Question

Drivers who are members of the teamsters Union earn an average of $17.15 per hour (U.S. News & World Report). Assume that available data indicate wages are normally distributed with a standard deviation of $2.25. 1) What is the probability that wages are between $15.00 and $20.00 per hours?

Answer 1

Answer:

$P(15<X<20)=P(\frac{15-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{20-\mu}{\sigma})=P(\frac{15-17.15}{2.25}<Z<\frac{20-17.15}{2.25})=P(-0.96<z<1.27)$

And we can find the probability with this difference and using the normal standard table:

$P(-0.96<z<1.27)=P(z<1.27)-P(z<-0.96)= 0.898-0.169 = 0.729$

Step-by-step explanation:

Let X the random variable that represent the wages, and for this case we know the distribution for X is given by:

$X \sim N(17.15,2.25)$  

Where $\mu=17.15$ and $\sigma=2.25$

We want to find this probability:

$P(15<X<20)$

And we can use the z score formula given by:

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

Using this formula we have:

$P(15<X<20)=P(\frac{15-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{20-\mu}{\sigma})=P(\frac{15-17.15}{2.25}<Z<\frac{20-17.15}{2.25})=P(-0.96<z<1.27)$

And we can find the probability with this difference and using the normal standard table:

$P(-0.96<z<1.27)=P(z<1.27)-P(z<-0.96)= 0.898-0.169 = 0.729$