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  • Norma Distribution problem

    A random sample of size 6 is selected from a normal distribution with mean 10 and variance 2. Calculate the probability that 2 of the sample values are less than 9, and the remaining 4 sample values are less than 9.

    I'm skeptical of the and they gave.

    I said Pr(X<9) U Pr(X>9) .
    therefore, Pr(X<9) +Pr(X>9)=0( empty set)
    hence 2Pr(X<9) +4Pr(X>9)
    Why is my reasoning wrong?

    I have a problem similar to the above.

    A variable X is distributed normally with mean 10 and standard deviation 2.

    If three values of X are taken at random,calculate the probability that two of them are less than 12 and the other is greater than 12.

    The ans to this problem says 14.74; 0.337 ?

  • #2
    The width of a bolt of fabric is normally distributed with mean 950mm and
    standard deviation 10 mm. What is the probability that a randomly chosen
    bolt has a width between 947 and 950 mm?

    A college has an enrollment of 3264 female students. Records show that
    the mean height of these students is 64.4 inches and the standard deviation
    is 2.4 inches. Since the shape of the relative histogram of sample college
    students approximately normally distributed, we assume the total population
    distribution of the height X of all the female college students follows the
    normal distribution with the same mean and the standard deviation. Find
    P(66 less than or equal to X less than or equal to 68):

    Comment


    • #3
      Originally posted by Dabsy View Post
      A random sample of size 6 is selected from a normal distribution with mean 10 and variance 2. Calculate the probability that 2 of the sample values are less than 9, and the remaining 4 sample values are less than 9.

      I'm skeptical of the and they gave.

      I said Pr(X<9) U Pr(X>9) .
      therefore, Pr(X<9) +Pr(X>9)=0( empty set)
      hence 2Pr(X<9) +4Pr(X>9)
      Why is my reasoning wrong?

      I have a problem similar to the above.

      A variable X is distributed normally with mean 10 and standard deviation 2.

      If three values of X are taken at random,calculate the probability that two of them are less than 12 and the other is greater than 12.

      The ans to this problem says 14.74; 0.337 ?
      Minor nitpick, you can't use "union" on probabilities, "union" is used for sets not numbers i.e. 3 U 5 is meaningless. Now back to your question... why are you saying Pr(X < 9) + Pr(X > 9) = 0? seems like it'd = 1?

      Comment


      • #4
        Originally posted by NoMoreExams View Post
        Minor nitpick, you can't use "union" on probabilities, "union" is used for sets not numbers i.e. 3 U 5 is meaningless. Now back to your question... why are you saying Pr(X < 9) + Pr(X > 9) = 0? seems like it'd = 1?

        Pr(X < 9) + Pr(X > 9) = empty set.

        On that union thing. If you have schaum's outline. Probability,Random Varibles,& Random Process. Page 70 problem 2.47

        You will see the following:
        A={ X <900} U {X> 1100} since {X<900} intersect {X>1100} =empty set.

        Then Pr(A)= Pr(X<900) + Pr(X> 1100) = ...

        Comment


        • #5
          Originally posted by Dabsy View Post
          Pr(X < 9) + Pr(X > 9) = empty set.

          On that union thing. If you have schaum's outline. Probability,Random Varibles,& Random Process. Page 70 problem 2.47

          You will see the following:
          A={ X <900} U {X> 1100} since {X<900} intersect {X>1100} =empty set.

          Then Pr(A)= Pr(X<900) + Pr(X> 1100) = ...
          Once again, why is it empty set. 1. it's not a set since it's a probability not a set but disregarding that, why isn't it 1.

          As far as your 2nd point, yes A is a set which is a union of 2 other sets, that doesn't mean Pr OF A is not a set.

          Comment


          • #6
            Originally posted by Dabsy View Post
            The width of a bolt of fabric is normally distributed with mean 950mm and
            standard deviation 10 mm. What is the probability that a randomly chosen
            bolt has a width between 947 and 950 mm?

            A college has an enrollment of 3264 female students. Records show that
            the mean height of these students is 64.4 inches and the standard deviation
            is 2.4 inches. Since the shape of the relative histogram of sample college
            students approximately normally distributed, we assume the total population
            distribution of the height X of all the female college students follows the
            normal distribution with the same mean and the standard deviation. Find
            P(66 less than or equal to X less than or equal to 68):
            First one:
            P(947<x<950)
            x~950,10

            Pz<950 = .5
            P<947 = 3/10 = Z< .3 = 1-.6179, .38

            .5-.38 = .12

            Second:
            66<x<68
            X~(64.4, 2.4)

            P(x<68) = I(1.5) = .93
            P(x<66) = I(.66) = .75

            .93-.75 = .18

            Or you can integrate, see attached
            Attached Files

            Comment


            • #7
              Originally posted by Dabsy View Post
              A random sample of size 6 is selected from a normal distribution with mean 10 and variance 2. Calculate the probability that 2 of the sample values are less than 9, and the remaining 4 sample values are less than 9.

              I'm skeptical of the and they gave.

              I said Pr(X<9) U Pr(X>9) .
              therefore, Pr(X<9) +Pr(X>9)=0( empty set)
              hence 2Pr(X<9) +4Pr(X>9)
              Why is my reasoning wrong?

              I have a problem similar to the above.

              A variable X is distributed normally with mean 10 and standard deviation 2.

              If three values of X are taken at random,calculate the probability that two of them are less than 12 and the other is greater than 12.

              The ans to this problem says 14.74; 0.337 ?
              Correct me if I'm wrong, haven't checked the answer for the first one, but here's how I did the first problem:

              The z-value for X = 9 is -0.5. So, use the z-table to look up the probability P(x<-0.5) = 0.3085, and the complement is 1-P = 0.6915. Now you can use the binomial formula to calculate the probability that exactly 2 of the 6 are less than 9 and the other 4 are greater. P = 6C2 * 0.3085^2 * 0.6915^4 = 0.3264. Is that the answer they gave?

              The second problem is very similar - you can solve it the same way. I did it and got 0.337, which is what they got too.
              Last edited by Trajork; April 28 2010, 04:09 PM.

              Comment


              • #8
                [QUOTE=weezy;64647]First one:
                P(947<x<950)
                x~950,10

                Pz<950 = .5
                P<947 = 3/10 = Z< .3 = 1-.6179, .38

                .5-.38 = .12

                Second:
                66<x<68
                X~(64.4, 2.4)

                P(x<68) = I(1.5) = .93
                P(x<66) = I(.66) = .75

                .93-.75 = .18
                GOT YOU
                Last edited by Dabsy; May 5 2010, 09:58 AM.

                Comment

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