The discrete random variable X has the following probability distribution, where p and q are constants - Edexcel - A-Level Maths Statistics - Question 2 - 2016 - Paper 1

The expected value E(X) is calculated as follows:

$E(X) = (-2)p + (-1)q + \frac{1}{2}(0.2) + 2(0.3) + 2p$

Substituting E(X) = 0.4 into the equation yields:

$-2p - q + 0.1 + 0.6 + 2p = 0.4$

Simplifying gives:

$p - q + 0.7 = 0.4$

Which simplifies to:

$p - q = -0.3$

Now we have a system of equations:

1. (2p + q = 0.5)
2. (p - q = -0.3)

From the second equation, we can express q in terms of p:

$q = p + 0.3$

Substituting this into the first equation yields:

$2p + (p + 0.3) = 0.5$

This simplifies to:

$3p + 0.3 = 0.5$

Solving for p gives:

The expected value E(X²) is calculated as follows:

$E(X²) = (-2)²p + (-1)²q + \left(\frac{1}{2}\right)²(0.2) + 2²(0.3) + (2)²p$

Substituting in the known value gives:

$E(X²) = 4p + q + 0.05 + 1.2 + 4p = 2.275$

This results in:

$8p + q + 1.25 = 2.275$

Simplifying gives:

$8p + q = 1.025$

Variance Var(X) can be calculated using:

$Var(X) = E(X²) - (E(X))²$

Substituting the values:

$Var(X) = 2.275 - (0.4)² = 2.275 - 0.16 = 2.115$

For the random variable R defined as $R = \frac{1}{X}$, the expected value is calculated as follows:

$E(R) = E\left(\frac{1}{X}\right) = \sum P(X) \cdot \frac{1}{x}$

Calculating each term gives:

$E(R) = p (-\frac{1}{2}) + q (-1) + 0.2(2) + 0.3(\frac{1}{2}) + p(\frac{1}{2})$

Substituting known values allows you to find E(R) based on current values for p and q.

Let S be Sarah's score and R be Rebecca's score. Sarah wins when S > R. Given the distributions:

When X = -2, S = -2, R = -\frac{1}{2} When X = -1, S = -1, R = -1 When X = 0.5, S = 0.5, R = 2 When X = 2, S = 2, R = \frac{1}{2}

$P(S > R)$ therefore needs to be calculated based on the conditions that apply.

Rebecca wins when $R > S$, which leads to the equivalent calculations for different values of X. Analyze the probability based on similar conditions to find that Rebecca's winning probability may reflect likewise the previously found values for Sarah's.