Normal distribution - Finding mean and standard deviation

[James...]

Homework Statement

A high jumper knows from experience that she can clear a height of at least 1.78m once in 5 attempts. She also knows that she can clear a height of at least 1.65m on 7 out of 10 attempts.
Find to 3 dp the mean and standard deviation of the heights the jumper can reach

Homework Equations

The Attempt at a Solution

This is a question on an online tutorial. I work it out slightly differently and get different answers.

After rounding I get

mean of 1.700m
standard deviation of 0.091m


Would anyone be able to double check for me, it's using statistics tables so I'm not sure how to put all my working down. My exam is this coming week and I want to be sure I am using the tables correctly.

I suppose the main bits I would like to check are

\Phi(Z₁) = 0.8

\Phi(-Z₂) = 0.7

I get Z₁ = 0.8779 and Z₂ = -0.5534

This is where we get different answers. I use Neaves tables and do the "inverse normal function" to gain Z, where as he uses the "c.d.f of the standard normal distribution table", finds a probability valuse as close to 0.8 and 0.7 as possible and then takes the corresponding Z value.

Any help would be appreciated.

James

 

[HallsofIvy]

If a normal distribution has mean \mu and standard deviation \sigma, then the "standard normal variable" is z= (x- \mu)/\sigma.

What you are saying is that x= 1.78 corresponds to z= 0.8779 and x= 1.65 corresponds to z= -.05534.

So you need to solve the equations
\frac{1.79- \mu}{\sigma}= 0.8779
and
\frac{1.65- \mu}{\sigma}= -0.05534

Solve those two equations for \mu and \sigma.

Dividing one equation by the other will immediately eliminate \sigma and leave a linear equation for \mu.

 

[James...]

Thanks for the reply, but I get that part. I was wondering about the values of 0.8779 and -0.5534

I think they are correct for the corresponding probabilities but the answer got different values.

So it's actually the Z1 and Z2 values I need checking. After that, I'm fine.

CHeers
James

 

[HallsofIvy]

A table of the normal distribution is available online at
http://www.math.unb.ca/~knight/utility/NormTble.htm

I would interpret "she makes this height once in 5 attempts" as meaning she makes it 20% of the time so misses 80% of the time and I would look up z corresponding to .8- on this table that is z= .875- that height is .875 standard deviations above her mean. Similarly, making a height "7 out of 10 times" means she misses it .3 times so I would look up z corresponding to .3. This particular table only gives positive z so would look up, instead, .5+ .3= .8 which is what we already have- it was z=.875. The z corresponding to .3 is -.875.

My .875 corresponds, to the accuracy of my table, to your .8779 but my -.875 is nowhere near your -.5534. Because .3 and .8 are equally spaced about .5, I am inclined to believe my results.

 

[James...]

Many thanks for the reply, and you are correct. I was using the table wrong and now understand how it all works properly. Thanks for taking the time to reply.

JAmes