# Help me answer this question.....? (question in picture below)

Nov 15, 2017

(i) $c = \frac{1}{10} , \text{ " d=1/6" }$ (ii) $\text{Var} \left(X\right) = \frac{16}{3}$

(iii) $\text{P"(X_1+X_2=10)=11/100" }$ (iv) $\text{E} \left(Y\right) = 240 ,$

$\text{Var"(Y)=256" }$ (v) $\text{P} \left(220 \le Y \le 260\right) \approx 0.7888$

#### Explanation:

(i) $\text{P"(X < 6) = "P} \left(X \ge 6\right)$

and by total probability

$\text{P"(X < 6) + "P} \left(X \ge 6\right) = 1$

so

$\text{P"(X < 6) = 1/2 = "P} \left(X \ge 6\right)$

Since there are 5 outcomes less than 6, and all 5 outcomes have the same probability, we divide that $\frac{1}{2}$ into 5 equal pieces to give to each of the 5 outcomes: $c = \left(\frac{1}{5}\right) \left(\frac{1}{2}\right) = \frac{1}{10}$.

Similarly, the outcomes 6, 7, and 8 are equally likely, and their probabilities sum to (the other) $\frac{1}{2}$, so $d = \left(\frac{1}{3}\right) \left(\frac{1}{2}\right) = \frac{1}{6.}$

(ii) "E"(X)=sum_(x=1)^8[x * "P"(X=x)]

$= {\sum}_{x = 1}^{5} \left[x \cdot \text{P"(X=x)]+sum_(x=6)^8[x * "P} \left(X = x\right)\right]$
$= \frac{1}{10} {\sum}_{x = 1}^{5} x + \frac{1}{6} {\sum}_{x = 6}^{8} x$

$= \frac{15}{10} + \frac{21}{6}$

$= \frac{45 + 105}{30} = \frac{150}{30} = 5$

"Var"(X)="E"(X^2)-{"E"(X)}^2

$= {\sum}_{x = 1}^{8} \left[{x}^{2} \text{P} \left(X = x\right)\right] - {5}^{2}$
$= \frac{1}{10} {\sum}_{x = 1}^{5} {x}^{2} + \frac{1}{6} {\sum}_{x = 6}^{8} {x}^{2} - 25$

$= \frac{55}{10} + \frac{149}{6} - 25$

$= \frac{165 + 745 - 750}{30} = \frac{160}{30} = \frac{16}{3}$

(iii) Let ${X}_{1}$ and ${X}_{2}$ be the scores for the first roll and second roll, respectively. Then

$\text{P} \left({X}_{1} + {X}_{2} = 10\right)$

$= \left(\left(\text{P"(X_1=2, X_2=8)),(+"P"(X_1=3, X_2=7)),(+"P"(X_1=4, X_2=6)))+"P"(X_1=5, X_2=5)+(("P"(X_1=6, X_2=4)),(+"P"(X_1=7, X_2=3)),(+"P} \left({X}_{1} = 8 , {X}_{2} = 2\right)\right)\right)$

SInce all 6 terms in the big brackets have the same probability (due to independence of ${X}_{1}$ and ${X}_{2}$), we have

$= 6 \left[\text{P"(X_1=2)"P"(X_2=8)] + ["P"(X_1=5)"P} \left({X}_{2} = 5\right)\right]$

$= 6 \left[\frac{1}{10} \times \frac{1}{6}\right] + \left[\frac{1}{10} \times \frac{1}{10}\right]$

$= \frac{1}{10} + \frac{1}{100}$

$= \frac{10 + 1}{100} = \frac{11}{100}$

(iv) $Y = {\sum}_{i = 1}^{48} {X}_{i}$, where the ${X}_{i} \text{'s}$ are independent. So

$\text{E"(Y)="E} \left({\sum}_{i = 1}^{48} {X}_{i}\right)$
color(white)("E"(Y))=sum_(i=1)^48 "E"(X_i)" " (by independence)
$\textcolor{w h i t e}{\text{E} \left(Y\right)} = {\sum}_{i = 1}^{48} 5$

$\textcolor{w h i t e}{\text{E} \left(Y\right)} = 48 \times 5 = 240$

Similarly,

$\text{Var"(Y)="Var} \left({\sum}_{i = 1}^{48} {X}_{i}\right)$
color(white)("Var"(Y))=sum_(i=1)^48 "Var"(X_i)" " (by independence)
$\textcolor{w h i t e}{\text{Var} \left(Y\right)} = {\sum}_{i = 1}^{48} \frac{16}{3}$

$\textcolor{w h i t e}{\text{Var} \left(Y\right)} = 48 \times \frac{16}{3} = 256$

(v) Assuming $Y$ is normally distributed, we have

$Y \text{ ~ " "N} \left(\mu = 240 , {\sigma}^{2} = 256\right)$

So

$\text{P} \left(220 \le Y \le 260\right)$

$= \text{P} \left(\frac{220 - \mu}{\sigma} \le \frac{Y - \mu}{\sigma} \le \frac{260 - \mu}{\sigma}\right)$

Using the standard normal $\left(Z\right)$ equivalence, this is

$= \text{P} \left(\frac{220 - 240}{16} \le Z \le \frac{260 - 240}{16}\right)$

="P"("–"1.25 <= Z <= 1.25)

="P"(Z <= 1.25) - "P"(Z < "–"1.25)

$= \Phi \left(1.25\right) - \Phi \left(\text{–} 1.25\right)$

These values can be found by lookup in a $z$-table (or using software). By table lookup:

$= 0.8944 - 0.1056 \text{ } = 0.7888$