| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution parameters found then approximation applied |
| Difficulty | Standard +0.3 Part (i) is a standard inverse normal calculation requiring one z-score lookup and simple algebra. Part (ii) applies normal approximation to binomial with given parameters (n=300, p=0.2), requiring continuity correction and standardization. Part (iii) tests understanding of np>5 and nq>5 conditions. This is a routine multi-part question slightly easier than average due to straightforward setup and standard techniques throughout. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 1800) = 0.96\), so \(P\!\left(Z > \frac{1800-2000}{\sigma}\right) = 0.96\) | B1 | \(\pm 1.75\) seen |
| \(\Phi\!\left(\frac{200}{\sigma}\right) = 0.96\), \(\quad \frac{200}{\sigma} = 1.751\) | M1 | \(z = \pm\frac{1800-2000}{\sigma}\), allow cc, allow sq rt, allow sq equated to a z-value |
| \(\sigma = 114\) | A1 | Correct final answer www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= 300 \times 0.2 = 60\) and variance \(= 300 \times 0.2 \times 0.8 = 48\) | B1 | Correct unsimplified mean and variance |
| \(P(X < 70) = P\!\left(Z > \frac{69.5 - 60}{\sqrt{48}}\right)\) | M1 | \(Z = \pm\frac{x - \text{their } 60}{\sqrt{\text{their } 48}}\) |
| \(= \Phi(1.371)\) | M1 | \(69.5\) or \(70.5\) seen in an attempted standardisation expression as cc |
| \(= 0.915\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(np = 60\), \(nq = 240\): both \(> 5\), (so normal approximation holds) | B1 | Both parts evaluated are required |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 1800) = 0.96$, so $P\!\left(Z > \frac{1800-2000}{\sigma}\right) = 0.96$ | B1 | $\pm 1.75$ seen |
| $\Phi\!\left(\frac{200}{\sigma}\right) = 0.96$, $\quad \frac{200}{\sigma} = 1.751$ | M1 | $z = \pm\frac{1800-2000}{\sigma}$, allow cc, allow sq rt, allow sq equated to a z-value |
| $\sigma = 114$ | A1 | Correct final answer www |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 300 \times 0.2 = 60$ and variance $= 300 \times 0.2 \times 0.8 = 48$ | B1 | Correct unsimplified mean and variance |
| $P(X < 70) = P\!\left(Z > \frac{69.5 - 60}{\sqrt{48}}\right)$ | M1 | $Z = \pm\frac{x - \text{their } 60}{\sqrt{\text{their } 48}}$ |
| $= \Phi(1.371)$ | M1 | $69.5$ or $70.5$ seen in an attempted standardisation expression as cc |
| $= 0.915$ | A1 | Correct final answer |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $np = 60$, $nq = 240$: both $> 5$, (so normal approximation holds) | B1 | Both parts evaluated are required |
---6 The lifetimes, in hours, of a particular type of light bulb are normally distributed with mean 2000 hours and standard deviation $\sigma$ hours. The probability that a randomly chosen light bulb of this type has a lifetime of more than 1800 hours is 0.96 .\\
(i) Find the value of $\sigma$.\\
New technology has resulted in a new type of light bulb. It is found that on average one in five of these new light bulbs has a lifetime of more than 2500 hours.\\
(ii) For a random selection of 300 of these new light bulbs, use a suitable approximate distribution to find the probability that fewer than 70 have a lifetime of more than 2500 hours.\\
(iii) Justify the use of your approximate distribution in part (ii).\\
\hfill \mbox{\textit{CAIE S1 2018 Q6 [8]}}