That is true. Have a great day
Answer:
A:30
B: -10
C: -4
Step-by-step explanation:
Answer:
Step-by-step explanation:
let us suppose that each call out of the 20 is indepent from each other. We have that the probability of having a fax message is 0.25. Let X the number of calls among the 20 that involve a fax message.
Then, the random variable X is distributed as a binomial random variable.
Recall the following for binomial random variables
and that the standar deviation is the square root of the variance. Then,
a) ![E[X] = np = 20\cdot0.25 = 5](https://tex.z-dn.net/?f=E%5BX%5D%20%3D%20np%20%3D%2020%5Ccdot0.25%20%3D%205)
b) ![\sqrt[]{np(1-p)} = \sqrt[]{20\cdot 0.25 \cdot 0.75} = 1.936](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%5B%5D%7B20%5Ccdot%200.25%20%5Ccdot%200.75%7D%20%3D%201.936)
c) We want
![P(X-E[X]>1.936*2) = P(X>8.87) = P(X\geq 9) = 1-P(X](https://tex.z-dn.net/?f=P%28X-E%5BX%5D%3E1.936%2A2%29%20%3D%20P%28X%3E8.87%29%20%3D%20P%28X%5Cgeq%209%29%20%3D%201-P%28X%3C9%29%20%3D%201-%20P%28X%5Cleq%208%29%20)
We have that
![P(X\leq 8 ) = \sum_{k=0}^8 \binom{20}{k} 0.25^k 0.75^{20-k} =0.959](https://tex.z-dn.net/?f=P%28X%5Cleq%208%20%29%20%3D%20%5Csum_%7Bk%3D0%7D%5E8%20%5Cbinom%7B20%7D%7Bk%7D%200.25%5Ek%200.75%5E%7B20-k%7D%20%3D0.959%20)
Then, the desired probabilty is 0.041.