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

    Find, to 3 decimal places, the mean and the standard deviation of the heights the high jumper can reach

    mark scheme: (1.78-u)/o^2 = 0.8416
    (1.65-u)/o^2 = -0.52
    44


    can anyone tell me how the bolded numbers were obtained? i follow everything else

    thanks in advance
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    We did this the other day in class.

    The mean is somewhere inbetween. So you know greater than 1.78 is 20% or 0.2. From this you can find the z value.

    You also know less than 1.65 is 0.3. You can also find the z value for this.

    From this you form a simultaneous equation and solve to find the standard deviation and the mean.
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    (Original post by Thrug)
    We did this the other day in class.

    The mean is somewhere inbetween. So you know greater than 1.78 is 20% or 0.2. From this you can find the z value.

    You also know less than 1.65 is 0.3. You can also find the z value for this.

    From this you form a simultaneous equation and solve to find the standard deviation and the mean.
    how do we know that >1.78 is 20% and <1.65 is 30%?
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    (Original post by the maths guy)
    how do we know that >1.78 is 20% and <1.65 is 30%?
    It tells you in the question. You need to be able to pick up on "once in 5 attempts" is 1/5 and hence 0.2. Same rule applies for the 7/10. Except in this case greater than 1.65 is 0.7, so less than would be 0.3.
 
 
 
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