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Math Help - exam question help on variance

  1. #1
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    exam question help on variance

    hello can someone please help me the question's below. i have the answers to question, but i just cannot figure out how they got the answer. I referred to my textbook for help but still was confused.




    1. An athlete believes that her times for running 200 metres in races are normally distributed
    with a mean of 22.8 seconds.
    (a) Given that her time is over 23.3 seconds in 20% of her races, calculate the variance of
    her times.
    (5 marks)
    (b) The record over this distance for women at her club is 21.82 seconds. According to her
    model, what is the chance that she will beat this record in her next race?
    (3 marks)



    Any help is appreichated

    Thank You

    Kind Regards
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by rpatel View Post
    hello can someone please help me the question's below. i have the answers to question, but i just cannot figure out how they got the answer. I referred to my textbook for help but still was confused.




    1. An athlete believes that her times for running 200 metres in races are normally distributed
    with a mean of 22.8 seconds.
    (a) Given that her time is over 23.3 seconds in 20% of her races, calculate the variance of
    her times.
    (5 marks)
    Look up the value of z that is exceeded 20% of the time for a standard
    normal distribution. This is ~0.84.

    Now if X~N(mu,sigma^2), then z=(x-mu)/sigma has a standard normal distribution. So (x-mu)/sigma > 0.84 20% of the time, rearrange:

    x>mu+(0.84*sigma)

    20% of the time. But we are told that the value exceeded 20% of the time is
    23.3, so:

    23.3=mu+0.84*sigma,

    and mu=22.8, so:

    23.3=22.8+0.84*sigma,

    so:

    sigma=0.5/0.84=0.5952..

    Which is the standard deviation, so the variance is sigma^2~=0.3543

    (b) The record over this distance for women at her club is 21.82 seconds. According to her
    model, what is the chance that she will beat this record in her next race?
    (3 marks)
    With these figures the z corresponding to 21.82 is:

    z=(21.82-22.8)/0.5952 ~= -1.647

    Looking this up in the normal table the probability of logging a time less than 21.82
    is ~=0.0495.

    RonL
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