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  1. #1
    JM08357
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    statistics question

    AT ONE HIGH SCHOOL, STUDENTS CAN RUN THE 100 YARD DASH IN AN AVERAGE OF 15.2 SECONDS WITH A STANDARD DIVIATION OF 0.9 SECONDS. THE TIMES ARE CLOSELY APPROXIMATED BY A NORMAL CURVE...

    1) find the percent of students whose times are between 16.1 seconds and 17 seconds.

    2) find the percent of the students whose times are between 14 seconds and 16.1 seconds

    3) if there are 500 students in the school how many of them can rum the 100 year dash in less than 17 seconds
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  2. #2
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    Quote Originally Posted by JM08357 View Post
    AT ONE HIGH SCHOOL, STUDENTS CAN RUN THE 100 YARD DASH IN AN AVERAGE OF 15.2 SECONDS WITH A STANDARD DIVIATION OF 0.9 SECONDS. THE TIMES ARE CLOSELY APPROXIMATED BY A NORMAL CURVE...

    1) find the percent of students whose times are between 16.1 seconds and 17 seconds.
    P(16.1<x<17)=P(15.2<x<17)-P(15.2<x<16.1)
    But,
    P(15.2<x<17)=\mbox{ 2 standard deviations }=.475
    P(15.2<x<16.1)=\mbox{ 1 standard deviation }=.34
    Thus,
    P(16.1<x<17)=.475-.34=?

    2) find the percent of the students whose times are between 14 seconds and 16.1 seconds

    3) if there are 500 students in the school how many of them can rum the 100 year dash in less than 17 seconds[/QUOTE]
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  3. #3
    Junior Member F.A.P's Avatar
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    Quote Originally Posted by JM08357 View Post
    AT ONE HIGH SCHOOL, STUDENTS CAN RUN THE 100 YARD DASH IN AN AVERAGE OF 15.2 SECONDS WITH A STANDARD DIVIATION OF 0.9 SECONDS. THE TIMES ARE CLOSELY APPROXIMATED BY A NORMAL CURVE...

    1) find the percent of students whose times are between 16.1 seconds and 17 seconds.

    2) find the percent of the students whose times are between 14 seconds and 16.1 seconds

    3) if there are 500 students in the school how many of them can rum the 100 year dash in less than 17 seconds
    We have (closely) that the running time X is N(15.2,0.9^2).
    If we put Z = (X-15.2)/0.9, then Z is standard normal N(0,1).

    1) P(16.1 < X < 17) = P(X > 16.1) - P(X > 17) =
    = P(Z > (16.1-15.2)/0.9) - P(Z > (17-15.2)/0.9) =
    = P(Z > 1) - P(Z > 2) =
    = 0.158655 - 0.022750 (from a standard normal table)

    2) P(14 < X < 16.1) = P(X > 14) - P(X > 16.1) =.... continue as above

    3) The number of students that can run the 100 yards in less than 17 seconds is the integer part of

    500*P(X < 17), where

    P(X < 17) = 1 - P(X > 17) = 1 - P(Z > (17-15.2)/0.9) = 1 - P(Z > 2)
    Last edited by F.A.P; December 18th 2006 at 04:38 PM.
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