Question

A stock's price fluctuations are approximately normally distributed with a mean of $104.50 and a standard deviation of $23.62. You decide to purchase whenever the price reaches its lowest 10% of values. What is the most you would be willing to pay for the stock?

a) $80.88
b) $74.23
c) $84.62
d) $134.77

Answer 1

`P(z<(a-\mu)/(\sigma))=0.10`

and we can set up the following equation

tex]z=-1.282<\frac{a-104.5}{23.62}[/tex]

And if we solve for a we got

`a=104.5 -1.282*23.62=74.22`

And the best answer for this case would be:

b) $74.23

Explanation:

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

`X \sim N(104.5,23.62)`

Where
`\mu=104.5` and
`\sigma=23.62`

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

`P(X>a)=0.90` (a)

`P(X<a)=0.10` (b)

As we can see on the figure attached the z value that satisfy the condition with 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.282. On this case P(Z<-1.282)=0.10 and P(z>-1.282)=0.90

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.10`

`P(z<(a-\mu)/(\sigma))=0.10`

and we can set up the following equation

tex]z=-1.282<\frac{a-104.5}{23.62}[/tex]

And if we solve for a we got

`a=104.5 -1.282*23.62=74.22`

And the best answer for this case would be:

b) $74.23