| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring routine z-score calculations and inverse normal lookups. Part (a) needs P(W≥70), part (b) requires finding k from P(80≤W<k)=0.35, and part (c) involves conditional probability with P(W>y|W<66)=0.75. While multi-part, each step uses familiar techniques with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Small | Large | |||
| Tiny | Petite | Extra | Jumbo | Mega |
| \(w < 66\) | \(66 \leqslant w < 70\) | \(70 \leqslant w < 80\) | \(80 \leqslant w < k\) | \(w \geqslant k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(W>70) = P\left(Z > \frac{70-80}{8}[=-1.25]\right)\) | M1 | Standardising with 70, 80 and 8 (allow \(\pm\)) |
| \(= P(Z > -1.25)\) or \(P(Z < 1.25)\) | A1 | \(1^{st}\) A1: \(z = \pm 1.25\) |
| \(= 0.8944\) awrt 0.894 | A1 | \(2^{nd}\) A1: awrt 0.894 (calc 0.894350...) NB do not ISW so 0.1056 is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(W < k) = 0.85\) or \(P(W > k) = 0.15\) | B1 | Either correct probability statement. Allow "," for \(<\) and \(>\) |
| \(\pm\left(\frac{k-80}{8}\right) = \mathbf{1.0364}\) | M1 B1 | M1 standardising with 80, 8 and equating to \(z\) where \(1 < |
| \(k = 88.29\ldots\) awrt 88.3 | A1 | NB awrt 88.3 implies \(1^{st}\) B1 and M1 but not \(2^{nd}\) B1 → B1M1B0A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(W < 66) = P\left(Z < \frac{66-80}{8}[=-1.75]\right)[= 0.0401\) (calc 0.040059...)] | M1 | \(1^{st}\) M1: standardising with 66, 80 and 8 (allow \(\pm\)) or seeing awrt 0.0401 |
| \(0.25 \times P(Z < -1.75)[= 0.010025\) (calc 0.0100147...)] or \(0.25 \times (1 - P(Z < 1.75))\) | dM1 | \(2^{nd}\) dM1 (dep on \(1^{st}\) M1): \(0.25\times\)"their \(P(Z<-1.75)\)" or \(0.0401 - 0.030075\) or seeing \([0.75\times0.0401 + 0.9599=]\ 0.9899\ldots\) |
| \(\frac{y-80}{8} = \mathbf{-2.32}(63)\) | M1 A1 | \(3^{rd}\) M1: standardising and equating to \(z\) where \( |
| \(y = 61.389\ldots\) awrt 61.4 | A1 | \(2^{nd}\) A1: awrt 61.4 (allow awrt 61.3) |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(W>70) = P\left(Z > \frac{70-80}{8}[=-1.25]\right)$ | M1 | Standardising with 70, 80 and 8 (allow $\pm$) |
| $= P(Z > -1.25)$ or $P(Z < 1.25)$ | A1 | $1^{st}$ A1: $z = \pm 1.25$ |
| $= 0.8944$ awrt **0.894** | A1 | $2^{nd}$ A1: awrt 0.894 (calc 0.894350...) NB do not ISW so 0.1056 is A0 |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(W < k) = 0.85$ or $P(W > k) = 0.15$ | B1 | Either correct probability statement. Allow "," for $<$ and $>$ |
| $\pm\left(\frac{k-80}{8}\right) = \mathbf{1.0364}$ | M1 B1 | M1 standardising with 80, 8 and equating to $z$ where $1 < |z| < 2$; B1 $z = \pm 1.0364$ or better |
| $k = 88.29\ldots$ awrt **88.3** | A1 | NB awrt 88.3 implies $1^{st}$ B1 and M1 but not $2^{nd}$ B1 → B1M1B0A1 |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(W < 66) = P\left(Z < \frac{66-80}{8}[=-1.75]\right)[= 0.0401$ (calc 0.040059...)] | M1 | $1^{st}$ M1: standardising with 66, 80 and 8 (allow $\pm$) or seeing awrt 0.0401 |
| $0.25 \times P(Z < -1.75)[= 0.010025$ (calc 0.0100147...)] or $0.25 \times (1 - P(Z < 1.75))$ | dM1 | $2^{nd}$ dM1 (dep on $1^{st}$ M1): $0.25\times$"their $P(Z<-1.75)$" or $0.0401 - 0.030075$ or seeing $[0.75\times0.0401 + 0.9599=]\ 0.9899\ldots$ |
| $\frac{y-80}{8} = \mathbf{-2.32}(63)$ | M1 A1 | $3^{rd}$ M1: standardising and equating to $z$ where $|z|>2$; $1^{st}$ A1: correct standardisation equation with compatible signs and 2.32, $|z|$, 2.34 |
| $y = 61.389\ldots$ awrt **61.4** | A1 | $2^{nd}$ A1: awrt 61.4 (allow awrt 61.3) |
---\begin{enumerate}
\item The weights, $W$ grams, of kiwi fruit grown on a farm are normally distributed with mean 80 grams and standard deviation 8 grams.
\end{enumerate}
The table shows the classifications of the kiwi fruit by their weight, where $k$ is a positive constant.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{Small} & \multicolumn{3}{|c|}{Large} \\
\hline
Tiny & Petite & Extra & Jumbo & Mega \\
\hline
$w < 66$ & $66 \leqslant w < 70$ & $70 \leqslant w < 80$ & $80 \leqslant w < k$ & $w \geqslant k$ \\
\hline
\end{tabular}
\end{center}
One kiwi fruit is selected at random from those grown on the farm.\\
(a) Find the probability that this kiwi fruit is Large.
35\% of the kiwi fruit are Jumbo.\\
(b) Find the value of $k$ to one decimal place.
75\% of Tiny kiwi fruit weigh more than $y$ grams.\\
(c) Find the value of $y$ giving your answer to one decimal place.
\hfill \mbox{\textit{Edexcel S1 2022 Q5 [12]}}