 352 \times 0.14 = 49.28 \]
Standard deviation ($\sigma$): \[ \sigma = \sqrt{np(1-p)} = \sqrt{352 \times 0.14 \times 0.86} \approx \sqrt{42.7888} \approx 6.54 \]

To use the normal approximation, we already checked that both $np \geq 5$ and $n(1-p) \geq 5$ hold (where $np = 49.28$ and $n(1-p) = 303.52$).

Step 4: Calculate the Z-score

Now we want to find the Z-score corresponding to the point where $X < 33$ (i.e., the number of individuals holding multiple jobs).

Continuity correction

Using continuity correction, for $X < 33$: \[ Z = \frac{X - \mu}{\sigma} = \frac{32.5 - 49.28}{6.54} \approx \frac{-16.78}{6.54} \approx -2.57 \]

Step 5: Find the probability associated with the Z-score

Using the standard normal distribution table or a calculator: \[ P(Z < -2.57) \approx 0.0051 \]

Step 6: Interpretation

The probability of observing less than 9.5% of individuals in the sample holding multiple jobs is approximately 0.0051, or 0.51%.

Conclusion

Since this probability is significantly low (typically, a probability less than $0.05$ is considered unusual), it would indeed be unusual if less than 9.5% of the individuals in the sample of 352 held multiple jobs.