The score, X, for a biased spinner is given by the probability distribution | x | 0 | 3 | 6 | |----|-----|-----|-----| | P(X=x) | 1/12 | 2/3 | 1/4 | Find (a) E(X) (b) Var(X) A biased coin has one face labelled 2 and the other face labelled 5 - Edexcel - A-Level Maths: Statistics - Question 6 - 2017 - Paper 1

To find the variance Var(X), we first need to compute E(X^2):

$E(X^2) = \sum_{i} x_i^2 \cdot P(X = x_i)$

Calculating:

$E(X^2) = 0^2 \cdot \frac{1}{12} + 3^2 \cdot \frac{2}{3} + 6^2 \cdot \frac{1}{4}$

Evaluating this:

$E(X^2) = 0 + 6 + 9 = 15$

Now use the variance formula:

$Var(X) = E(X^2) - (E(X))^2$

Thus:

$Var(X) = 15 - (3.5)^2 = 15 - 12.25 = 2.75$

For the biased coin, we have: $P(Y=2) + P(Y=5) = 1$ Substituting in the probabilities, we get: $1 - p + p = 1$ Solving for p yields: $p = \frac{1}{3}$

The probability distribution for Y is:

To find P(S = 30), we need to consider: $P(S = XY = 30)$ Which happens under the scenario when X = 6 and Y = 5:

$P(X=6) \cdot P(Y=5) = \frac{1}{4} \cdot p$ Substituting p = \frac{1}{3} yields: $P(S = 30) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$

The probability distribution for S can be determined by calculating:

• For X=0, S = Y^2 (Y values: 2, 5)
• For X=3, S = XY (X=3, Y=2 or Y=5)
• For X=6, S = XY (X=6, Y=2 or Y=5)

Compile probabilities:

To find the expected value E(S), use:

$E(S) = \sum_{s} s \cdot P(S = s)$

Calculate each term and sum:

After calculating:

$E(S) \approx 11.416$

Charlotte's score is based solely on X^2, while Sam calculates his score based on both X and Y. Since the expectation of X^2 will generally be higher than the product XY, Charlotte should achieve the higher total score due to less variability and a different scoring mechanism.