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Math Help - Normal Distribution

  1. #1
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    Normal Distribution

    Would really appreciate if someone could help !!!

    1) If a large number of samples of size n are taken from B(20,0.2) and approximately 90% of the sample means are less than 4.354, estimate n.

    2) i. The random variable X has the distribution N(50, 8^2). Given that  X_{1} and  X_{2} are to independent observations of  X , find  P( X_{1} > X_{2} + 15)
    ii. The random variable  Y is related to  X by the formula  Y = aX + b , where  a and  b are constants with  a > 0 . Given that  P(Y < 74) = P(Y > 146) = 0.00668 , find the values of  E(Y) and  Var(Y) , and hence find the values of a and b.

    thanks!!
    Last edited by teddybear67; April 2nd 2011 at 06:48 AM.
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  2. #2
    Senior Member Sambit's Avatar
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    1) If X is the corresponding random variable, then P(\bar{X}<4.354)=0.90

    => P(\frac{\bar{X}-20}{\sqrt{\frac{0.2}{n}}}<\frac{4.354-20}{\sqrt{\frac{0.2}{n}}})=0.90

    => \Phi(\frac{4.354-20}{\sqrt\frac{0.2}{n}})=0.90

    Find the inverse cdf of a standard normal distribution having cumulative probability 0.90; that will be equal to \frac{4.354-20}{\sqrt\frac{0.2}{n}}. Then solve for n.


    2)i) HINT: You have to find P(X_1-X_2>15) where (X_1-X_2)\sim N(50-50 , 8^2+8^2)

    ii) HINT: The distribution of Y is symmetric about \frac{74+146}{2}. So the mean of Y is \frac{74+146}{2}. For variance, replace Y by X and proceed.
    Last edited by Sambit; April 3rd 2011 at 09:58 AM. Reason: little mistake in (1)
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  3. #3
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    for 1), i did as you taught me to, but the answer given for n is 42 whereas i got 0.00134182 oO

    and as for the rest, i got them! thanks!!
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  4. #4
    Senior Member Sambit's Avatar
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    Horrible mistake! I read it normal instead of binomial! The LHS will be

    \Phi(\frac{4.354-(20*0.2)}{\sqrt{\frac{20*0.2*(1-0.2)}{n}}})
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  5. #5
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    Ok! I got it!! Thanks!!
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