Question

In a poll about work, 60% of respondents said that their jobs were sometimes or always stressful. Fourteen workers are chosen at random.
(a) What is the mean number who find their jobs stressful in a sample of 14 workers? Round the answer to two decimal places.
(b) What is the standard deviation of the number who find their jobs stressful in a sample of 14 workers? Round the answer to four decimal places.

Answer 1

To solve these problems, we need to use the properties of a binomial distribution, since we are dealing with a fixed number of trials (14 workers) and two possible outcomes (finding the job stressful or not).

### (a) Mean Number of Workers Who Find Their Jobs Stressful

The mean \(\mu\) of a binomial distribution can be calculated using the formula:

\[
\mu = n \cdot p
\]

where:
- \(n\) is the number of trials (workers selected)
- \(p\) is the probability of success (in this case, the probability that a worker finds their job stressful)

Here, we have:
- \(n = 14\)
- \(p = 0.60\) (60% of respondents find their jobs stressful)

Now we can calculate the mean:

\[
\mu = 14 \cdot 0.60 = 8.4
\]

### Answer for (a):
The mean number of workers who find their jobs stressful in a sample of 14 workers is **8.40**.

### (b) Standard Deviation of the Number Who Find Their Jobs Stressful

The standard deviation \(\sigma\) of a binomial distribution can be calculated using the formula:

\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

Using the values already provided:
- \(n = 14\)
- \(p = 0.60\)
- \(1 - p = 0.40\)

We can now calculate the standard deviation:

\[
\sigma = \sqrt{14 \cdot 0.60 \cdot 0.40}
\]

Calculating inside the square root first:

\[
\sigma = \sqrt{14 \cdot 0.60 \cdot 0.40} = \sqrt{14 \cdot 0.24} = \sqrt{3.36}
\]

Now, taking the square root:

\[
\sigma \approx 1.8330
\]

### Answer for (b):
The standard deviation of the number who find their jobs stressful in a sample of 14 workers is approximately **1.8330**.