## Part (i)
$$\int_0^{\infty} f(x)dx = \int_0^{\infty} \lambda e^{-\lambda x} dx = \left[-e^{-\lambda x}\right]_0^{\infty} = (0 - (-e^0)) = 1$$

| M1 | Integration of $f(x)$. |
| M1 | Use of limits or the given result. |
| A1 | Convincingly obtained (Answer given.) |

Curve, with negative gradient, in the first quadrant only. Must intersect the y-axis.

| G1 | |

$(0, 2)$ labelled; asymptotic to x-axis.

| G1 | [5]

## Part (ii)
$$E(X) = \int_0^{\infty} \lambda x e^{-\lambda x} dx = \lambda \cdot \frac{1}{\lambda^2} = \frac{1}{\lambda}$$

| M1 | Correct integral. |
| A1 | c.a.o. (using given result) |

$$E(X^2) = \int_0^{\infty} \lambda x^2 e^{-\lambda x} dx = \lambda \cdot \frac{2}{\lambda^3} = \frac{2}{\lambda^2}$$

| M1 | Correct integral. |
| A1 | c.a.o. (using given result) |

$$\text{Var}(X) = E(X^2) - E(X)^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}$$

| M1 | Use of $E(X^2) - E(X)^2$ |
| A1 | |

[6]

## Part (iii)
$\mu = 6$; $\lambda = \frac{1}{6}$

| B1 | Obtained $\lambda$ from the mean. |
| B1 | Normal. |
| B1 | Mean. It c's $\lambda$. |
| B1 | Variance. It c's $\lambda$. |

$$\bar{X} \sim \text{(approx)} N\left(6, \frac{6^2}{50}\right)$$

[4]

## Part (iv)
**EITHER** can argue that 7.8 is more than 2 SDs from $\mu$.

$(6 + 2\sqrt{0.72} = 7.697;$

must refer to SD($\bar{X}$), not SD(X))

i.e. outlier.

$\Rightarrow$ doubt.

| M1 | |
| | A 95% C.I would be (6.1369, 9.4631). |
| M1 | |
| A1 | |

**OR** formal significance test:

$$\frac{7.8 - 6}{\sqrt{0.72}} = 2.121, \text{ refer to } N(0,1), \text{ sig at (eg) 5%}$$

$\Rightarrow$ doubt.

| M1 | |
| M1 | Depends on first M, but could imply it. $P(|Z| > 2.121) = 0.0339$ |
| A1 | |

[3]