Question

The average life of a certain type of small motor is 8 years with a standard deviation of 2 years. The manufacturer replaces free all motors that fail while under warranty. If she is willing to provide a 5-year and the distribution of the motor lives is normal, calculate the percentage of motors that will have to be replaced. Use 3 decimal places and enter your answer as a decimal value

Answer 1

Answer:

$P(X<5)=P(\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(Z<\frac{5-8}{2})=P(z<-1.5)$

And we can find this probability using the normal standard distribution or Excel:

$P(z<-1.5)=0.067$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the average life of a population, and for this case we know the distribution for X is given by:

$X \sim N(8,2)$  

Where $\mu=8$ and $\sigma=2$

We are interested on this probability

$P(X<5)$

Since all the motors with an average life of lesss than 5 years satisfy the warranty period

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X<5)=P(\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(Z<\frac{5-8}{2})=P(z<-1.5)$

And we can find this probability using the normal standard distribution or Excel:

$P(z<-1.5)=0.067$