Question

Find the probability that the proportion of individuals in the sample of 225 who hold multiple jobs is between 0.14 and 0.18. Round the answer to at least four decimal places. The probability that the proportion of individuals in the sample of 225 hold multiple jobs is between 0.14 and 0.18 i

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: number of individuals that hold multiple jobs in a sample of 225.

And you need to calculate the probability of the proportion of individuals being between 0.14 and 0.18.

Considering that the parameter of interest is the proportion and the sample is large enough you can use the standard normal approximation to calculate this interval.

$Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }$

Unfortunately, there is no given value for the population proportion of individuals that have multiple jobs. Let's say, for example, that his proportion is 10%.

You can symbolize the probabilities as:

P(0.14≤X≤0.18)= P(X≤0.18)-P(X≤0.14)

Using the approximation of the standard normal you can standardize these proportion values:

P(Z≤(0.18-0.1)/√[(0.1*0.9)/225])-P(Z≤(0.14-0.1)/√[(0.1*0.9)/225])

P(Z≤4)-P(Z≤2)

Now you have to look for the corresponding values in the table of the Z distribution (right table, positive numbers of Z)

P(Z≤4)-P(Z≤2)= 1 - 0.97725= 0.02275

I hope it helps!

Answer 2

Answer:

0.1276

Step-by-step explanation:

Let sample space = S

∴ n(S) = 225

Probability of multiple jobs A, that is, P(A) = 0.14

Probability of multiple jobs B, that is, P(B) = 0.18

Hence, P(A) = n(A)/n(S)

∴ 0.14 = n(A)/225

n(A) = 0.14 X 225 = 31.5%

Also, P(B) = n(B)/n(S)

∴ 0.18 = n(B)/225

n(B) = 0.18 X 225 = 40.5%

The probability that the multiple jobs, P( A2 ∩ B2) = P(A2) X P(B2)

P(A2) = 0.315 and P(B2) = 0.405

∴ P(A2 ∩ B2) =  P(A2) X P(B2) = 0.315 X 0.405 = 0.1276