The random variable X has probability distribution given in the table below - Edexcel - A-Level Maths Statistics - Question 3 - 2008 - Paper 2
Question 3
The random variable X has probability distribution given in the table below.
| x | -1 | 0 | 1 | 2 | 3 |
|-----|------|-----|-----|-----|-----|
| P(X=x) ... show full transcript
Worked Solution & Example Answer:The random variable X has probability distribution given in the table below - Edexcel - A-Level Maths Statistics - Question 3 - 2008 - Paper 2
Step 1
the value of p and the value of q.
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Answer
To find the values of p and q, we use two main properties of probabilities: the sum of all probabilities in the distribution must equal 1, and the expected value E(X) is given by:
Sum of Probabilities:p+q+0.2+0.15+0.15=1
Simplifying, we have:
p+q+0.5=1
Thus,
p+q=0.5(1)
Expected Value Calculation:
Given that E(X) = 0.55:
E(X)=(−1)p+0(q)+1(0.2)+2(0.15)+3(0.15)
This simplifies to:
E(X)=−1p+0+0.2+0.3+0.45
Therefore:
−1p+0.95=0.55
Rearranging gives:
−1p=0.55−0.95(2)−1p=−0.4⇒p=0.4
Substituting p into equation (1):
Since we know p = 0.4:
0.4+q=0.5⇒q=0.1
Final Values:
Therefore, the final values are:
p=0.4andq=0.1
Step 2
Var(X).
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Answer
To find the variance Var(X), we use the formula:
Var(X)=E(X2)−(E(X))2
Finding E(X^2):
We calculate E(X^2) as follows:
E(X2)=(−1)2p+02q+12(0.2)+22(0.15)+32(0.15)
This leads to:
E(X2)=1(0.4)+0+0.2+0.6+1.35
Summing these values gives:
E(X2)=1+0.2+0.6+1.35=3.15
Substituting into the Variance Formula:
We now have:
Var(X)=E(X2)−(E(X))2
We calculated E(X) = 0.55, so:
Var(X)=3.15−(0.55)2
Calculating (E(X))2:
(0.55)2=0.3025
Thus:
Var(X)=3.15−0.3025=2.8475
Rounding to four significant figures, we get:
Var(X)extisapproximately2.85
Step 3
E(2X - 4).
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Answer
To compute E(2X - 4), we use the linearity of expectation:
E(2X−4)=2E(X)−4
Substituting the known value:
Using E(X) = 0.55, we find:
E(2X−4)=2(0.55)−4
This calculates to:
=1.1−4=−2.9
Therefore, the expected value E(2X - 4) is:
E(2X−4)=−2.9
