Mid-points 1.7, 1.9, 2.1, 2.3, 2.5

$\bar{x} = \frac{206.2}{100} = 2.062 \text{ (kg)}$ | M1 A1 | With no working shown allow only correct answers (to 3 sf or better). Use of mid-points seen or implied. c.a.o.

$s = \sqrt{\frac{431.16}{100} - 2.062^2}$ | M1 B1 | Use of correct formula for standard deviation; may be implied. Correct $\Sigma fx^2$ s.o.i.

$\therefore s = \sqrt{0.059756} = 0.244(45...) \text{ (kg)}$ | A1 | c.a.o. Allow unbiased estimator (0.24568…) for full marks. 2.06 used for sd (gives 0.2607… or unbiased 0.2620…) gets max M1 B1 A0.

## Question 2(i):

$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.3$ | M1 A1 | Probability rule applied, s.o.i. c.a.o. Accept solutions based on Venn diagrams.

## Question 2(ii):

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.6} = 0.5$ | M1 A1 | Conditional probability rule applied, s.o.i. ft (i) provided both $P(A \cap B)$ and $P(B)$ lie between 0 and 1.

## Question 2(iii):

$A$ and $B$ are independent since $P(B|A) = P(B) = 0.5$ | B1 | ft (ii). Must be supported by explicit numerical evidence. Accept alternatives, e.g. $P(A \cap B) = P(A) \times P(B)$, with evidence.

## Question 3(i):

$p = 1 - (0.4 + 0.3 + 0.1) = 0.2$ | B1 | 

## Question 3(ii):

$(1 \times 0.4) + (2 \times 0.3) + (n \times 0.2) + (7 \times 0.1) = 2.5$

$\therefore 0.2n + 1.7 = 2.5$

$\therefore 0.2n = 0.8$

$\therefore n = 4$ | M1 A1 | Use of formula for $E(X)$ s.o.i. to set up an equation in $n$. c.a.o.

## Question 3(iii):

$E(X^2) = (1^2 \times 0.4) + (2^2 \times 0.3) + (4^2 \times 0.2) + (7^2 \times 0.1) = 9.7$

$\text{Var}(X) = 9.7 - 2.5^2 = 3.45$ | B1 M1 A1 | Correct expression for $E(X^2)$ s.o.i. ft c's $n$. Use of formula for Var$(X)$ s.o.i. c.a.o.

## Question 4(i):

$E(X) = 20 \times 0.4 = 8$ | B1 |

## Question 4(ii):

State or imply Bin$(20, 0.4)$

$P(X = 8) = 0.5956 - 0.4159 = 0.1797$ | B1 M1 A1 | May be awarded elsewhere if not here. Use of tables for $P(X \leq 8) - P(X \leq 7)$ or formula for $P(X = 8)$. c.a.o

## Question 4(iii):

$P(X \geq 8) = 1 - 0.4159 = 0.5841$ | M1 A1 | Attempt $1 - P(X \leq 7)$ c.a.o

## Question 5(i):

Recognise combination problem.

$^{15}C_2 = \frac{15!}{11!4!} = 1365$ | M1 A1 | c.a.o.

## Question 5(ii):

Recognise implication of "no restrictions".

$15^4 = 50625$ | M1 A1 | c.a.o.

## Question 5(iii):

$\frac{15 \times 14 \times 13 \times 12}{15^4} = \frac{32760}{50625} = \frac{728}{1125} = 0.647(11...)$ | M1 M1 A1 | Correct numerator. Correct denominator; ft (ii). c.a.o.

## Question 6(i):

$D \sim N(8.3, 0.20^2)$

$P(8.1 < D < 8.3) = P\left(\frac{8.1 - 8.3}{0.20} < Z < \frac{8.5 - 8.3}{0.20}\right)$ | M1 M1 | Standardising, either term. Relevant difference of 2 terms s.o.i.

$= \Phi(1.0) - \Phi(-1.0) = 0.8413 - (1 - 0.8413) = 0.6826$ | B1 M1 A1 | Correct table look-up: 0.8413 seen. $1 - \ldots$ to deal with negative z value. c.a.o.

## Question 6(ii):

Now $D \sim N(\mu, \sigma^2)$

$P(D < 8.5) = 0.88 \Rightarrow \frac{8.5 - \mu}{\sigma} = 1.175$

$P(D < 8.1) = 0.10 \Rightarrow \frac{8.1 - \mu}{\sigma} = -1.282$ | M1 B1 A1 | Set up at least 1 equation for $\mu$ and $\sigma$. 1.175 and/or (−)1.282 seen. Both equations correct.

$\therefore \mu + 1.175\sigma = 8.5$ and $\mu - 1.282\sigma = 8.1$

$\therefore 2.457\sigma = 0.4$

$\therefore \sigma = 0.1628(0...)$

$\therefore \mu = 8.5 - 1.175 \times 0.1628 = 8.3087$

or $\mu = 8.1 + 1.282 \times 0.1628$ | M1 A1 A1 | Attempt to eliminate either $\mu$ or $\sigma$. One of $\sigma$ or $\mu$ found. c.a.o. The other found. c.a.o. Allow 0.163 used and a.w.r.t. 8.31