# Thread: Can any one help with Normal Distribution. dont have much time left!!

1. ## Can any one help with Normal Distribution. dont have much time left!!

i have a last Q left to do for my maths coursework and it has completely stumped me. can anyone help. here is the whole Q:

A scientist was measuring oak trees in a woodland and noticed that their height was normally distributed. Unfortunately a sudden gust of wind caught her by surprise and most of her results were blown away. All she had left was the knowledge that she measured the height of 200 oak trees and found that 12 trees were less than 8 metres tall and 36 trees were less than 10 metres tall.

Fortunately the scientist knew the properties of normal distribution to work out the mean and standard deviation of the height of the oak trees. Do you?

By sketching a graph of the normal distribution and indicating appropraite areas, determine the mean height of the oak trees and their standard deviation.

If anyone can help quickly that would be great.

Thanks

2. Originally Posted by wicca16
i have a last Q left to do for my maths coursework and it has completely stumped me. can anyone help. here is the whole Q:

A scientist was measuring oak trees in a woodland and noticed that their height was normally distributed. Unfortunately a sudden gust of wind caught her by surprise and most of her results were blown away. All she had left was the knowledge that she measured the height of 200 oak trees and found that 12 trees were less than 8 metres tall and 36 trees were less than 10 metres tall.

Fortunately the scientist knew the properties of normal distribution to work out the mean and standard deviation of the height of the oak trees. Do you?

By sketching a graph of the normal distribution and indicating appropraite areas, determine the mean height of the oak trees and their standard deviation.

If anyone can help quickly that would be great.

Thanks
Pr(X < 8) = 12/200 = 0.06.

Pr(X < 10) = 36/100 = 0.18.

$z_{0.06} = -1.5548$

$z_{0.18} = -0.9154$

(correct to four decimal places).

$Z = \frac{X-\mu}{\sigma}$. Therefore:

$-1.5548 = \frac{8-\mu}{\sigma}$ .... (1)

$-0.9154 = \frac{10-\mu}{\sigma}$ .... (2)

Solve equations (1) and (2) simultaneously.

3. ## help a bit more please!

-0.5 = 8 - mue / s.d.

8 - mue = -0.5

B = 0.14 z= 0.0557

0.0557 = 10 - mue / s.d.

10 - mue = 0.0557

0.0557 x 100 = 5.57

dont know if this is right or not and we are quite stuck on how else to do it. could you help any more?

4. Originally Posted by wicca16
-0.5 = 8 - mue / s.d.

8 - mue = -0.5

B = 0.14 z= 0.0557

0.0557 = 10 - mue / s.d.

10 - mue = 0.0557

0.0557 x 100 = 5.57

dont know if this is right or not and we are quite stuck on how else to do it. could you help any more?
I have no idea what you've done here.

And I have no inclination to explain why it's wrong - all I can see is that you've completely ignored the approach I suggested.