| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Binomial from normal with unknown parameter |
| Difficulty | Standard +0.3 This is a straightforward S1 normal distribution question requiring standard z-score calculations and understanding of independent events. Part (a) is routine standardization, part (b) requires working backwards from P(X>32)^2 = 0.16 using inverse normal tables, and part (c) uses complement rule with binomial probability. All techniques are standard S1 material 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Scheme | Marks |
| (a) | \(P(X < 37) = P\left(Z < \frac{37-40}{2.4}\right) = P(Z < -1.25) = 1 - 0.8944\) | M1, M1; A1 (3) |
| awrt 0.106 | ||
| (b) | \(P(\text{one value is greater than 32}) = \sqrt{0.16} [= 0.4]\) | M1 |
| \(\frac{32-m}{2.4} = 0.2533\) | M1 B1 | 2nd M1 standardising 32 with \(m\) and 2.4 and setting equal to \(z\) value \(0.2 < \ |
| B1 for \(z = \pm 0.2533\) or better (calc gives 0.2533470931…..) used in a linear equation for \(m\) | ||
| awrt 31.4 | A1 (4) | A1 awrt 31.4 or better |
| SC [using 0.16]Allow M0M1 B0 A0 for \(\frac{32-m}{2.4} = z\) where \(0.99 \leq z \leq 1.04\) | ||
| (c) | \(P(Y < 0) = P\left(Z < \frac{0-4}{8}\right) = P(Z < -0.5) [= 0.3085]\) | M1 |
| Let \(X\) be the number of negative values | M1 | 2nd M1 realising they need to find \(1 - P(X = 0)\) i.e writing or using \(1 - P(\text{no negative values})\) o.e. May be implied by \(1 - (0.6915)^5\) |
| \(P(X \geq 1) = 1 - P(X = 0)\) o.e | ||
| \(= 1 - (0.6915)^5 = 0.84188\ldots\) | M1, A1 (4) | M1 \(= 1 - (0.6915)^5\). A1 awrt 0.842 (tables: 0.8418894…. calculator: 0.84193233….) |
| awrt 0.842 | ||
| [11] | NB If they use Binomial and get 0.842 full marks, and get 0.125 then award M1M1M0A0, otherwise send to Review |
| Part | Scheme | Marks | Guidance |
|------|--------|-------|----------|
| (a) | $P(X < 37) = P\left(Z < \frac{37-40}{2.4}\right) = P(Z < -1.25) = 1 - 0.8944$ | M1, M1; A1 (3) | 1st M1 standardising 37 (or 43) with 40 and 2.4 (allow ±). 2nd M1 for $1 - p$ (where $0.88 < p < 0.90$) Implied by correct answer. A1 for awrt 0.106 (calc. 0.105649…) |
| | awrt **0.106** | | |
| (b) | $P(\text{one value is greater than 32}) = \sqrt{0.16} [= 0.4]$ | M1 | 1st M1 correct expression for one value > 32 (may be implied by sight of 0.2533…). Allow any value between 0.25 and 0.26 inclusive |
| | $\frac{32-m}{2.4} = 0.2533$ | M1 B1 | 2nd M1 standardising 32 with $m$ and 2.4 and setting equal to $z$ value $0.2 < \|z\| < 0.3$ |
| | | | B1 for $z = \pm 0.2533$ or better (calc gives 0.2533470931…..) used in a linear equation for $m$ |
| | awrt **31.4** | A1 (4) | A1 awrt 31.4 or better |
| | | | **SC** [using 0.16]Allow M0M1 B0 A0 for $\frac{32-m}{2.4} = z$ where $0.99 \leq z \leq 1.04$ |
| (c) | $P(Y < 0) = P\left(Z < \frac{0-4}{8}\right) = P(Z < -0.5) [= 0.3085]$ | M1 | 1st M1 standardising 0 with 4 and 8 (allow ±) or seeing 0.3085 or 0.6915 |
| | Let $X$ be the number of negative values | M1 | 2nd M1 realising they need to find $1 - P(X = 0)$ i.e writing or using $1 - P(\text{no negative values})$ o.e. May be implied by $1 - (0.6915)^5$ |
| | $P(X \geq 1) = 1 - P(X = 0)$ o.e | | |
| | $= 1 - (0.6915)^5 = 0.84188\ldots$ | M1, A1 (4) | M1 $= 1 - (0.6915)^5$. A1 awrt 0.842 (tables: 0.8418894…. calculator: 0.84193233….) |
| | awrt **0.842** | | |
| | | [11] | **NB** If they use Binomial and get 0.842 full marks, and get 0.125 then award M1M1M0A0, otherwise send to Review |
---\begin{enumerate}
\item Jia writes a computer program that randomly generates values from a normal distribution. He sets the mean as 40 and the standard deviation as 2.4\\
(a) Find the probability that a particular value generated by the computer program is less than 37
\end{enumerate}
Jia changes the mean to $m$ but leaves the standard deviation as 2.4\\
The computer program then randomly generates 2 independent values from this normal distribution.
The probability that both of these values are greater than 32 is 0.16\\
(b) Find the value of $m$, giving your answer to 2 decimal places.
Jia now changes the mean to 4 and the standard deviation to 8\\
The computer program then randomly generates 5 independent values from this normal distribution.\\
(c) Find the probability that at least one of these values is negative.
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\hfill \mbox{\textit{Edexcel S1 2022 Q5 [11]}}