Question

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standarddeviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability thatthe mean daily revenue for the next 30 days will be between $7000 and $7500

Answer 1

Answer:

$P(7000<\bar X<7500)$

And we can use the z score formula to solve this problem

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

And replacing we got:

$z = \frac{7000-7200}{\frac{1200}{\sqrt{30}}}= -0.913$

$z = \frac{7500-7200}{\frac{1200}{\sqrt{30}}}= 1.369$

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

$P(-0.913<z<1.369)=P(z<1.369)-P(z<-0.913)= 0.9145 -0.1806= 0.7339$

Step-by-step explanation:

Let X the random variable that represent the daily revenue of a population, and for this case we know the distribution for X is given by:

$X \sim N(7200,1200)$  

Where $\mu=7200$ and $\sigma=1200$

We are interested on this probability

$P(7000<\bar X<7500)$

And we can use the z score formula to solve this problem

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

And replacing we got:

$z = \frac{7000-7200}{\frac{1200}{\sqrt{30}}}= -0.913$

$z = \frac{7500-7200}{\frac{1200}{\sqrt{30}}}= 1.369$

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

$P(-0.913<z<1.369)=P(z<1.369)-P(z<-0.913)= 0.9145 -0.1806= 0.7339$