Question

A manufacturer of personal computers sets tests competing brands and finds that the amounts of energy they require are 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 sets?a. [269.76 300.24]b. [278.90, 288.55]c. [280.22 289.78]d. [280.22 300.24]

Answer 1

Answer:

b. [278.90, 288.55]

See the explanation below.

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

$X \sim N(285,9.1)$  

Where $\mu=285$ and $\sigma=9.1$

LOWER LIMIT

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

$P(X>a)=0.75$   (a)

$P(X$   (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:

$P(X$  

$P(z$

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

$z=-0.674$

And if we solve for a we got

$a=285 -0.674*9.1=278.90$

So the value of height that separates the bottom 25% of data from the top 75% is 278.90.  

UPPER LIMIT

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

$P(X>a)=0.3$   (a)

$P(X$   (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:

$P(X$  

$P(z$

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

$z=0.524$

And if we solve for a we got

$a=285 +0.524*9.1=289.78$

So the value of height that separates the bottom 70% of data from the top 30% is 289.78.  

With the procedure we got for the limits [278.9 , 289.78]

And since any of the options are on the list we take th most similar for this case:

b. [278.90, 288.55]