Question

In a certain city, the hourly wage of workers on temporary employment contracts is normally distributed. The mean is $15 and the standard deviation is $3. What percentage of temporary workers earn less than $12 per hour?

Answer 1

Answer:

15.87%

Step-by-step explanation:

We have been given that in a certain city, the hourly wage of workers on temporary employment contracts is normally distributed. The mean is $15 and the standard deviation is $3.

Let us find the z-score of 12, using z-score formula.

$z=\frac{x-\mu}{\sigma}$, where,

$z$ = z-score,

$x$ = Random sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

Upon substituting our given values in z-score formula we will get,

$z=\frac{12-15}{3}$

$z=\frac{-3}{3}$

$z=-1$

Let us find P(z<-1) using normal distribution table.

$P(z$

Therefore, probability of a temporary worker earns less than $12 per hour is 0.15866. Let us convert our given probability in percentage by multiplying by 100.

$\text{The percentage of temporary workers earn less than \ 12 per hour}=0.15866\times 100$

$\text{The percentage of temporary workers earn less than \ 12 per hour}=15.866\approx 15.87$

Therefore, 15.87% of temporary workers earn less than $12 per hour.