| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | October |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question with routine calculations: parts (a)-(b) involve basic z-score lookups, (c) applies binomial expectation to a profit context, (d) uses inverse normal with two equations to find μ and σ, and (e) requires simple comparison. All techniques are textbook exercises 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 transformation5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(L > 4.3) = P\left(Z > \frac{4.3 - 4.1}{0.125}\right)\) | M1 | 1st M1: standardising. Allow use of 0.125² |
| \(= P(Z > 1.6)\) or \(1 - P(Z < 1.6)\) or \(1 - 0.9452\) | M1 | |
| \(= 0.0548\) | A1 | |
| (3) | ||
| (b) \(P(3.9 < L < 4.3) = P(Z < 1.6) - P(Z < -1.6)\) or \(2P(Z < 1.6) - 0.5\) | B1cso (1) | B1cso: sight of 0.8904 or better (calc: 0.8904041212...) or a correct subtraction |
| \(= 0.9452 - 0.0548 = 0.8904\) or \(= 2(0.9452 - 0.5)\) | ||
| \(= 0.8904\) | ||
| (c) Number of unusable bolts \(= (1 - 0.89) \times 500 = [55]\) | M1oe | 1st M1: \((1 - "0.89") \times 500\) or \(0.89 \times 9 + 0.11 \times 1\) |
| Value of bolts \(= "445" \times 9 + "55" \times 1\) or profit \(= "445" \times 9 + "55" \times 1 - 500 \times 5\) | M1oe | SC: think 55 scrap loses 1p. 1st M1 for sight of 55. B1 for answer of awrt 1450 p |
| Profit from bolts \(= 1560\) pence | M1oe | Score as: M1M0M0A1 |
| A1 | For awrt £15.60 or 1560 pence(p) [need units] | |
| (4) | ||
| (d) \(\frac{4.198 - \mu}{\sigma} = 1.96\) or \(4.198 - \mu = 1.96\sigma\) oe | M1A1 | 1st M1: Forming either equation – must have z value but allow \(\pm\) z where \( |
| \(\frac{4.065 - \mu}{\sigma} = -0.7\) or \(4.065 - \mu = -0.7\sigma\) oe | A1 | 2nd A1: correct equation \(4.065 - \mu = -0.7\sigma\) - any form (or allow \(z =\) awrt \(-0.700\)) |
| \(0.133 = 2.66\sigma\) | M1 | 2nd M1: eliminating \(\mu\) or \(\sigma\) (method seen leading to equation in 1 variable) |
| \(\sigma = 0.05\) (or awrt 0.0500) | A1 | |
| \(\mu = 4.1\) (or awrt 4.10 dep on 1st or 2nd A1) | A1 | |
| (6) | ||
| (e) The mean the same but the st. dev. decreased or \(P(3.9 < L < 4.3)\) increased. So the profit will increase | B1ft, dB1ft | 1st B1ft: if \(\mu = 4.1\) then ft \(\sigma\); if \(\mu < 3.9\) (allow any \(\sigma\)) otherwise need to see \(P(3.9 < L < 4.3)\) calc. 2nd dB1ft: therefore profit will increase (o.e.). [\(\sigma < 0\) is B0B0] |
| (2) | ||
| NB Use of \(+ 0.7\) in (c) \(\rightarrow \mu = 3.99, \sigma = 0.106\), prob \(\approx 0.80 \rightarrow\) profit down | Total 16 |
**(a)** $P(L > 4.3) = P\left(Z > \frac{4.3 - 4.1}{0.125}\right)$ | M1 | 1st M1: standardising. Allow use of 0.125²
| $= P(Z > 1.6)$ or $1 - P(Z < 1.6)$ or $1 - 0.9452$ | M1 |
| $= 0.0548$ | A1 |
| | (3) |
**(b)** $P(3.9 < L < 4.3) = P(Z < 1.6) - P(Z < -1.6)$ or $2P(Z < 1.6) - 0.5$ | B1cso (1) | B1cso: sight of 0.8904 or better (calc: 0.8904041212...) or a correct subtraction
| $= 0.9452 - 0.0548 = 0.8904$ or $= 2(0.9452 - 0.5)$ | |
| | | $= 0.8904$ |
**(c)** Number of unusable bolts $= (1 - 0.89) \times 500 = [55]$ | M1oe | 1st M1: $(1 - "0.89") \times 500$ or $0.89 \times 9 + 0.11 \times 1$
| Value of bolts $= "445" \times 9 + "55" \times 1$ or profit $= "445" \times 9 + "55" \times 1 - 500 \times 5$ | M1oe | SC: think 55 scrap loses 1p. 1st M1 for sight of 55. B1 for answer of awrt 1450 p
| Profit from bolts $= 1560$ pence | M1oe | Score as: M1M0M0A1
| | A1 | For awrt £15.60 or 1560 pence(p) [need units]
| | (4) |
**(d)** $\frac{4.198 - \mu}{\sigma} = 1.96$ or $4.198 - \mu = 1.96\sigma$ oe | M1A1 | 1st M1: Forming either equation – must have z value but allow $\pm$ z where $|z| > 0.6$. 1st A1: correct equation $4.198 - \mu = 1.96\sigma$ - any form (or allow $z =$ awrt 1.960)
| $\frac{4.065 - \mu}{\sigma} = -0.7$ or $4.065 - \mu = -0.7\sigma$ oe | A1 | 2nd A1: correct equation $4.065 - \mu = -0.7\sigma$ - any form (or allow $z =$ awrt $-0.700$)
| $0.133 = 2.66\sigma$ | M1 | 2nd M1: eliminating $\mu$ or $\sigma$ (method seen leading to equation in 1 variable)
| $\sigma = 0.05$ (or awrt 0.0500) | A1 |
| $\mu = 4.1$ (or awrt 4.10 dep on 1st or 2nd A1) | A1 |
| | (6) |
**(e)** The mean the same but the st. dev. decreased or $P(3.9 < L < 4.3)$ increased. So the profit will increase | B1ft, dB1ft | 1st B1ft: if $\mu = 4.1$ then ft $\sigma$; if $\mu < 3.9$ (allow any $\sigma$) otherwise need to see $P(3.9 < L < 4.3)$ calc. 2nd dB1ft: therefore profit will increase (o.e.). [$\sigma < 0$ is B0B0]
| | (2) |
| NB Use of $+ 0.7$ in (c) $\rightarrow \mu = 3.99, \sigma = 0.106$, prob $\approx 0.80 \rightarrow$ profit down | **Total 16** |\begin{enumerate}
\item A machine makes bolts such that the length, $L \mathrm {~cm}$, of a bolt has distribution $L \sim \mathrm {~N} \left( 4.1,0.125 ^ { 2 } \right)$
\end{enumerate}
A bolt is selected at random.\\
(a) Find the probability that the length of this bolt is more than 4.3 cm .\\
(b) Show that $\mathrm { P } ( 3.9 < L < 4.3 )$ is 0.890 correct to 3 decimal places.
The machine makes 500 bolts.\\
The cost to make each bolt is 5 pence.\\
Only bolts with length between 3.9 cm and 4.3 cm can be used. These are sold for 9 pence each. All the bolts that cannot be used are recycled with a scrap value of 1 pence each.\\
(c) Calculate an estimate for the profit made on these 500 bolts.
Following adjustments to the machine, the length of a bolt, $B \mathrm {~cm}$, made by the machine is such that $B \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$
Given that $\mathrm { P } ( B > 4.198 ) = 0.025$ and $\mathrm { P } ( B < 4.065 ) = 0.242$\\
(d) find the value of $\mu$ and the value of $\sigma$\\
(e) State, giving a reason, whether the adjustments to the machine will result in a decrease or an increase in the profit made on 500 bolts.
\hfill \mbox{\textit{Edexcel S1 2018 Q6 [16]}}