Yes and thank you for the points
Answer:
a)
b) ![P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452](https://tex.z-dn.net/?f=P%284%5Cleq%20X%5Cleq%208%29%3D0.1954%2B0.1563%2B0.1042%2B0.0595%2B0.0298%3D0.5452)
c) ![P(X \geq 8) = 1-P(X](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%201-P%28X%3C8%29%20%3D%201-P%28X%5Cleq%207%29%3D1-%5B0.0183%2B0.0733%2B%200.1465%2B0.1954%2B0.1954%2B0.1563%2B%200.1042%2B0.0595%5D%3D0.0511)
d) ![P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559](https://tex.z-dn.net/?f=P%284%5Cleq%20X%20%5Cleq%206%29%3D0.1954%2B0.1563%2B0.1042%3D0.4559)
Step-by-step explanation:
Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that
The probability mass function for the random variable is given by:
And f(x)=0 for other case.
For this distribution the expected value is the same parameter
,
, ![Sd(X)=2](https://tex.z-dn.net/?f=Sd%28X%29%3D2)
a. Compute both P(X≤4) and P(X<4).
Using the pmf we can find the individual probabilities like this:
![P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E1%7D%7B1%21%7D%3D0.0733)
![P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465](https://tex.z-dn.net/?f=P%28X%3D2%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E2%7D%7B2%21%7D%3D0.1465)
![P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D3%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E3%7D%7B3%21%7D%3D0.1954)
![P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D4%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E4%7D%7B4%21%7D%3D0.1954)
b. Compute P(4≤X≤ 8).
![P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)](https://tex.z-dn.net/?f=P%284%5Cleq%20X%5Cleq%208%29%3DP%28X%3D4%29%2BP%28X%3D5%29%2B%20P%28X%3D6%29%2BP%28X%3D7%29%2BP%28X%3D8%29)
![P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D4%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E4%7D%7B4%21%7D%3D0.1954)
![P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563](https://tex.z-dn.net/?f=P%28X%3D5%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E5%7D%7B5%21%7D%3D0.1563)
![P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042](https://tex.z-dn.net/?f=P%28X%3D6%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E6%7D%7B6%21%7D%3D0.1042)
![P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595](https://tex.z-dn.net/?f=P%28X%3D7%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E7%7D%7B7%21%7D%3D0.0595)
![P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298](https://tex.z-dn.net/?f=P%28X%3D8%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E8%7D%7B8%21%7D%3D0.0298)
![P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452](https://tex.z-dn.net/?f=P%284%5Cleq%20X%5Cleq%208%29%3D0.1954%2B0.1563%2B%200.1042%2B0.0595%2B0.0298%3D0.5452)
c. Compute P(8≤ X).
![P(X \geq 8) = 1-P(X](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%201-P%28X%3C8%29%20%3D%201-P%28X%5Cleq%207%29%3D1-%5BP%28X%3D0%29%2BP%28X%3D1%29%2B%20P%28X%3D2%29%2BP%28X%3D3%29%2BP%28X%3D4%29%2BP%28X%3D5%29%2B%20P%28X%3D6%29%2BP%28X%3D7%29%5D)
![P(X \geq 8) = 1-P(X](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%201-P%28X%3C8%29%20%3D%201-P%28X%5Cleq%207%29%3D1-%5B0.0183%2B0.0733%2B%200.1465%2B0.1954%2B0.1954%2B0.1563%2B%200.1042%2B0.0595%5D%3D0.0511)
d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?
The mean is 4 and the deviation is 2, so we want this probability
![P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)](https://tex.z-dn.net/?f=P%284%5Cleq%20X%20%5Cleq%206%29%3DP%28X%3D4%29%2BP%28X%3D5%29%2BP%28X%3D6%29)
![P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954](https://tex.z-dn.net/?f=P%28X%3D4%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E4%7D%7B4%21%7D%3D0.1954)
![P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563](https://tex.z-dn.net/?f=P%28X%3D5%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E5%7D%7B5%21%7D%3D0.1563)
![P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042](https://tex.z-dn.net/?f=P%28X%3D6%29%3D%5Cfrac%7Be%5E%7B-4%7D%204%5E6%7D%7B6%21%7D%3D0.1042)
![P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559](https://tex.z-dn.net/?f=P%284%5Cleq%20X%20%5Cleq%206%29%3D0.1954%2B0.1563%2B0.1042%3D0.4559)
we have
![V(x)=x^3-12x^2+48x-64](https://tex.z-dn.net/?f=V%28x%29%3Dx%5E3-12x%5E2%2B48x-64)
we know that
The volume of a cube is equal to
![V=b^{3}](https://tex.z-dn.net/?f=V%3Db%5E%7B3%7D)
where
b is the length side of a cube
Using a graph tool------> find the roots of the polynomial
see the attached figure
x=4
so
![x^3-12x^2+48x-64=(x-4)^{3}](https://tex.z-dn.net/?f=x%5E3-12x%5E2%2B48x-64%3D%28x-4%29%5E%7B3%7D)
therefore
<u>The answer is</u>
the length side of a cube is equal to (x-4)
This problem is asking you to find the common multiple
So if you make a list of all the multiples of each number (9*1 9*2 9*3)
The number that each has in common is 36
The answer is 36