Answer:
x < 25
Step-by-step explanation:
Answer:
24
Step-by-step explanation:
24 cubes. Reasoning varies. Sample reasoning: The length of the box
can fit or 5/4 : 5/12 =3 cubes. The width of the box can fit or 5/3 : 5/12 = 4 cubes. The height of the box can fit 5/6 : 5/12 = 2 cubes. The box can fit
or 24 cubes.
Answer:
a) ![P(X=3) = 0.1](https://tex.z-dn.net/?f=%20P%28X%3D3%29%20%3D%200.1)
b) ![P(X\geq 3) =1-P(X](https://tex.z-dn.net/?f=%20P%28X%5Cgeq%203%29%20%3D1-P%28X%3C3%29%20%3D%201-P%28X%5Cleq%202%29%20%3D%201-%5BP%28X%3D0%29%20%2BP%28X%3D1%29%2BP%28X%3D2%29%5D)
And replacing we got:
![P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-%20%5B0.2%2B0.3%2B0.1%5D%3D%200.4)
c) ![P(X=4) = 0.3](https://tex.z-dn.net/?f=%20P%28X%3D4%29%20%3D%200.3)
d) ![P(X=0) = 0.2](https://tex.z-dn.net/?f=%20P%28X%3D0%29%20%3D%200.2)
e) ![E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2](https://tex.z-dn.net/?f=%20E%28X%29%20%3D0%2A0.2%20%2B1%2A0.3%2B2%2A0.1%20%2B3%2A0.1%20%2B4%2A0.3%3D%202)
f) ![E(X^2)= \sum_{i=1}^n X^2_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%5E2%29%3D%20%5Csum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
And replacing we got:
![E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D0%5E2%2A0.2%20%2B1%5E2%2A0.3%2B2%5E2%2A0.1%20%2B3%5E2%2A0.1%20%2B4%5E2%2A0.3%3D%206.4)
And the variance would be:
![Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4](https://tex.z-dn.net/?f=%20Var%28X0%20%3DE%28X%5E2%29-%20%5BE%28X%29%5D%5E2%20%3D%206.4%20-%282%5E2%29%3D%202.4)
And the deviation:
![\sigma =\sqrt{2.4} = 1.549](https://tex.z-dn.net/?f=%20%5Csigma%20%3D%5Csqrt%7B2.4%7D%20%3D%201.549)
Step-by-step explanation:
We have the following distribution
x 0 1 2 3 4
P(x) 0.2 0.3 0.1 0.1 0.3
Part a
For this case:
![P(X=3) = 0.1](https://tex.z-dn.net/?f=%20P%28X%3D3%29%20%3D%200.1)
Part b
We want this probability:
![P(X\geq 3) =1-P(X](https://tex.z-dn.net/?f=%20P%28X%5Cgeq%203%29%20%3D1-P%28X%3C3%29%20%3D%201-P%28X%5Cleq%202%29%20%3D%201-%5BP%28X%3D0%29%20%2BP%28X%3D1%29%2BP%28X%3D2%29%5D)
And replacing we got:
![P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-%20%5B0.2%2B0.3%2B0.1%5D%3D%200.4)
Part c
For this case we want this probability:
![P(X=4) = 0.3](https://tex.z-dn.net/?f=%20P%28X%3D4%29%20%3D%200.3)
Part d
![P(X=0) = 0.2](https://tex.z-dn.net/?f=%20P%28X%3D0%29%20%3D%200.2)
Part e
We can find the mean with this formula:
![E(X)= \sum_{i=1}^n X_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%29%3D%20%5Csum_%7Bi%3D1%7D%5En%20X_i%20P%28X_i%29)
And replacing we got:
![E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2](https://tex.z-dn.net/?f=%20E%28X%29%20%3D0%2A0.2%20%2B1%2A0.3%2B2%2A0.1%20%2B3%2A0.1%20%2B4%2A0.3%3D%202)
Part f
We can find the second moment with this formula
![E(X^2)= \sum_{i=1}^n X^2_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%5E2%29%3D%20%5Csum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
And replacing we got:
![E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D0%5E2%2A0.2%20%2B1%5E2%2A0.3%2B2%5E2%2A0.1%20%2B3%5E2%2A0.1%20%2B4%5E2%2A0.3%3D%206.4)
And the variance would be:
![Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4](https://tex.z-dn.net/?f=%20Var%28X0%20%3DE%28X%5E2%29-%20%5BE%28X%29%5D%5E2%20%3D%206.4%20-%282%5E2%29%3D%202.4)
And the deviation:
![\sigma =\sqrt{2.4} = 1.549](https://tex.z-dn.net/?f=%20%5Csigma%20%3D%5Csqrt%7B2.4%7D%20%3D%201.549)
Answer:
probability sample
Step-by-step explanation: