Respuesta :
Answer:
[tex]a=2.25 -0.739*0.15=2.139[/tex]
So the value of height that separates the bottom 23% of data from the top 77% is 2.139.
So we can conclude that 77% of the apples will contain at least 2.139 ounces
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 heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.25,0.15)[/tex]
Where [tex]\mu=2.25[/tex] and [tex]\sigma=0.15[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.77[/tex] (a)
[tex]P(X<a)=0.23[/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 0.23 of the area on the left and 0.77 of the area on the right it's z=-0.739. On this case P(Z<-0.739)=0.23 and P(z>-0.739)=0.7 7
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.23[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.23[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.739<\frac{a-2.25}{0.15}[/tex]
And if we solve for a we got
[tex]a=2.25 -0.739*0.15=2.139[/tex]
So the value of height that separates the bottom 23% of data from the top 77% is 2.139.
So we can conclude that 77% of the apples will contain at least 2.139 ounces
