# Statistics Probability Help

I just began to take this stats course in HS and I'm a little stuck on these 2 problems below. Can anybody please help me out with the solutions? Thank you. Anything is appreciated.

1. Let $Y$ be a random variable. Define a function $Q$ by $Q(m) = E[(Y − m)^{2}]$

(a) Write $\mu E[Y]$. How do you show that $Q(m) = \text{var}[Y] + (m − \mu)^{2}$. I was thinking it had something to do with $Y −m=(Y −\mu)−(m−\mu)$.

(b) Also, how do you show that $Q(m)$ is minimized at $m = E[Y]$. I was thinking that it had something to do with checking the sign of $Q(m) − Q(\mu)$.

2. Let $Y$ be a random variable distributed with $N(3, 16)$. Find $Pr[Y \gt 9.2]$.

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hint: for 1) just apply the definition of $var(Y)=E(Y-\mu)^2$ or simply add $(E[Y])^2$ to the expansion of $(Y-m)^2$. For 2), it simply means that the random variable is a normal or Gaussian random variable and the terms in brackets are the mean and variance respectively. I am sure they must have thought you how to calculate $P(Y>y)$? – jay-sun Jan 19 '13 at 0:31
I used to get help from [tutorteddy](tutorteddy.com/) for stat and probability problems. – user60115 Jan 29 '13 at 16:22

You are right about the approach. We have $Y-m=(Y-\mu)+(\mu-m)$ and therefore $$(Y-m)^2=(Y-\mu)^2+2(\mu-m)(Y-\mu)+(\mu-m)^2.$$ Now take the expectation, noting that $E(Y-\mu)^2=\text{Var}(Y)$. By linearity we get $$E(Y-m)^2=\text{Var}(Y)+2(\mu-m))E(Y-\mu)+E((\mu-m)^2).$$ Now we are finished, because $E(Y)=\mu$, so $E(Y-\mu)=0$, and $\mu -m$ is constant, so $E((\mu-m)^2)=(\mu-m)^2$.

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2.)Let $Y$ be a random variable distributed with $N(3, 16)$. Find $Pr[Y \gt 9.2]$.

First, you have to standardize and find the $z-$score of $9.2$

We get: $\frac{9.2-3}{16} = 0.3875$, then we take

$1 - F(0.3875) \approx 1-0.6517$ = $0.3483$

I used $z = .39$ from this table

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