Question

In a certain city , the hourly wage of workers on temporary employment crontracts is normally distributed. The mean is $15 and the standar deviation is $3. What percentage of temporary workers earn less than $12 per hour?
A) 6%
B) 16%
C) 26%
D) 36%

Answer 1

Answer:

16%

Step-by-step explanation:

The mean is $15 and the standar deviation is $3.

mean = 15  and SD = 3

We need to find percentage less than 12 per hour

P(x<12)= P(x=12)

to find P(x=12) we find z-score

$z= \frac{x-mean}{SD} =\frac{12-15}{3} =-1$

Now use z-score table . z-score = 0.1587

P(x=12)=0.1587

To get percentage we multiply by 100

0.1587 * 100 = 15.87 = 16%