Normal Distribution

**teddybear67** · April 2nd 2011, 06:07 AM

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!!

**Sambit** · April 3rd 2011, 05:34 AM

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.

**teddybear67** · April 4th 2011, 05:11 AM

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!!

**Sambit** · April 4th 2011, 05:31 AM

Horrible mistake! I read it normal instead of binomial! The LHS will be

$\Phi(\frac{4.354-(20*0.2)}{\sqrt{\frac{20*0.2*(1-0.2)}{n}}})$

**teddybear67** · April 4th 2011, 08:48 PM

Ok! I got it!! Thanks!!

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