Question

Please make sure to display your thought process? It is imperative to be able to follow how the answer was deduced. Please be as thorough as possible. Please address all parts of this question.

The effective life of a component in an aircraft engine is modeled as a random variable with mean value 5000 hours and standard deviation 40 hours. The engine manufacturer introduced an improvement into the manufacturing process for this component that increases the mean life to 5050 hours and decreases the standard deviation to 30 hours. A random sample of n₁ = 64 components is selected from the previously used process, and a random sample of n₂ = 100 components is selected from the improved process. We assume that the previously produced components and the improved components form independent populations, and that the distribution of effective lifetime is close to a normal distribution in both cases.

A) What is the probability that the difference between the sample means ?̅₂ − ?̅₁ is at least 25 hours?

B) What is the probability that the difference between the sample means ?̅₂ − ?̅₁ is at least 50 hours?

Answer 1

Let x₁ be the effective lifetime of a previously produced component and  x₂ be the effective lifetime of a improved produced component

x₁ has mean ( µ₁ ) = 5000 and standard deviation (σ₁) = 40

x₂ has mean ( µ₂ ) = 5050 and standard deviation (σ₂) = 30

n₁ = 64 and n₂ = 100

So distribution of  ₂ - ₁ follows approximately normal distribution with mean µ = µ₂ - µ₁ = 5050 - 5000 = 50 and  

Standard deviation σ = = = 5.8310

A) We are asked to find P( ₂ - ₁ ≥ 25 )

=

= P( z ≥ -4.29 )

= 1 - P( z < -4.29 )

= 1 - 0 --- ( table value for -4.29 is 0 from z score table )

= 1

Probability that the difference between the sample means ?̅₂ − ?̅₁ is at least 25 hours is 1

B)

We are asked to find P( ₂ - ₁ ≥ 50 )

=

= P( z ≥ 0 )

= 1 - P( z < 0 )

= 1 - 0.5 --- ( table value for 0 is 0.5 from z score table )

= 0.5

Probability that the difference between the sample means ?̅₂ − ?̅₁ is at least 50 hours is 0.5