the whole statement is true, make sure you read all of your choices next time and list them below.
Answer:
P = 0.0215 = 2.15%
Step-by-step explanation:
First we need to convert the values of 900 and 975 to standard scores using the equation:

Where z is the standard value, x is the original value,
is the mean and
is the standard deviation. So we have that:
standard value of 900: 
standard value of 975: 
Now, we just need to look at the standard distribution table (z-table) for the values of z = 2 and z = 3:
z = 2 -> p_2 = 0.9772
z = 3 -> p_3 = 0.9987
We want the interval between 900 and 975 hours, so we need the interval between z = 2 and z = 3, so we just need to subtract their p-values:
P = p_3 - p_2 = 0.9987 - 0.9772 = 0.0215
So the probability is 0.0215 = 2.15%
Answer: E(Y) = 1.6 and Var(Y)=1.12
Step-by-step explanation:
Since we have given that
X 0 1 2
P(X) 0.4 0.4 0.2
Here, number of games = 2
So, 
Since
are independent variables.
so, ![E[Y]=2E[X]\\\\Var[Y]=2Var[X]](https://tex.z-dn.net/?f=E%5BY%5D%3D2E%5BX%5D%5C%5C%5C%5CVar%5BY%5D%3D2Var%5BX%5D)
So, we get that
![E(X)=0.4\times 0+0.4\times 1+0.2\times 2=0.8\\\\and Var[x]=E[x^2]-(E[x])^2\\\\E[x^2]=0\times 0.4+1\times 0.4+4\times 0.2=1.2\\\\So, Var[x]=1.2-(0.8)^2\\\\Var[x]=1.2-0.64=0.56](https://tex.z-dn.net/?f=E%28X%29%3D0.4%5Ctimes%200%2B0.4%5Ctimes%201%2B0.2%5Ctimes%202%3D0.8%5C%5C%5C%5Cand%20Var%5Bx%5D%3DE%5Bx%5E2%5D-%28E%5Bx%5D%29%5E2%5C%5C%5C%5CE%5Bx%5E2%5D%3D0%5Ctimes%200.4%2B1%5Ctimes%200.4%2B4%5Ctimes%200.2%3D1.2%5C%5C%5C%5CSo%2C%20Var%5Bx%5D%3D1.2-%280.8%29%5E2%5C%5C%5C%5CVar%5Bx%5D%3D1.2-0.64%3D0.56)
So, E[y]=2×0.8=1.6
and Var[y]=2×0.56=1.12
Hence, E(Y) = 1.6 and Var(Y)=1.12
The probability is 7/20.
There are 20 outcomes in the sample space:
2(1) 2(3) 2(5) 2(7) 2(9)
3(1) 3(3) 3(5) 3(7) 3(9)
4(1) 4(3) 4(5) 4(7) 4(9)
5(1) 5(3) 5(5) 5(7) 5(9)
Out of these, 14 are two digit products. Out of those 14, only 7 are odd. Thus the probability is 7/20.