# The question in the picture below..?

Nov 19, 2017

(i) $\text{P} \left(C \ge 200\right) = 0.8849$
(ii) $\text{P} \left(400 \le D \le 405\right) = 0.3451$
(iii) $a = 0.6125$

#### Explanation:

(i) Let $C$ be the mass of coffee in one random jar. Then $C \text{ ~ N} \left(\mu = 203 , {\sigma}^{2} = {2.5}^{2}\right) .$

$\text{P"(C>=200)="P} \left(\frac{C - \mu}{\sigma} \ge \frac{200 - \mu}{\sigma}\right)$
color(white)("P"(C>=200))="P"(Z >= (200-203)/2.5)
color(white)("P"(C>=200))="P"(Z >= –1.2)
color(white)("P"(C>=200))="P"(Z < 1.2)
$\textcolor{w h i t e}{\text{P} \left(C \ge 200\right)} = \Phi \left(1.2\right)$
color(white)("P"(C>=200))=0.8849" "=88.49%

(ii) Let ${C}_{1}$ and ${C}_{2}$ be the masses of coffee in two independently chosen random jars, and let $D = {C}_{1} + {C}_{2}$. Then
$D \text{ ~ N} \left(\mu = 2 \times 203 , {\sigma}^{2} = 2 \times {2.5}^{2}\right) .$
$\textcolor{w h i t e}{D} = \text{N} \left(406 , 12.5\right)$

$\text{P} \left(400 \le D \le 405\right)$
="P((400-406)/sqrt(12.5)<=(D-mu)/sigma <= (405-406)/sqrt(12.5))
~~ "P(–1.70 <= Z <= –0.28)
=Phi(–0.28)-Phi(–1.70)
$= 0.3897 - 0.0446$
=0.3451" "=34.51%

(iii) $\overline{C} \text{ ~ N"(mu=203, sigma^2=(2.5^2)/20)="N} \left(203 , 0.3125\right)$

$\text{P} \left(\left\mid \overline{C} - 203 \right\mid < a\right) = 0.95$
$\implies 2 \left\{\text{P} \left(0 < \left[\overline{C} - 203\right] < a\right)\right\} = 0.95$
$\implies 2 \left\{\text{P} \left(0 < \frac{\overline{C} - 203}{0.3125} < \frac{a}{0.3125}\right)\right\} = 0.95$
$\implies \text{P} \left(0 < Z < \frac{a}{0.3125}\right) = 0.475$

$\implies \Phi \left(\frac{a}{0.3125}\right) - \Phi \left(0\right) = 0.475$

$\implies \Phi \left(\frac{a}{0.3125}\right) - 0.5 = 0.475$

$\implies \Phi \left(\frac{a}{0.3125}\right) = 0.975$

=>a/0.3125 = Phi^(–1)(0.975)

$\implies \frac{a}{0.3125} = 1.96$

$\implies a = 0.6125$

Note: The value of $a$ is equal to ${z}_{0.025} \times \sigma ,$ where ${z}_{\alpha / 2}$ is the $z$-coordinate of the standard normal curve $Z$ that has an area of $\alpha / 2$ to its right. Here, $\alpha = 1 - 0.95 ,$ so $\alpha / 2 = 0.025 .$

Also, the $\sigma$ used here is the standard deviation of $\overline{C}$. You may see a similar term used to help calculate confidence intervals: ${z}_{\alpha / 2} \times \frac{\sigma}{\sqrt{n}} .$ In this form, $\sigma$ is the standard deviation of a single observation, thus $\frac{\sigma}{\sqrt{n}}$ is the standard deviation of the mean of $n$ observations.