Answer:
i) ![P(9.5 < X](https://tex.z-dn.net/?f=%20P%289.5%20%3C%20X%3C11%29)
And we can solve this problem using the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability with this difference:
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
ii) The z scores for this case are:
![z_1 = \frac{9.5-10}{1.414}= -0.354](https://tex.z-dn.net/?f=%20z_1%20%3D%20%5Cfrac%7B9.5-10%7D%7B1.414%7D%3D%20-0.354)
![z_2 = \frac{11-10}{1.414}= 0.707](https://tex.z-dn.net/?f=%20z_2%20%3D%20%5Cfrac%7B11-10%7D%7B1.414%7D%3D%200.707)
And we can check the answer with the following excel code:
=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)
iii)
(a)
(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 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.
iv) ![z = \frac{7.228-10}{1.414}= -1.96](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B7.228-10%7D%7B1.414%7D%3D%20-1.96)
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 blade length of a population, and for this case we know the distribution for X is given by:
Where
and
Part i
For this case we want this probability:
![P(9.5 < X](https://tex.z-dn.net/?f=%20P%289.5%20%3C%20X%3C11%29)
And we can solve this problem using the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability with this difference:
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
Part ii
The z scores for this case are:
![z_1 = \frac{9.5-10}{1.414}= -0.354](https://tex.z-dn.net/?f=%20z_1%20%3D%20%5Cfrac%7B9.5-10%7D%7B1.414%7D%3D%20-0.354)
![z_2 = \frac{11-10}{1.414}= 0.707](https://tex.z-dn.net/?f=%20z_2%20%3D%20%5Cfrac%7B11-10%7D%7B1.414%7D%3D%200.707)
And we can check the answer with the following excel code:
=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)
Part iii
For this part we want to find a value a, such that we satisfy this condition:
(a)
(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 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.
Part iv
The z score for this value is given by:
![z = \frac{7.228-10}{1.414}= -1.96](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B7.228-10%7D%7B1.414%7D%3D%20-1.96)