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The probability function of a discrete random variable X is given by p(x) = kx^2 where k is a positive constant - Edexcel - A-Level Maths Statistics - Question 5 - 2010 - Paper 1
Question 5
The probability function of a discrete random variable X is given by p(x) = kx^2 where k is a positive constant. (a) Show that k = \frac{1}{14} Find (b) ... show full transcript
Worked Solution & Example Answer:The probability function of a discrete random variable X is given by p(x) = kx^2 where k is a positive constant - Edexcel - A-Level Maths Statistics - Question 5 - 2010 - Paper 1
Step 1
Show that k = \frac{1}{14}
Answer
To show that ( k = \frac{1}{14} ), we need to ensure that the total probability sums to 1. We calculate the probabilities for the values of X:
[ \begin{align*} P(X=1) & = k(1^2) = k, \ P(X=2) & = k(2^2) = 4k, \ P(X=3) & = k(3^2) = 9k. \end{align*} ]
The total probability is: [ P(X=1) + P(X=2) + P(X=3) = k + 4k + 9k = 14k. ]
Setting this equal to 1: [ 14k = 1 \Rightarrow k = \frac{1}{14}. ]
Step 2
Step 3
E(X)
Answer
The expected value E(X) is calculated as:
[ E(X) = \sum_{x=1}^{3} x \cdot P(X=x) = 1 \cdot k + 2 \cdot 4k + 3 \cdot 9k. ]
Substituting k: [ E(X) = 1 \cdot \frac{1}{14} + 2 \cdot 4 \times \frac{1}{14} + 3 \cdot 9 \times \frac{1}{14} = \frac{1}{14} + \frac{8}{14} + \frac{27}{14} = \frac{36}{14} = 2.57. ]
Step 4
Var(1 - X)
Answer
To find the variance of the random variable (1 - X):
We first need E(X^2):
[ E(X^2) = \sum_{x=1}^{3} x^2 \cdot P(X=x) = 1^2 \cdot k + 2^2 \cdot 4k + 3^2 \cdot 9k. ]
Now substituting: [ E(X^2) = 1 \cdot \frac{1}{14} + 4 \cdot 4 \times \frac{1}{14} + 9 \cdot 9 \times \frac{1}{14} = \frac{1}{14} + \frac{16}{14} + \frac{81}{14} = \frac{98}{14} = 7. ]
Variance is given by: [ Var(1 - X) = Var(X) = E(X^2) - (E(X))^2 = 7 - (2.57)^2 = 7 - 6.6049 \approx 0.3951. ]