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

1) If X is the corresponding random variable, then $\displaystyle P(\bar{X}<4.354)=0.90$
=> $\displaystyle P(\frac{\bar{X}20}{\sqrt{\frac{0.2}{n}}}<\frac{4.35420}{\sqrt{\frac{0.2}{n}}})=0.90$
=> $\displaystyle \Phi(\frac{4.35420}{\sqrt\frac{0.2}{n}})=0.90$
Find the inverse cdf of a standard normal distribution having cumulative probability $\displaystyle 0.90$; that will be equal to $\displaystyle \frac{4.35420}{\sqrt\frac{0.2}{n}}$. Then solve for $\displaystyle n$.
2)i) HINT: You have to find $\displaystyle P(X_1X_2>15)$ where $\displaystyle (X_1X_2)\sim N(5050 , 8^2+8^2)$
ii) HINT: The distribution of Y is symmetric about $\displaystyle \frac{74+146}{2}$. So the mean of $\displaystyle Y$ is $\displaystyle \frac{74+146}{2}$. For variance, replace $\displaystyle Y$ by $\displaystyle X$ and proceed.

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!!

Horrible mistake! I read it normal instead of binomial! The LHS will be
$\displaystyle \Phi(\frac{4.354(20*0.2)}{\sqrt{\frac{20*0.2*(10.2)}{n}}})$
