Question

1.    A country town installs 2000 new electric lights in a new housing estate. These lamps have an average life of 1000 hours with a standard deviation of 200 hours. Hint: Draw and label a normal “bell-shaped” curve to help answer this?

 

a)   What percentage of bulbs would be expected to fail between 800 hours and 1200 hours?

b)   How many bulbs would this be?

c) How many bulbs would be expected to last longer than 1600 hours

Answer 1

Let $X$ be a random variable denotes the life of electric bulb.

Then by the given conditions everage life of electric bulb is $1000$ hours and standard deviation of $200$ hours.

Therefore $\mu =1000$ and $\sigma =200$

Here $X$ is normally distributed.

Let $Z=\frac{X-\mu}{\sigma}$ .Then $Z=\frac{X-1000}{200}$.

(a) First we have to find the probability of the life of electric bulb lie between $800$ hours and $1200$ hours.

$\therefore P(800<X<1200)=P(\frac{800-100}{200}<Z<\frac{1200-100}{200})$

$=P(-1<Z<1)$

$=2×P(0<Z<1)$

$=2×0.3413=0.6826$ [from normal distribution table]

Therefore probability of bulbs expected to fail between $800$ hours and $1000$ hours is $=(1-0.6876)=0.3174$

$\therefore$ Percentage of bulbs would be expected to fail between $800$ hours and $1000$ hours is $=32$ (approximately)

(b) There are $2000$ new bulbs.

Therefore the number of bulbs would be expected to fail between $800$ hours and $1000$ hours is $=(2000×0.3174)=63$ (approximately).

(c) Probability of life of bulbs expected to last longer than 1600 hours is $P(X>1600).$

$\therefore P(X>1600)=P(Z>\frac{1600-1000}{200})$

$=P(Z>3)$

$=0.5-P(0\leq Z\leq 3)$

$=0.5-0.4987$

$=0.0013$ (approximately)

Therefore number of bulbs would be expected to last longer than $1600$ hours is $=(2000×0.0013)=3$ (approximately)