## Part (a)
**Answer:** Bell shaped, must have inflexions. 154, 172 on axis. 5% and 30%

**Marks:** B1 B1 B1 | (3)

**Guidance:**
- 2nd B1 for 154 and 172 marked but 154 must be < $\mu$ and 172 > $\mu$. But $\mu$ need not be marked. Allow for $\frac{154 - \mu}{\sigma}$ and $\frac{172 - \mu}{\sigma}$ marked on appropriate sides of the peak
- 3rd B1 the 5% and 30% should be clearly indicated in the correct regions i.e. LH tail and RH tail

## Part (b)
**Answer:** $P(X < 154) = 0.05$

$\frac{154 - \mu}{\sigma} = -1.6449$ or $\frac{\mu - 154}{\sigma} = 1.6449$

$\mu = 154 + 1.6449\sigma$ **given**

**Marks:** M1 B1 A1 cso | (3)

**Guidance:**
- M1 for $\pm \frac{154 - \mu}{\sigma} = z$ value ($z$ must be recognizable e.g. 1.64, 1.65, 1.96 but NOT 0.5199 etc)
- B1 for $\pm$ 1.6449 seen in a line before the final answer
- A1cso for no incorrect statements (in $\mu$, $\sigma$) equating a z value and a probability or incorrect signs. e.g. $\frac{\sigma}{z} = 0.05$ or $\frac{\mu - 154}{\sigma} = 1.6449$ or $P(Z < \frac{\sigma}{z}) = 1.6449$

## Part (c)
**Answer:** $172 - \mu = 0.5244\sigma$ or $\frac{172 - \mu}{\sigma} = 0.5244$ (allow $z = 0.52$ or better here but must be in an equation)

Solving gives $\sigma = 8.2976075$ (awrt 8.30) and $\mu = 167.64873$ (awrt 168)

**Marks:** B1 M1 A1 A1 | (4)

**Guidance:**
- B1 for a correct 2nd equation (NB 172 – $\mu$ = 0.525$\sigma$ is B0, since $z$ is incorrect)
- M1 for solving their two linear equations leading to $\mu = ...$ or $\sigma = ...$
- 1st A1 for $\sigma = $ awrt 8.30
- 2nd A1 for $\mu = $ awrt 168 [NB the 168 can come from false working. These A marks require use of correct equation from (b), and a z value for "0.5244" in (c)]
- NB use of $z = 0.52$ will typically get $\sigma = 8.31$ and $\mu = 167.67...$ and score B1M1A0A1
- No working and both correct scores 4/4, only one correct scores 0/4

## Part (d)
**Answer:** $P(\text{Taller than } 160\text{cm}) = P\left(Z > \frac{160 - \mu}{\sigma}\right) = P(Z < 0.9217994) = 0.8212$

**Marks:** M1 B1 A1 | (3)

**Guidance:**
- M1 for attempt to standardise with 160, their $\mu$ and their $\sigma$ (> 0). Even allow with symbols $\mu$ and $\sigma$
- B1 for $z = $ awrt $\pm$ 0.92
- No working and a correct answer can score 3/3 provided $\sigma$ and $\mu$ are correct to 2sf

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# Guidance Notes

**Question 1 Special Case - With Replacement** (This oversimplifies so do not apply Mis-Read: max mark 2/5)

(a) B1 for 3 branches followed by 3, 3, 3 with correct labels and probabilities of $\frac{1}{4}, \frac{1}{4}, \frac{1}{2}$ on each.

(b) M1 for identifying 2, possibly correct cases and adding 2 products of probabilities but A0 for wrong answer. $\left[\left(\frac{1}{4} \times \frac{1}{4}\right) + \left(\frac{1}{4} \times \frac{1}{4}\right)\right]$ will be sufficient for M1A0 here but $\frac{1}{4} \times \frac{1}{2} + ...$ would score M0