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%

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

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%