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.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z[/tex] = z-score,
[tex]x[/tex] = Random sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
Upon substituting our given values in z-score formula we will get,
[tex]z=\frac{12-15}{3}[/tex]
[tex]z=\frac{-3}{3}[/tex]
[tex]z=-1[/tex]
Let us find P(z<-1) using normal distribution table.
[tex]P(z<-1)=0.15866[/tex]
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.
[tex]\text{The percentage of temporary workers earn less than \$12 per hour}=0.15866\times 100[/tex]
[tex]\text{The percentage of temporary workers earn less than \$12 per hour}=15.866\approx 15.87[/tex]
Therefore, 15.87% of temporary workers earn less than $12 per hour.
