Respuesta :
Answer:
[tex] IQR= 6.7392-5.6608=1.0784[/tex]
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:
[tex]X \sim N(6.2,0.8)[/tex]
Where [tex]\mu=6.2[/tex] and [tex]\sigma=0.8[/tex]
For this case we want to find the interquartile range. From definition the IQR is a measure of dispersion defined as:
[tex] IQR =Q_3 -Q_1[/tex]
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:
[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 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-6.2}{0.8}[/tex]
And if we solve for a we got
[tex]a=6.2 -0.674*0.8=5.6608[/tex]
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:
[tex]P(X>a)=0.25[/tex] (a)
[tex]P(X<a)=0.75[/tex] (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
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.75[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.75[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.674<\frac{a-6.2}{0.8}[/tex]
And if we solve for a we got
[tex]a=6.2 +0.674*0.8=6.7392[/tex]
So the value of height that separates the bottom 75% of data from the top 25% is 6.7392.
And then the [tex] IQR= 6.7392-5.6608=1.0784[/tex]
