orking two jobs: About 5\% of employed adults in the United States held multiple jobs. A random sample of 68 employed adults is chosen. Use the $$ \pi-84 $$ is calculator as needed. Part 1 of 5 (a) Is it appropriate to use the normal approximation to find the probability that less than $$ 6.5 \% $$ of the individuals in the sample hold multiple jobs? If so, find the probability. If not, explain why not. It $$ \square $$ is not appropriate to use the normal curve, since $$ n p= $$ $$ \square $$ 3.4 $$ \square $$ 10. Part 2 of 5 (b) A new sample of 336 employed adults is chosen. Find the probability that less than $$ 6.5 \% $$ of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places as needed. The probability that less than $$ 6.5 \% $$ of the individuals in this sample hold multiple jobs is 0.8964 . Part 3 of 5 (c) Find the probability that more than $$ 6.5 \% $$ of the individuals in the sample of 336 hold multiple jobs. Round the answer to at least four decimal places as needed. The probability that more than $$ 6.5 \% $$ of individuals in the sample of 336 hold multiple jobs is 0.1036 . Part: $$ 3 / 5 $$ Part 4 of 5 (d) Find the probability that the proportion of individuals in the sample of 336 who hold multiple jobs is between 0.054 and 0.10 . Round the answer to at least four decimal places as needed. The probability that the proportion of individuals in the sample of 336 hold multiple jobs is between 0.054 and 0.10 is $$ \square $$ .

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We are told that about 5% (p = 0.05) of employed adults hold multiple jobs.

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Part 1 (a):

For a sample of n = 68, we first check if the normal approximation is appropriate. The usual rule is that both
  np ≥ 10  and  n(1 – p) ≥ 10.
Here, 
  np = 68 × 0.05 = 3.4  (less than 10)  and  n(1 – p) = 68 × 0.95 = 64.6.
Since np = 3.4 is less than 10, it is not appropriate to use the normal approximation.

Answer to Part (a):
 It is not appropriate to use the normal curve since np = 3.4, which is less than 10.

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Parts 2 & 3 use a larger sample.

For a sample of n = 336,
  np = 336 × 0.05 = 16.8  and  n(1 – p) = 336 × 0.95 = 319.2,
so the normal approximation is acceptable.

The standard deviation (the standard error) of the sample proportion is
  σₚ̂ = √[p(1 – p)/n] = √[0.05 × 0.95 / 336] ≈ √[0.0475/336] ≈ √(0.00014137) ≈ 0.01189.

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Part 2 (b):

We are to find the probability that less than 6.5% of the individuals in the sample hold multiple jobs, i.e.,
  P(p̂ < 0.065).

Calculate the z‐score:
  z = (0.065 – 0.05) / 0.01189 ≈ 0.015 / 0.01189 ≈ 1.261.

Using the normal distribution, 
  P(Z < 1.261) ≈ 0.8964.

Thus the probability is about 0.8964.

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Part 3 (c):

The probability that more than 6.5% hold multiple jobs is the complement:
  P(p̂ > 0.065) = 1 – P(p̂ < 0.065) = 1 – 0.8964 = 0.1036.

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Part 4 (d):

We need the probability that the sample proportion is between 0.054 and 0.10, that is,
  P(0.054 < p̂ < 0.10).

Find the z-scores for the endpoints.

For the lower bound:
  z₁ = (0.054 – 0.05) / 0.01189 = 0.004 / 0.01189 ≈ 0.336.

For the upper bound:
  z₂ = (0.10 – 0.05) / 0.01189 = 0.05 / 0.01189 ≈ 4.206.

Now,
  P(0.054 < p̂ < 0.10) = P(z₁ < Z < z₂).

From the standard normal table:
  P(Z < 0.336) ≈ 0.6317  (this is the area to the left of z₁)
  P(Z < 4.206) ≈ 1.0000  (since 4.206 is extremely high).

Thus,
  P(0.054 < p̂ < 0.10) ≈ 1.0000 – 0.6317 = 0.3683.

Rounded to four decimal places, the probability is approximately 0.3683.

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Summary of Answers:
(a) It is not appropriate to use the normal curve since np = 3.4, which is less than 10.
(b) The probability that less than 6.5% hold multiple jobs for n = 336 is approximately 0.8964.
(c) The probability that more than 6.5% hold multiple jobs for n = 336 is approximately 0.1036.
(d) The probability that the proportion is between 0.054 and 0.10 for n =336 is approximately 0.3683.
**Part 1 (a):** It is not appropriate to use the normal approximation because $$ np = 3.4 $$ is less than 10. **Part 2 (b):** The probability that less than 6.5% of the individuals in the sample hold multiple jobs is approximately 0.8964. **Part 3 (c):** The probability that more than 6.5% of the individuals in the sample hold multiple jobs is approximately 0.1036. **Part 4 (d):** The probability that the proportion of individuals in the sample who hold multiple jobs is between 0.054 and 0.10 is approximately 0.3683.

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