| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard table lookups and basic binomial probability. Part (a) is routine standardization, part (b) is inverse normal lookup, and part (c) applies binomial probability with n=3. All techniques are standard textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(z = \dfrac{53-50}{2}\) | M1 | M1 for using 53, 50 and 2, either way around on numerator |
| \(P(X>53) = 1 - P(Z < 1.5)\) | B1 | B1: \(1-\) any probability for mark; \(1-\) probability required can be implied |
| \(= 1 - 0.9332\) | ||
| \(= 0.0668\) | A1 | A1: 0.0668 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(P(X \leq x_0) = 0.01\) | M1 | M1 can be implied or seen in a diagram, or equivalent with correct use of 0.01 or 0.99 |
| \(\dfrac{x_0 - 50}{2} = -2.3263\) | M1B1 | M1 attempt to standardise with 50 and 2; B1 for \(\pm 2.3263\) |
| \(x_0 = 45.3474\) | M1A1 | M1 equate expression with 50 and 2 to a \(z\) value with consistent signs and attempt to solve; A1 awrt 45.3 or 45.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(P(\text{2 weigh more than 53kg and 1 less}) = 3 \times 0.0668^2(1-0.0668)\) | B1M1A1ft | B1 for 3; M1 \(p^2(1-p)\) for any value of \(p\); A1ft for \(p\) being their answer to part (a) without 3 |
| \(= 0.012492487\ldots\) | A1 | A1 awrt 0.012 or 0.0125 |
## Question 7:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $z = \dfrac{53-50}{2}$ | M1 | M1 for using 53, 50 and 2, either way around on numerator |
| $P(X>53) = 1 - P(Z < 1.5)$ | B1 | B1: $1-$ any probability for mark; $1-$ probability required can be implied |
| $= 1 - 0.9332$ | | |
| $= 0.0668$ | A1 | A1: 0.0668 cao |
**[3 marks]**
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $P(X \leq x_0) = 0.01$ | M1 | M1 can be implied or seen in a diagram, or equivalent with correct use of 0.01 or 0.99 |
| $\dfrac{x_0 - 50}{2} = -2.3263$ | M1B1 | M1 attempt to standardise with 50 and 2; B1 for $\pm 2.3263$ |
| $x_0 = 45.3474$ | M1A1 | M1 equate expression with 50 and 2 to a $z$ value with consistent signs and attempt to solve; A1 awrt 45.3 or 45.4 |
**[5 marks]**
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $P(\text{2 weigh more than 53kg and 1 less}) = 3 \times 0.0668^2(1-0.0668)$ | B1M1A1ft | B1 for 3; M1 $p^2(1-p)$ for any value of $p$; A1ft for $p$ being their answer to part (a) without 3 |
| $= 0.012492487\ldots$ | A1 | A1 awrt 0.012 or 0.0125 |
**[4 marks] — Total: 12**7. A packing plant fills bags with cement. The weight $X \mathrm {~kg}$ of a bag of cement can be modelled by a normal distribution with mean 50 kg and standard deviation 2 kg .
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 53 )$.
\item Find the weight that is exceeded by $99 \%$ of the bags.
Three bags are selected at random.
\item Find the probability that two weigh more than 53 kg and one weighs less than 53 kg .
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2008 Q7 [12]}}