| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | October |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question covering routine techniques: standardization to z-scores, using tables/inverse normal, and solving simultaneous equations with two percentiles. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim N(210, 25^2)\); \(P(X<240) = P\!\left(Z < \frac{240-210}{25}\right) [= P(Z<1.2)]\) | M1 | For standardising using 240, 210 and 25 |
| \(= 0.8849\) | A1* | Cao; as answer is given no incorrect working should be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
\(P(190| M1 |
For standardising using 190/230, 210 and 25 and subtracting from 0.8849; may be implied by \(\Phi(1.2)+\Phi(0.8)-1\) or \(0.8849+0.7881-1\) |
|
| \(0.8849 - 0.2119 = 0.673\) | A1 | awrt 0.673 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{210+k-210}{25} = 1.96\) or \(\frac{210-k-210}{25} = -1.96\) | M1 | For standardising and setting equal to a \(z\) value where \(1.9 < |
| B1 | For \( | z |
| \(k = 49\) | A1 | awrt 49 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
\(P(X| M1 B1 |
For standardising using \(S\) and setting equal to \(z\) value where \(1 < |
|
| \(S = 184.09\) | A1 | awrt 184 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(Y \sim N(\mu, \sigma^2)\); \(P(Y<152) = 0.05 \Rightarrow \frac{152-\mu}{\sigma} = -1.6449\) | M1 A1 | For correct method to form equation in \(\mu\) and \(\sigma\) set equal to \(z\) value where \(-1.6 < z < -1.7\) or \(0.2 < z < 0.3\); signs must be compatible; for correct equation for \(P(Y<152)\) |
| \(P(Y>180) = 0.40 \Rightarrow \frac{180-\mu}{\sigma} = 0.2533\) | A1 | For correct equation for \(P(Y>180)\) |
| \(28 = 1.8982\sigma\) | dM1 | Dependent on previous M; for solving 2 equations simultaneously; if answers incorrect then working must be shown; may be implied by \(\sigma =\) awrt 14.8 and \(\mu =\) awrt 176 |
| \(\sigma = 14.75...\), \(\mu = 176.26...\) | A1 | For \(\sigma =\) awrt 14.8 and \(\mu =\) awrt 176 |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim N(210, 25^2)$; $P(X<240) = P\!\left(Z < \frac{240-210}{25}\right) [= P(Z<1.2)]$ | M1 | For standardising using 240, 210 and 25 |
| $= 0.8849$ | A1* | Cao; as answer is given no incorrect working should be seen |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(190<X<240) = 0.8849 - P\!\left(Z<\frac{190-210}{25}\right) [= 0.8849 - P(Z<-0.8)]$ | M1 | For standardising using 190/230, 210 and 25 and subtracting from 0.8849; may be implied by $\Phi(1.2)+\Phi(0.8)-1$ or $0.8849+0.7881-1$ |
| $0.8849 - 0.2119 = 0.673$ | A1 | awrt 0.673 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{210+k-210}{25} = 1.96$ or $\frac{210-k-210}{25} = -1.96$ | M1 | For standardising and setting equal to a $z$ value where $1.9 < |z| < 2$ |
| | B1 | For $|z| = 1.96$ or better |
| $k = 49$ | A1 | awrt 49 |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X<S) = 0.15 \Rightarrow \frac{S-210}{25} = -1.0364$ | M1 B1 | For standardising using $S$ and setting equal to $z$ value where $1 < |z| < 1.1$; for $z = -1.0364$ |
| $S = 184.09$ | A1 | awrt 184 |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $Y \sim N(\mu, \sigma^2)$; $P(Y<152) = 0.05 \Rightarrow \frac{152-\mu}{\sigma} = -1.6449$ | M1 A1 | For correct method to form equation in $\mu$ and $\sigma$ set equal to $z$ value where $-1.6 < z < -1.7$ or $0.2 < z < 0.3$; signs must be compatible; for correct equation for $P(Y<152)$ |
| $P(Y>180) = 0.40 \Rightarrow \frac{180-\mu}{\sigma} = 0.2533$ | A1 | For correct equation for $P(Y>180)$ |
| $28 = 1.8982\sigma$ | dM1 | Dependent on previous M; for solving 2 equations simultaneously; if answers incorrect then working must be shown; may be implied by $\sigma =$ awrt 14.8 and $\mu =$ awrt 176 |
| $\sigma = 14.75...$, $\mu = 176.26...$ | A1 | For $\sigma =$ awrt 14.8 and $\mu =$ awrt 176 |
---\begin{enumerate}
\item The weights, $X$ grams, of a particular variety of fruit are normally distributed with
\end{enumerate}
$$X \sim \mathrm {~N} \left( 210,25 ^ { 2 } \right)$$
A fruit of this variety is selected at random.\\
(a) Show that the probability that the weight of this fruit is less than 240 grams is 0.8849\\
(b) Find the probability that the weight of this fruit is between 190 grams and 240 grams.\\
(c) Find the value of $k$ such that $\mathrm { P } ( 210 - k < X < 210 + k ) = 0.95$
A wholesaler buys large numbers of this variety of fruit and classifies the lightest $15 \%$ as small.\\
(d) Find the maximum weight of a fruit that is classified as small.
You must show your working clearly.
The weights, $Y$ grams, of a second variety of this fruit are normally distributed with
$$Y \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$$
Given that $5 \%$ of these fruit weigh less than 152 grams and $40 \%$ weigh more than 180 grams,\\
(e) calculate the mean and standard deviation of the weights of this variety of fruit.
\hfill \mbox{\textit{Edexcel S1 2023 Q5 [15]}}