Normal Distribution and High Jumping Heights: Probability and Standard Deviation

[thomas49th]

Homework Statement

From expirence a higher jumper knows he can claer a height of at least 1.78m once in 5 attempts. He also knows that he can clear a height of at least 1.65m on 7 out of 10 attempts

Assuming that the heights the high jumper can reach follow a Normal Distribution,

a) Draw a sketch to illustrate the information above

b) find, to 3 decimal places, the mean and the standard deviation of the heights the higher-jumper can reach

c) Calculate the probability that he can jump atleast 1.74m

Homework Equations

You'll need the distribution table

The Attempt at a Solution

Part a)

Well I can form the equations

P(h >= 1.78) =0.2
P(h >= 1.65) = 0.7

so what we're saying is the probability the height will be atleast 1.78m is 0.2
and
so what we're saying is the probability the height will be atleast 1.65m is 0.7

okay

so how do i go about drawing these on a normal distrubution. I think I have to do the flip around thing but I am not sure why?

Thanks :)

 

[HallsofIvy]

First thing you need to do is find the z values from a table of the standard normal distribution corresponding to P(z)= 0.2 and P(z)= 0.7. I'll call those z_0.2 and z_0.7.
Here's a good table: http://www.itl.nist.gov/div898/handbook/eda/section3/eda3671.htm

The formula for changing from a normal distribution with mean $\mu$ and standard deviation $\sigma$ is $z= (x-\mu)/\sqrt{\sigma}$. Solve $z_{0.2}= (1.64- \mu)/\sqrt{\sigma}$ and $z_{0.7}= (1.78- \mu)/\sqrt{z}$ for $\mu$ and $\sigma$.

Use that information to draw the appropriate "bell shaped curve" with correct mean and standard deviation and to find $P(h\ge 1.74)$.

 

[thomas49th]

Ahh yes, I know about the standardising using the formula, ill need that for part B

so P(z) = 0.2

z = 0.07926

P(z) = 0.7
z = 1 - phi(0.7)
= 1 - 0.11791
= 0.8209

Is that right?

Thanks :)