A manufacturer of personal computers tests competing brands and finds that the amount of energy they require is normally distributed with a mean of 285 kwh and a standard deviation of 9.1 kwh. If the lowest 25% and the highest 30% are not included in a second round of tests, what are the upper and lower limits for the energy amounts of the remaining computers?

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

[tex]X \sim N(285,9.1)[/tex]

Where [tex]\mu=285[/tex] and [tex]\sigma=9.1[/tex]

Lowest 25%

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-285}{9.1}[/tex]

And if we solve for a we got

[tex]a=285 -0.674*9.1=278.87[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 278.87.

Highest 30%

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

[tex]P(X>a)=0.30[/tex] (a)

[tex]P(X<a)=0.70[/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 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524. On this case P(Z<0.524)=0.7 and P(z>0.524)=0.3

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

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.7[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.7[/tex]

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

[tex]z=0.524<\frac{a-285}{9.1}[/tex]

And if we solve for a we got

[tex]a=285 +0.524*9.1=289.77[/tex]

So the value of height that separates the bottom 70% of data from the top 30% is 289.77.