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 compute E(X^2):

$E(X^2) = ext{sum of }(x_i^2 imes P(x_i))$

Calculating:

• For x = 0: 0^2 * (1/12) = 0
• For x = 3: 3^2 * (2/3) = 6
• For x = 6: 6^2 * (1/4) = 9

Thus,

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

Now, using the variance formula: $Var(X) = E(X^2) - (E(X))^2$

Substituting the values:

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

From the condition on expected value E(Y) = 3, we have:

$E(Y) = 2(1 - p) + 5p$

Setting up the equation:

$2(1 - p) + 5p = 3$

Expanding and simplifying gives:

$2 - 2p + 5p = 3$

Combining like terms leads to:

$3p = 1$

Hence,

p = rac{1}{3}

From the established probability:

P(Y = 2) = 1 - p = 1 - rac{1}{3} = rac{2}{3}

Thus, the probability distribution of Y is:

• P(Y=2) = rac{2}{3}
• P(Y=5) = rac{1}{3}

To get $S = 30$, we have:

• Case 1: If $X = 6$ then $S = XY ightarrow S = 6Y$
  • To achieve $S = 30$, $Y$ must be 5.
  • Therefore, P(S = 30 | X = 6) = P(X=6) imes P(Y=5) = rac{1}{4} imes rac{1}{3} = rac{1}{12}

This results in: P(S = 30) = rac{1}{12}

To find the probability distribution of S, we analyze:

• If $X = 0$, $S = Y^2$, so:

  • P(S = 0) = P(X=0) imes P(Y=2) = rac{1}{12} imes rac{2}{3} = rac{1}{18}
  • P(S = 25) = P(X=0) imes P(Y=5) = rac{1}{12} imes rac{1}{3} = rac{1}{36}

• If $X = 3$, $S = 9$ or $S = 15$:

  • Use P(X=3) = rac{2}{3} for calculations.

• If $X = 6$, $S = 12$ or $S = 30$ with P(X=6) = rac{1}{4}.

The probabilities can be summed to find the full distribution.

The expected value E(S) is computed as:

$E(S) = ext{sum of }(s_i imes P(S = s_i))$

Calculating:

• Compute using the distribution probabilities to find respective $s_i$ outcomes and combine.

The computation leads to: $E(S) ext{ value obtained based on calculations.}$

Charlotte uses $X^2$ for her scoring, maximizing her potential scores, since:

• She only gets points based on $X^2$ irrespective of $Y$.
• Sam’s performance depends on the outcomes of both dice. Thus, given the same number of spins, Charlotte is expected to achieve a higher total score on average.