Question

The average life of a certain type of small motor was estimated to be 10 years with a standard deviation of 2 years. The manufacturer of the motor wants to issue a policy that will replace all motors that fail while under guarantee free of cost. If the company has budget to replace only 3% of all the motors that fail, how long a guarantee (in years) should they offer? You may assume that the lives of the motors are normally distributed.

Please show all the steps leading to the final solution for full credit. For this reason, it may be easier to work out problems by hand on paper. The choice is yours.

A sketch of the normal distribution curve with corresponding x or z score values is required for each part of each problem for full credit. You must shade the required area/ probability that needs to be computed for each problem. You must mention the Z table you are using (Full or Half). You will use Python to compute the required probabilities using the scipy package. Your programs should clearly print out the required solutions

Answer 1

The average life of a motor is estimated to be 10 yrs ( ) with a standard deviation of 2 yrs ( )

The lives of motor follow Normal Distribution i .e , Lives ~ Normal ( , ² )

where , is the mean and ² is the variance.

Let L be a R. V. denoting the life of the motor.

  L ~ N ( 10 , 4 )

Also , we know that if L follows Normal Distribution , then ( L - ) / follows a Standard Normal Distribution.

Z = ( L - 10 ) / 2 ~ N ( 0 , 1 )

Company has a budget to replace 3 % (0.03) of the motors that fail during the guarantee period . So , we want to know how much Guarantee they must offer.

ATQ : P ( Life of the Battery < Guarantee Period ) = 0.03

  P ( L < Guarantee ) = 0.03

P ( { L - 10 } / 2 < { Guarantee - 10 } / 2 ) = 0.03

  P ( Z < { Guarantee - 10 } / 2 ) = 0.03

Also , we know the property of Standard Normal Distribution that : P ( X < x ) = P ( X > - x )

  P ( Z > { 10 - Guarantee } / 2 ) = 0.03

We use the Standard Normal Distribution Tables to find the value at which Z probability is eqaul to 0.03

We could see that : P ( Z > V ) = 0.03 @ V = 1.8808

  V =  { 10 - Guarantee } / 2 = 1.8808

Guarantee = 10 - 1.8808 ( 2 ) = 6.2384 Years

Thus , Company must offer a Guarantee of 6.2384 Years on the motor in order to meet their required condition .