Answer to Question #153240 in Statistics and Probability for samantha

Question

An Urban council has installed 2000 lamps with mercury bulbs in the   streets of town area. The lifetimes of these bulbs are normally distributed with a mean of 1200 burning hours and having a standard deviation of 200 hours. 

(1) How many of these bulbs can be expected to fail in the first 800 burning hours? 

(2)  How many of these bulbs can be expected to fail in Between 900 and 1750 burning hours? 

(3) After what number of burning hours would you expect that 10 % of the bulbs would fail?

(4) After what number of burning hours would you expect that 150 bulbs are still in good condition?

Answer 1

Let X denote the random variable for the lifetime of bulbs.

X~N(1200, 200²)

z=$\frac{X-\mu} {\sigma}$

1) P(X<800)

z=$\frac{800-1200}{200}$ =-2

P(z<-2)=0.02275 from the z tables.

0.02275 *2000=45.5

$\approx 46 \text {bulbs}$

2)P(900<X<1750)

Z1=$\frac{1750-1200}{200}$ =2.75

P(z1<2.75)=0.99702

Z2=$\frac {900-1200}{200}$ =-1.5

P(z2<-1.5)=0.06681

P(z1<z<z2)=0.99702-0.06681= 0.93021,

Number of bulbs=0.93021*2000=1860.42

$\approx 1860 \text{bulbs}$

3)P(X<x)=0.1

$\phi^{-1}(0.1)=-1.2816$

$-1.2816 =\frac{x-1200}{200}$

x=943.68 hours

After about 944 hours, 10% of the bulbs would fail.

4)150 bulbs

$\frac{150}{1200}=0.075$

150 bulbs is equivalent to 7.5% of the total number of bulbs.

P(X<x) =(1-0.075) =0.925

$\Phi ^{-1}(0.925)=1.44$

$1.44=\frac{x-1200}{200}$

x=1488 hours.

After 1488 hours, we expect that 150 bulbs are still in good condition.