| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution with standard procedures: finding P(X > a), finding an inverse normal value for a given percentile, and scaling a probability to estimate frequency. All three parts use direct calculator/table methods with no conceptual challenges or multi-step reasoning beyond basic standardization. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(x > 10.2) = P\!\left(z > \dfrac{10.2 - 9.5}{1.3}\right)\) | M1 | Standardising, allow cc, sq rt, sq |
| \(= P(z > 0.53846) = 1 - 0.7046 = 0.295\) | M1 A1 [3] | \(1 - \Phi\) final solution attempt |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = -1.282\) | B1 | \(\pm\) rounding to 1.28 seen |
| \(-1.282 = \dfrac{t - 9.5}{1.3}\) | M1 | Standardising correctly, can be \(\pm z\) value here |
| \(t = 7.83\) | A1 [3] | Correct answer from \(z = -1.282\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(x < 8.8) = 0.2954\) by symmetry | B1 | oe method, FT *their 0.2954 from (i)* |
| Days \(= 365 \times 0.2954 = 107\) or \(108\) | M1 A1 [3] | Mult a probability \(<1\) by 365. Correct answer (no decimals) |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(x > 10.2) = P\!\left(z > \dfrac{10.2 - 9.5}{1.3}\right)$ | M1 | Standardising, allow cc, sq rt, sq |
| $= P(z > 0.53846) = 1 - 0.7046 = 0.295$ | M1 A1 [3] | $1 - \Phi$ final solution attempt |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = -1.282$ | B1 | $\pm$ rounding to 1.28 seen |
| $-1.282 = \dfrac{t - 9.5}{1.3}$ | M1 | Standardising correctly, can be $\pm z$ value here |
| $t = 7.83$ | A1 [3] | Correct answer from $z = -1.282$ only |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(x < 8.8) = 0.2954$ by symmetry | B1 | oe method, FT *their 0.2954 from (i)* |
| Days $= 365 \times 0.2954 = 107$ or $108$ | M1 A1 [3] | Mult a probability $<1$ by 365. Correct answer (no decimals) |
---6 The time in minutes taken by Peter to walk to the shop and buy a newspaper is normally distributed with mean 9.5 and standard deviation 1.3.\\
(i) Find the probability that on a randomly chosen day Peter takes longer than 10.2 minutes.\\
(ii) On $90 \%$ of days he takes longer than $t$ minutes. Find the value of $t$.\\
(iii) Calculate an estimate of the number of days in a year ( 365 days) on which Peter takes less than 8.8 minutes to walk to the shop and buy a newspaper.
\hfill \mbox{\textit{CAIE S1 2016 Q6 [9]}}