Question

If you get a sneak peak at the salary list of your job and realize you are in the lowest bracket, which is the lowest 1%, what is your salary if you know that the mean salary is $48,000 and it has a standard deviation of $5500?

Answer 1

Answer:

$z=-2.33<\frac{a-48000}{5500}$

And if we solve for a we got

$a=48000 -2.33*5500=35185$

And for this case the answer would be 35185 the lowest 1% for the salary

Step-by-step explanation:

Let X the random variable that represent the salary, and for this case we can assume that the distribution for X is given by:

$X \sim N(48000,5500)$  

Where $\mu=48000$ and $\sigma=5500$

And we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.99$   (a)

$P(X<a)=0.01$   (b)

We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. On this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.99

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.01$  

$P(z<\frac{a-\mu}{\sigma})=0.01$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-2.33<\frac{a-48000}{5500}$

And if we solve for a we got

$a=48000 -2.33*5500=35185$

And for this case the answer would be 35185 the lowest 1% for the salary