Question

A manufacturer of light bulbs says that the life of a light bulb is a normally distriubted random variable with mean of 2000 hours and a standard deviation of 110 hours.

a. What beyond what value will only 20% of the bulbs last?

b. Find the first quartile

c. The company wishes to guarantee the light bulb for full replacement so that on 8% of them will be replaced. What value is this?

Answer 1

a)

X ~ N ( µ = 2000 , σ = 110 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.2 = 0.8
To find the value of x
Looking for the probability 0.8 in standard normal table to calculate critical value Z = 0.8416
Z = ( X - µ ) / σ
0.8416 = ( X - 2000 ) / 110
X = 2092.58

b)

First quartile = P(X < x) = 0.25

X ~ N ( µ = 2000 , σ = 110 )
P ( X < x ) = 25% = 0.25
To find the value of x
Looking for the probability 0.25 in standard normal table to calculate critical value Z = -0.6745
Z = ( X - µ ) / σ
-0.6745 = ( X - 2000 ) / 110
X = 1925.81

c)

X ~ N ( µ = 2000 , σ = 110 )
P ( X < x ) = 8% = 0.08
To find the value of x
Looking for the probability 0.08 in standard normal table to calculate critical value Z = -1.4051
Z = ( X - µ ) / σ
-1.4051 = ( X - 2000 ) / 110
X = 1845.44