A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance life expectancy," 2013). Assume the life of a dishwasher is normally distributed. Find the number of years that the bottom 25% of dishwasher would last.

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 mean life of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(12,1.25)[/tex]

Where [tex]\mu=12[/tex] and [tex]\sigma=1.25[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.75[/tex] (a)

[tex]P(X<a)=0.25[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.674<\frac{a-12}{1.25}[/tex]

And if we solve for a we got

[tex]a=12 -0.674*1.25=11.157[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 11.157.