| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear relationship μ = kσ |
| Difficulty | Standard +0.8 This question requires understanding the relationship μ = 4σ, using inverse normal tables to find z-scores, solving simultaneous equations, then applying binomial approximation to normal distribution. It combines multiple statistical concepts with algebraic manipulation, making it moderately harder than a standard normal distribution question but still within typical A-level scope. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = 1.036\) or \(1.037\) | B1 | \(\pm 1.036\) or \(\pm 1.037\) seen |
| \(1.036 = \frac{5-4s}{s}\) | B1 | \(\frac{5-4\sigma}{\sigma}\) seen or \(\frac{5-\mu}{\mu/4}\) oe |
| \(s = 0.993\) | M1 | One variable and sensible solving attempt, z-value not necessary |
| \(\mu = 3.97\) | A1 [4] | Both answers correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p = 0.85\), \(\mu = 200 \times 0.85 = 170\) | B1 | \(200 \times 0.85\ (170)\) and \(200 \times 0.85 \times 0.15\ (25.5)\) seen |
| \(\text{var} = 200 \times 0.85 \times 0.15 = 25.5\) | M1 | Standardising, sq rt and must have used 200 |
| \(P(\text{at least } 160) = P\!\left(z > \frac{159.5-170}{\sqrt{25.5}}\right)\) | M1 | Continuity correction 159.5 or 160.5 |
| \(= P(z > -2.079)\) | M1 | Correct area (\(> 0.5\)) must have used 200 |
| \(= 0.981\) | A1 [5] | Correct value |
## Question 4:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = 1.036$ or $1.037$ | B1 | $\pm 1.036$ or $\pm 1.037$ seen |
| $1.036 = \frac{5-4s}{s}$ | B1 | $\frac{5-4\sigma}{\sigma}$ seen or $\frac{5-\mu}{\mu/4}$ oe |
| $s = 0.993$ | M1 | One variable and sensible solving attempt, z-value not necessary |
| $\mu = 3.97$ | A1 [4] | Both answers correct |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $p = 0.85$, $\mu = 200 \times 0.85 = 170$ | B1 | $200 \times 0.85\ (170)$ and $200 \times 0.85 \times 0.15\ (25.5)$ seen |
| $\text{var} = 200 \times 0.85 \times 0.15 = 25.5$ | M1 | Standardising, sq rt and must have used 200 |
| $P(\text{at least } 160) = P\!\left(z > \frac{159.5-170}{\sqrt{25.5}}\right)$ | M1 | Continuity correction 159.5 or 160.5 |
| $= P(z > -2.079)$ | M1 | Correct area ($> 0.5$) must have used 200 |
| $= 0.981$ | A1 [5] | Correct value |
---4 The mean of a certain normally distributed variable is four times the standard deviation. The probability that a randomly chosen value is greater than 5 is 0.15 .\\
(i) Find the mean and standard deviation.\\
(ii) 200 values of the variable are chosen at random. Find the probability that at least 160 of these values are less than 5 .
\hfill \mbox{\textit{CAIE S1 2012 Q4 [9]}}