Answer:
a)
So the value of height that separates the bottom 75% of data from the top 25% is 236.121.
b) ![P(X \geq 3) = 1-P(X](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-P%28X%3C3%29%20%3D%201-P%28X%20%5Cleq%202%29%20%3D%201-%5BP%28X%3D0%29%2BP%28X%3D1%29%20%2BP%28X%3D2%29%5D%3D%201-0.5256%3D0.4744)
c) ![P(\bar X \geq 225)=1- P(\bar X](https://tex.z-dn.net/?f=P%28%5Cbar%20X%20%5Cgeq%20225%29%3D1-%20P%28%5Cbar%20X%20%3C225%29%20%3D%201-P%28Z%3C%5Cfrac%7B225-225%7D%7B%5Cfrac%7B16.5%7D%7B%5Csqrt%7B5%7D%7D%7D%29%20%3D%201-P%28Z%3C0%29%20%3D%201-0.5%20%3D%200.5)
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".
2) Part a
Let X the random variable that represent the cuts of a population, and for this case we know the distribution for X is given by:
Where
and
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 0.75 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:
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 75% of data from the top 25% is 236.121.
Part b
For this case we know that the individual probability of select one wheel with a cutting rate higher than the calculated value in part a is 0.25, and we select n =10 so then we can use the binomial distribution for this case:
![X\sim Bin(n=10, p=0.25)](https://tex.z-dn.net/?f=%20X%5Csim%20Bin%28n%3D10%2C%20p%3D0.25%29)
And we want this probability:
![P(X \geq 3) = 1-P(X](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-P%28X%3C3%29%20%3D%201-P%28X%20%5Cleq%202%29%20%3D%201-%5BP%28X%3D0%29%2BP%28X%3D1%29%20%2BP%28X%3D2%29%5D)
We can find the individual probabilities like this:
![P(X \geq 3) = 1-P(X](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-P%28X%3C3%29%20%3D%201-P%28X%20%5Cleq%202%29%20%3D%201-%5BP%28X%3D0%29%2BP%28X%3D1%29%20%2BP%28X%3D2%29%5D%3D%201-0.5256%3D0.4744)
Part c
For this case we know that the distribution for the sample mean is given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
And we want this probability:
![P(\bar X \geq 225)](https://tex.z-dn.net/?f=P%28%5Cbar%20X%20%5Cgeq%20225%29)
And for this case we can use the complement rule and the z score given by:
![z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=%20z%3D%20%5Cfrac%7B%5Cbar%20X%20-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
And if we replace we got:
![P(\bar X \geq 225)=1- P(\bar X](https://tex.z-dn.net/?f=P%28%5Cbar%20X%20%5Cgeq%20225%29%3D1-%20P%28%5Cbar%20X%20%3C225%29%20%3D%201-P%28Z%3C%5Cfrac%7B225-225%7D%7B%5Cfrac%7B16.5%7D%7B%5Csqrt%7B5%7D%7D%7D%29%20%3D%201-P%28Z%3C0%29%20%3D%201-0.5%20%3D%200.5)