| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a standard S1 normal distribution question requiring routine z-score calculations, inverse normal lookup, and basic binomial probability. All parts follow textbook procedures with no novel insight needed, making it slightly easier than average but still requiring multiple techniques across four parts. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([W = \text{weight of a package delivered to Susie } W \sim N(510, 45^2)]\) \(\text{P}(W < 450) = \text{P}\left(Z < \frac{450-510}{45}\right) \text{ or P}(Z < -1.3333)\) | M1 | For standardising 450 with 510 and 45 (allow +) |
| \(= 1 - 0.9082\) | M1 | For 1 – p (where 0.90 < p < 0.99) |
| \(= 0.0918 \quad \mathbf{[0.0912-0.0918]}\) | A1 | For answer in the range 0.0912 to 0.0918 inclusive (calc. 0.09121133…) |
| \([P(W > d) = 0.05 \text{ implies}] \frac{d - 510}{45} = 1.6449\) | M1B1 | For standardising their letter d with 510 and 45 and setting equal to z value \(1 < \ |
| \(d = 584.0205\ldots \quad \text{awrt } \mathbf{584}\) | A1 | For awrt 584 (calc 584.0184…) |
| \(\text{P}(W > 450 | W < \text{"584.02…"}) = \frac{\text{P}(450 < W < \text{"584.02…"})}{\text{P}(W < \text{"584.02…"})}\) | M1 |
| \(= \frac{0.95 - \text{"0.0918"}}{0.95} \quad \text{or} \quad \frac{\text{"0.9082"} - 0.05}{0.95}\) | M1A1 | For numerator of awrt 0.95 – their answer to (a). For a correct denominator of awrt 0.95 (dep on M1M1) |
| \(= 0.903368\ldots \quad \text{awrt } \mathbf{0.904 \text{ or } 0.903}\) | A1 | For awrt 0.904 or awrt 0.903 |
| \(\left(\frac{19}{20}\right)^4 \times \frac{1}{20} \times 5\) | M1dM1 | For k \(p^4(1 - p)\) for any positive integer k and any probability p (allow k = 1) |
| \(= 0.203626\ldots \quad \text{awrt } \mathbf{0.204}\) | A1 | For awrt 0.204 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) M1 for standardising their letter d with 510 and 45 and setting equal to z value \(1 < | z | < 2\) |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[W = \text{weight of a package delivered to Susie } W \sim N(510, 45^2)]$ $\text{P}(W < 450) = \text{P}\left(Z < \frac{450-510}{45}\right) \text{ or P}(Z < -1.3333)$ | M1 | For standardising 450 with 510 and 45 (allow +) |
| $= 1 - 0.9082$ | M1 | For 1 – p (where 0.90 < p < 0.99) |
| $= 0.0918 \quad \mathbf{[0.0912-0.0918]}$ | A1 | For answer in the range 0.0912 to 0.0918 inclusive (calc. 0.09121133…) |
| $[P(W > d) = 0.05 \text{ implies}] \frac{d - 510}{45} = 1.6449$ | M1B1 | For standardising their letter d with 510 and 45 and setting equal to z value $1 < \|z\| < 2$ |
| $d = 584.0205\ldots \quad \text{awrt } \mathbf{584}$ | A1 | For awrt 584 (calc 584.0184…) |
| $\text{P}(W > 450 | W < \text{"584.02…"}) = \frac{\text{P}(450 < W < \text{"584.02…"})}{\text{P}(W < \text{"584.02…"})}$ | M1 | For a correct ratio of probability expressions ft their answer to (b) where (b) > 450 |
| $= \frac{0.95 - \text{"0.0918"}}{0.95} \quad \text{or} \quad \frac{\text{"0.9082"} - 0.05}{0.95}$ | M1A1 | For numerator of awrt 0.95 – their answer to (a). For a correct denominator of awrt 0.95 (dep on M1M1) |
| $= 0.903368\ldots \quad \text{awrt } \mathbf{0.904 \text{ or } 0.903}$ | A1 | For awrt 0.904 or awrt 0.903 |
| $\left(\frac{19}{20}\right)^4 \times \frac{1}{20} \times 5$ | M1dM1 | For k $p^4(1 - p)$ for any positive integer k and any probability p (allow k = 1) |
| $= 0.203626\ldots \quad \text{awrt } \mathbf{0.204}$ | A1 | For awrt 0.204 |
**Notes:**
**Correct answer only in (a), (c) or (d) scores all the marks for that part**
**(a)** 1st M1 for standardising 450 with 510 and 45 (allow +)
2nd M1 for 1 – p (where 0.90 < p < 0.99)
A1 for answer in the range 0.0912 to 0.0918 inclusive (calc. 0.09121133…)
**(b)** M1 for standardising their letter d with 510 and 45 and setting equal to z value $1 < |z| < 2$
B1 for use of z = ± 1.6449 or better (calc 1.644853626…)
A1 for awrt 584 (calc 584.0184…)
**Ans only** [awrt 584.02 scores 3/3 584 scores M1B0A1]
**(c)** 1st M1 for a correct ratio of probability expressions ft their answer to (b) where (b) > 450
2nd M1 for numerator of awrt 0.95 – their answer to (a)
1st A1 for a correct denominator of awrt 0.95 (dep on M1M1)
NB a correct ratio of probabilities will score the 1st 3 marks
2nd A1 for awrt 0.904 or awrt 0.903
**(d)** 1st M1 for k $p^4(1 - p)$ for any positive integer k and any probability p (allow k = 1)
2nd dM1 for k = 5
A1 for awrt 0.204
---\begin{enumerate}
\item The weights of packages delivered to Susie are normally distributed with a mean of 510 grams and a standard deviation of 45 grams.\\
(a) Find the probability that a randomly selected package delivered to Susie weighs less than 450 grams.
\end{enumerate}
The heaviest 5\% of packages delivered to Susie are delivered by Rav in his van, the others are delivered by Taruni on foot.\\
(b) Find the weight of the lightest package that Rav would deliver to Susie.
Susie randomly selects a package from those delivered by Taruni.\\
(c) Find the probability that this package weighs more than 450 grams.
On Tuesday there are 5 packages delivered to Susie.\\
(d) Find the probability that 4 are delivered by Taruni and 1 is delivered by Rav.
\hfill \mbox{\textit{Edexcel S1 2019 Q4 [13]}}