Answer to Question #152819 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. 

(i) How many of these bulbs can be expected to fail.

 (a) In the first 800 burning hours? 

Answer 1

Let random variable X represents life time of the bulb

X~N($\mu$ , $\sigma$ ²) =N (1200 , 200²) then Z = (X - $\mu$ )/$\sigma$ = (X -1200 )/200 ~N (0 , 1)

P(X$\leq$  800) = P((X -1200 )/200$\leq$ (800 -1200 )/200)= P (Z$\leq$ -2 ) = 0.0228

Number of bulbs expected to fail in the first 800 burning hours

= P(X$\leq$ 800)*(total number of lamps)

= 0.0228 (2000) = 45.6

Answer : Number of bulbs expected to fail in the first 800 burning hours is approximately equal to 46.