Question

Q.2. A type of 100-watt light bulb has been found to have a mean life of 2000 hours and a standard deviation of 250 hours. What is the probability that a sample of 81 bulbs will have an average life (i) fewer than 1920 hours, (ii) more than 2010, (iii) between 2050 and 1950?

solve the above problem step by step in proper format

Answer 1

Solutions:
Given that,
n=81
μ =2000 hours , σ= 250 hours
By using the definition of central limit theorem,
For a random sample of size n , sample mean is equal to the population mean
i.e. μx̄ = μ and
sample standard deviation can be calculated as
σx̄=σ/(√n)
Therefore,
μx̄ = μ= 2000
And
σx̄=σ/(√n)
= 250/(√81)
= 27.7778
i)
P(x̄<1920)= P{[ (x̄-μx̄ )/σx̄]<[(1920 - 2000)/27.7778]}
= P (Z < -2.88) ( From. Standard Normal table)
= 0.0020

ii)
P(x̄>2010)= 1-P(x̄<=2010)
=1 - P{[ (x̄-μx̄ ) /σx̄ ]<=[(2010 - 2000)/27.7778]}
=1 - P (Z <= 0.36) ( From. Standard Normal table)
= 1- 0.6406
= 0.3594
iii)
P(1950<x̄< 2050)= p{[(1950- 2000)/27.7778]<[(x̄- μx̄)/σx̄]<[(2050- 2000)/27.7778]}
=P(-1.80<Z<1.80)
= p(Z<1.80) - p(Z< -1.80)
= 0.9641 - 0.0359 ( From. Standard Normal table)
=0.9282