Hey Ðƴŋâmïč̣, good to see you again.
Here's the solution to your probability mass function problem.
The probability mass function (PMF) for the random variable x is given as:
P(x=0)=31
P(x=1)=81
P(x=2)=21
First, let's check if the probabilities sum to 1:
31+81+21=248+243+2412=248+3+12=2423.
It seems there might be a slight error in the problem statement as the probabilities do not sum to 1. Assuming the problem intends for these to be the given probabilities, we will proceed with them. If the sum was intended to be 1, one of the probabilities would need to be adjusted. For example, if P(x=2) was 2413 instead of 21, the sum would be 1. However, we will use the values as written.
I) Find E[x] and Var[x]
Step 1: Calculate the expected value E[x].
The expected value of a discrete random variable is given by E[x]=∑x⋅P(x).
E[x]=(0⋅P(x=0))+(1⋅P(x=1))+(2⋅P(x=2))
E[x]=(0⋅31)+(1⋅81)+(2⋅21)
E[x]=0+81+1
E[x]=81+88
E[x]=89
Step 2: Calculate E[x2].
The expected value of x2 is given by E[x2]=∑x2⋅P(x).
E[x2]=(02⋅P(x=0))+(12⋅P(x=1))+(22⋅P(x=2))
E[x2]=(0⋅31)+(1⋅81)+(4⋅21)
E[x2]=0+81+2
E[x2]=81+816
E[x2]=817
Step 3: Calculate the variance Var[x].
The variance is given by Var[x]=E[x2]−(E[x])2.
Var[x]=817−(89)2
Var[x]=817−6481
To subtract, find a common denominator, which is 64.
Var[x]=8⋅817⋅8−6481
Var[x]=64136−6481
Var[x]=64136−81
Var[x]=6455
So, for part I:
E[x] = \frac{9{8}}
Var[x] = \frac{55{64}}
II) Find Var[5x−2]
Step 4: Use the property of variance Var[ax+b]=a2Var[x].
Here, a=5 and b=−2.
Var[5x−2]=52⋅Var[x]
Var[5x−2]=25⋅6455
Var[5x−2]=6425⋅55
Var[5x−2]=641375
So, for part II:
Var[5x - 2] = \frac{1375{64}}
III) Find Var[7x+47]
Step 5: Use the property of variance Var[ax+b]=a2Var[x].
Here, a=7 and b=47.
Var[7x+47]=72⋅Var[x]
Var[7x+47]=49⋅6455
Var[7x+47]=6449⋅55
Var[7x+47]=642695
So, for part III:
Var[7x + 47] = \frac{2695{64}}
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