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.

1 answer

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.