| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Conditional probability with normal |
| Difficulty | Standard +0.3 This is a straightforward S1 normal distribution question requiring standard z-score calculations and inverse normal lookup. Part (a) is routine standardization, part (b) requires recognizing conditional probability with truncated normal (though students might solve it approximately), and part (c) is a standard inverse normal problem. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \([W \sim N(64, 8^2)]\ \ P(W < 51) = P\!\left(Z < \frac{51-64}{8}\right)\) or \(P(Z < -1.625)\) | M1 | For standardising with 51 (or 77), 64 and 8 (allow \(\pm\)) |
| \(= 1 - 0.9484\) | M1 | For \(1 - p\) where \(0.9 < p < 1\) |
| \(= \text{awrt}\ \mathbf{0.052}\) | A1 | For awrt 0.052 |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Require: \(P(W > 49 \mid W < 51)\) | M1 | For a correctly stated conditional probability |
| \(= \dfrac{P(49 < W < 51)}{P(W < 51)}\) or \(\dfrac{P(-1.875 < Z < -1.625)}{P(Z < -1.625)}\) | M1 | For a correct ratio of probabilities |
| \(= \dfrac{0.021684...}{(\text{a})}\) | A1ft | Correct ratio with their (a) on denominator; numerator in range [0.0215, 0.0219] |
| \(= 0.4163...\ \ \ \text{awrt}\ \mathbf{0.42}\) | A1 | For awrt 0.42 |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[P(W > H) = 0.10 \Rightarrow\right]\ \dfrac{H - 64}{8} = 1.2816\) | M1B1 | M1 for standardising; B1 for using \(z = \pm 1.2816\) |
| \(H = 74.2528...\ \ \ \text{awrt}\ \mathbf{74.3}\) | A1 | For awrt 74.3 |
| (3) [10 marks] |
# Question 3:
## Part (a):
$[W \sim N(64, 8^2)]\ \ P(W < 51) = P\!\left(Z < \frac{51-64}{8}\right)$ or $P(Z < -1.625)$ | M1 | For standardising with 51 (or 77), 64 and 8 (allow $\pm$)
$= 1 - 0.9484$ | M1 | For $1 - p$ where $0.9 < p < 1$
$= \text{awrt}\ \mathbf{0.052}$ | A1 | For awrt 0.052
| | (3)
## Part (b):
Require: $P(W > 49 \mid W < 51)$ | M1 | For a correctly stated conditional probability
$= \dfrac{P(49 < W < 51)}{P(W < 51)}$ or $\dfrac{P(-1.875 < Z < -1.625)}{P(Z < -1.625)}$ | M1 | For a correct ratio of probabilities
$= \dfrac{0.021684...}{(\text{a})}$ | A1ft | Correct ratio with their (a) on denominator; numerator in range [0.0215, 0.0219]
$= 0.4163...\ \ \ \text{awrt}\ \mathbf{0.42}$ | A1 | For awrt 0.42
| | (4)
## Part (c):
$\left[P(W > H) = 0.10 \Rightarrow\right]\ \dfrac{H - 64}{8} = 1.2816$ | M1B1 | M1 for standardising; B1 for using $z = \pm 1.2816$
$H = 74.2528...\ \ \ \text{awrt}\ \mathbf{74.3}$ | A1 | For awrt 74.3
| | (3) **[10 marks]**
---\begin{enumerate}
\item The weights of women boxers in a tournament are normally distributed with mean 64 kg and standard deviation 8 kg .\\
(a) Find the probability that a randomly chosen woman boxer in the tournament weighs less than 51 kg .
\end{enumerate}
In the tournament, women boxers who weigh less than 51 kg are classified as lightweight. Ren weighs 49 kg and she has a match against another randomly selected, lightweight woman boxer.\\
(b) Find the probability that Ren weighs less than the other boxer.
In the tournament, women boxers who weigh more than $H \mathrm {~kg}$ are classified as heavyweight. Given that $10 \%$ of the women boxers in the tournament are classified as heavyweight,\\
(c) find the value of $H$.\\
\hfill \mbox{\textit{Edexcel S1 2019 Q3 [10]}}