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

To find P(X > 2), we can express the probability as:

$$P(X > 2) = 1 - P(X = 1) - P(X = 2).$$

Calculating the individual probabilities:

• For X = 1:
  $p(1) = k(1^2) = \frac{1}{14}(1) = \frac{1}{14}.$
• For X = 2:
  $p(2) = k(2^2) = \frac{1}{14}(4) = \frac{4}{14}.$
• For X = 3:
  $p(3) = k(3^2) = \frac{1}{14}(9) = \frac{9}{14}.$

Now substituting these values:

$$P(X > 2) = 1 - \frac{1}{14} - \frac{4}{14} = 1 - \frac{5}{14} = \frac{9}{14} \approx 0.92857.$$

To calculate the expected value E(X), we use the formula:

$E(X) = \sum_{x=1}^3 x p(x) = 1 \cdot p(1) + 2 \cdot p(2) + 3 \cdot p(3).$ Substituting the probabilities:

$$= 1 \cdot \frac{1}{14} + 2 \cdot \frac{4}{14} + 3 \cdot \frac{9}{14} = \frac{1}{14} + \frac{8}{14} + \frac{27}{14} = \frac{36}{14} = \frac{18}{7} \approx 2.57143.$$

To find Var(1 - X), we embrace the property of variance:

$Var(1 - X) = Var(X).$ We calculate Var(X) using its definition:

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Calculating E(X^2):

$E(X^2) = 1^2 \cdot p(1) + 2^2 \cdot p(2) + 3^2 \cdot p(3) = 1 \cdot \frac{1}{14} + 4 \cdot \frac{4}{14} + 9 \cdot \frac{9}{14}.$

Thus,

$$E(X^2) = \frac{1}{14} + \frac{16}{14} + \frac{81}{14} = \frac{98}{14} = 7.$$

Then,

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