Question

The weights of ripened peaches grown in the southeastern United States follow an approximately normal distribution with a mean of 6.2 ounces and a standard deviation of 0.8 ounces. Find the interquartile range (IQR) of the weights of these peaches

Answer 1

`IQR= 6.7392-5.6608=1.0784`

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 weights of ripened peaches grown in the southeastern United States of a population, and for this case we know the distribution for X is given by:

`X \sim N(6.2,0.8)`

Where
`\mu=6.2` and
`\sigma=0.8`

For this case we want to find the interquartile range. From definition the IQR is a measure of dispersion defined as:

`IQR =Q_3 -Q_1`

Where Q3 represent the 3 quartile and Q1 the first quartile. We need to find these values.

For this Q1 we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.75` (a)

`P(X<a)=0.25` (b)

Both conditions are equivalent on this case. We can use the z score 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<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.25`

`P(z<(a-\mu)/(\sigma))=0.25`

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

`z=-0.674<(a-6.2)/(0.8)`

And if we solve for a we got

`a=6.2 -0.674*0.8=5.6608`

So the value of height that separates the bottom 25% of data from the top 75% is 5.6608.

For this Q3 we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.25` (a)

`P(X<a)=0.75` (b)

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

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

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.75`

`P(z<(a-\mu)/(\sigma))=0.75`

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

`z=0.674<(a-6.2)/(0.8)`

And if we solve for a we got

`a=6.2 +0.674*0.8=6.7392`

So the value of height that separates the bottom 75% of data from the top 25% is 6.7392.

And then the
`IQR= 6.7392-5.6608=1.0784`