Respuesta :
Answer:
a) [tex]a=225 +0.674*16.5=236.121[/tex]
So the value of height that separates the bottom 75% of data from the top 25% is 236.121.
b) [tex] P(X \geq 3) = 1-P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]= 1-0.5256=0.4744[/tex]
c) [tex]P(\bar X \geq 225)=1- P(\bar X <225) = 1-P(Z<\frac{225-225}{\frac{16.5}{\sqrt{5}}}) = 1-P(Z<0) = 1-0.5 = 0.5[/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".
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:
[tex]X \sim N(225,16.5)[/tex]
Where [tex]\mu=225[/tex] and [tex]\sigma=16.5[/tex]
For this part 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 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:
[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-225}{16.5}[/tex]
And if we solve for a we got
[tex]a=225 +0.674*16.5=236.121[/tex]
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:
[tex] X\sim Bin(n=10, p=0.25)[/tex]
And we want this probability:
[tex] P(X \geq 3) = 1-P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
We can find the individual probabilities like this:
[tex]P(X=0)=(10C0)(0.25)^0 (1-0.25)^{10-0}=0.0563[/tex]
[tex]P(X=1)=(10C1)(0.25)^1 (1-0.25)^{10-1}=0.1877[/tex]
[tex]P(X=2)=(10C2)(0.25)^2 (1-0.25)^{10-2}=0.2816[/tex]
[tex] P(X \geq 3) = 1-P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]= 1-0.5256=0.4744[/tex]
Part c
For this case we know that the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we want this probability:
[tex]P(\bar X \geq 225)[/tex]
And for this case we can use the complement rule and the z score given by:
[tex] z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace we got:
[tex]P(\bar X \geq 225)=1- P(\bar X <225) = 1-P(Z<\frac{225-225}{\frac{16.5}{\sqrt{5}}}) = 1-P(Z<0) = 1-0.5 = 0.5[/tex]
