Q.The life of an electric bulb manufactured by a company is normally distributed with mean 2,000 hours and standard deviation 150 hours. What percentage of bulbs is expected to fail (burn out) before 1,800 hours?

(A) $\sigma=150$; $P(X<1000)=0.0918\Rightarrow z=-1.33$; $\mu=1000+1.33(150)=1199.5$. (B) $\mu=25,\sigma=10$: $Q_3=\mu+0.675\sigma=31.75$, $QD=0.675\sigma=6.75$, $MD=0.8\sigma=8$.

(A) For normal students. Profit $X\sim N(\mu,\ \sigma^2)$ with variance $22{,}500\Rightarrow\sigma=\sqrt{22500}=150$. Given $P(X<1000)=0.0918$.

Since $0.0918<0.5$, the value $1000$ lies below the mean. The area between $0$ and $|z|$ is $0.5-0.0918=0.4082$, which corresponds to $z=1.33$. As $1000$ is on the left, $z=-1.33$:

$\frac{1000-\mu}{150}=-1.33\ \Rightarrow\ 1000-\mu=-199.5\ \Rightarrow\ \mu=1199.5.$

So the average daily profit is about Rs. 1,199.5 ($\approx$ Rs. 1,200).

(B) For blind students. The density

$f(x)=\text{const}\cdot e^{-\frac12\left(\frac{x-25}{10}\right)^2}$ …

$\mu=55,\sigma=7$: $z_{41}=-2,z_{62}=1$; $P(-2<Z<1)=0.4772+0.3413=0.8185\Rightarrow409$; $P(Z<-2)=0.0228\Rightarrow11$. OR: $\mu=400,\sigma=30$; $D_4=392.4$, $P_{90}=438.4$.

Main part: $\mu=55$ kg, $\sigma=7$ kg, $N=500$.

(1) Between 41 and 62 kg:

$z_1=\frac{41-55}{7}=-2,\qquad z_2=\frac{62-55}{7}=1.$

$P(-2<Z<1)=P(0<Z<2)+P(0<Z<1)=0.4772+0.3413=0.8185.$

Number of persons $=500\times0.8185=409.25\approx409$.

(2) Less than 41 kg:

$z=\frac{41-55}{7}=-2,\qquad P(Z<-2)=0.5-0.4772=0.0228.$

Number of persons $=500\times0.0228=11.4\approx11$.

OR part: $\mu=400$, variance $=900\Rightarrow\sigma=30$.

• Fourth decile $D_4$ (area $0.40$ to its left, i.e. $0.10$ below the mean): $z=-0.253$. …