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

Answer:

$IQR= 6.7392-5.6608=1.0784$

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 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)

$P(X$   (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:

$P(X$  

$P(z$

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

$a=6.2 -0.674*0.8=5.6608$

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)

$P(X$   (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

$P(X$  

$P(z$

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

$a=6.2 +0.674*0.8=6.7392$

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

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