| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| 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 This is a straightforward multi-part normal distribution question requiring standard techniques: finding a parameter using inverse normal (part i), calculating a probability from the normal distribution (part ii), and applying normal approximation to binomial with continuity correction (part iii). All steps are routine applications of standard methods with no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Working | Mark | Guidance |
| \(z = -0.842\) | B1 | \(\pm\) rounding to \(0.84\) seen |
| \(P(x > 1.35) = P\!\left(z > \dfrac{1.35 - 1.9}{\sigma}\right)\) | M1 | \(\pm\dfrac{1.35-1.9}{\sigma}\) = a probability or a \(z\)-value NOT \(0.8\) or \(0.2\); allow a \(1-\ldots\) |
| \(-0.842 = -0.55/\sigma\) | ||
| \(\sigma = 0.653\) | A1 [3] | Correct answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(x < 2) = P\!\left(z < \dfrac{2 - 1.9}{0.6532}\right)\) | M1 | \(\pm\) standardising; no continuity correction; their \(\sigma\) |
| \(= P(z < 0.1531)\) | ||
| \(= 0.561\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(X \sim N(160, 32)\); \(P(162.5 < x < 173.5)\) | B1 | Unsimplified \(160\) and \(32\) seen |
| \(P\!\left(\dfrac{162.5 - 160}{\sqrt{32}} < z < \dfrac{173.5 - 160}{\sqrt{32}}\right)\) | M1 | Standardising; need square root |
| \(P(0.442 < z < 2.386)\) | M1 | Any of \(162.5, 163.5, 172.5, 173.5\) seen |
| \(= \Phi(2.386) - \Phi(0.442)\) | M1 | \(\Phi_2 - \Phi_1\) oe |
| \(= 0.9915 - 0.6707\) | A1 | One correct \(\Phi\) to 3sf |
| \(= 0.321\) | A1 [6] | Correct answer; accept \(0.320\) |
## Question 7:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $z = -0.842$ | B1 | $\pm$ rounding to $0.84$ seen |
| $P(x > 1.35) = P\!\left(z > \dfrac{1.35 - 1.9}{\sigma}\right)$ | M1 | $\pm\dfrac{1.35-1.9}{\sigma}$ = a probability or a $z$-value NOT $0.8$ or $0.2$; allow a $1-\ldots$ |
| $-0.842 = -0.55/\sigma$ | | |
| $\sigma = 0.653$ | A1 **[3]** | Correct answer from correct working |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(x < 2) = P\!\left(z < \dfrac{2 - 1.9}{0.6532}\right)$ | M1 | $\pm$ standardising; no continuity correction; their $\sigma$ |
| $= P(z < 0.1531)$ | | |
| $= 0.561$ | A1 **[2]** | Correct answer |
### Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $X \sim N(160, 32)$; $P(162.5 < x < 173.5)$ | B1 | Unsimplified $160$ and $32$ seen |
| $P\!\left(\dfrac{162.5 - 160}{\sqrt{32}} < z < \dfrac{173.5 - 160}{\sqrt{32}}\right)$ | M1 | Standardising; need square root |
| $P(0.442 < z < 2.386)$ | M1 | Any of $162.5, 163.5, 172.5, 173.5$ seen |
| $= \Phi(2.386) - \Phi(0.442)$ | M1 | $\Phi_2 - \Phi_1$ oe |
| $= 0.9915 - 0.6707$ | A1 | One correct $\Phi$ to 3sf |
| $= 0.321$ | A1 **[6]** | Correct answer; accept $0.320$ |7 The time Rafa spends on his homework each day in term-time has a normal distribution with mean 1.9 hours and standard deviation $\sigma$ hours. On $80 \%$ of these days he spends more than 1.35 hours on his homework.\\
(i) Find the value of $\sigma$.\\
(ii) Find the probability that, on a randomly chosen day in term-time, Rafa spends less than 2 hours on his homework.\\
(iii) A random sample of 200 days in term-time is taken. Use an approximation to find the probability that the number of days on which Rafa spends more than 1.35 hours on his homework is between 163 and 173 inclusive.
\hfill \mbox{\textit{CAIE S1 2014 Q7 [11]}}