Simple
like when you buy something at the store. the person is selling products to you
The x² term will be C(5,2)x²·2³ = 80x².
The coefficient is 80.
_____
C(n, k) = n!/(k!·(n-k)!)
C(5, 2) = 5·4/(2·1) = 10
Answer:
The probability of getting two consumers comfortable with drones is 0.3424.
Step-by-step explanation:
The probability that a consumer is comfortable having drones deliver their purchases is, <em>p</em> = 0.43.
A random sample of <em>n</em> = 5 consumers are selected, and exactly <em>x</em> = 2 of them are comfortable with the drones.
To compute the probability of getting two consumers comfortable with drones followed by three consumers not comfortable, we will use the Binomial distribution instead of the multiplication rule to find the probability.
This is because in this case we need to compute the number of possible combinations of two consumers who are comfortable with drones.
So, <em>X</em> = number of consumers comfortable with drones, follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.43.
Compute the probability of getting two consumers comfortable with drones as follows:
![P(X=x)={5\choose x}\ 0.43^{x}\ (1-0.43)^{5-x};\ x=0,1,2,3...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7B5%5Cchoose%20x%7D%5C%200.43%5E%7Bx%7D%5C%20%281-0.43%29%5E%7B5-x%7D%3B%5C%20x%3D0%2C1%2C2%2C3...)
![P(X=2)={5\choose 2}\ 0.43^{2}\ (1-0.43)^{5-2}](https://tex.z-dn.net/?f=P%28X%3D2%29%3D%7B5%5Cchoose%202%7D%5C%200.43%5E%7B2%7D%5C%20%281-0.43%29%5E%7B5-2%7D)
![=10\times 0.1849\times 0.185193\\=0.342421857\\\approx 0.3424](https://tex.z-dn.net/?f=%3D10%5Ctimes%200.1849%5Ctimes%200.185193%5C%5C%3D0.342421857%5C%5C%5Capprox%200.3424)
Thus, the probability of getting two consumers comfortable with drones is 0.3424.
Answer:
No evidence .because the configurations factors and failure mode are independent
Step-by-step explanation:
Determine if there is sufficient evidence to conclude that configuration affects the type of failure
Number of configurations = 3
Number of failures = 4
assuming <em>Pij </em>represents proportion of items in pop i of the category j
H0 : <em>Pij = Pi * </em>Pj<em>, </em>I = 1,2,-- I, j = 1,2,---J
Hence the expected frequencies will be
16.10 , 43.58 , 18, 12.31
7.5, 19.37, 8 , 5.47
10.73, 29.05, 12, 8.21
x^2 = 13.25
test statistic = 14.44. Hence the significance level where Null hypothesis will be acceptable will be = 2.5%