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Homework Help: Normal Distrubution Question

  1. Dec 23, 2008 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations

    You'll need the distribution table

    3. 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
    so what we're saying is the probability the height will be atleast 1.65m is 0.7


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

    Thanks :)
  2. jcsd
  3. Dec 23, 2008 #2


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    Science Advisor

    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 z0.2 and z0.7.
    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 [itex]\mu[/itex] and standard deviation [itex]\sigma[/itex] is [itex]z= (x-\mu)/\sqrt{\sigma}[/itex]. Solve [itex]z_{0.2}= (1.64- \mu)/\sqrt{\sigma}[/itex] and [itex]z_{0.7}= (1.78- \mu)/\sqrt{z}[/itex] for [itex]\mu[/itex] and [itex]\sigma[/itex].

    Use that information to draw the appropriate "bell shaped curve" with correct mean and standard deviation and to find [itex]P(h\ge 1.74)[/itex].
  4. Dec 23, 2008 #3
    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 :)
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