| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.8 Part (i) is a standard inverse normal calculation requiring z-tables. Part (ii) requires recognizing that 'within 1 standard deviation' gives P≈0.68 from normal distribution, then applying binomial distribution B(9,0.68) to find P(X>7), combining two distinct probability distributions in a non-routine way that goes beyond typical textbook exercises. |
| Spec | 2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = -1.282\) | B1 | Rounding to \(\pm 1.28\) seen |
| \(-1.282 = \dfrac{t - 6.5}{1.76}\) | M1 | Standardising, no cc, no sq or sq rt, \(z \neq \pm 0.9, \pm 0.1\) |
| \(t = 4.24\) | A1 | Correct answer, accept 4.25 — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(z < 1) = 0.8413\) | M1 | \(z=1\) used to find a probability |
| \(P(\text{within 1sd of mean}) = 2\Phi - 1 = 0.6826\) | B1 | Correct prob, accept answer rounding to 0.66, 0.67, 0.68, not from wrong working. If quoted, then implies first M1 |
| \(P(8,9) = {}^9C_8(0.6826)^8(0.3174) + (0.6826)^9\) | M1 | Binomial term \(p^r(1-p)^{9-r}{}^9C_r\), \({}^9C_r\) must be seen |
| M1 | Binomial expression for \(P(8)+P(9)\), any \(p\) | |
| \(= 0.167\) | A1 | Correct answer — Total: 5 |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = -1.282$ | B1 | Rounding to $\pm 1.28$ seen |
| $-1.282 = \dfrac{t - 6.5}{1.76}$ | M1 | Standardising, no cc, no sq or sq rt, $z \neq \pm 0.9, \pm 0.1$ |
| $t = 4.24$ | A1 | Correct answer, accept 4.25 — **Total: 3** |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(z < 1) = 0.8413$ | M1 | $z=1$ used to find a probability |
| $P(\text{within 1sd of mean}) = 2\Phi - 1 = 0.6826$ | B1 | Correct prob, accept answer rounding to 0.66, 0.67, 0.68, not from wrong working. If quoted, then implies first M1 |
| $P(8,9) = {}^9C_8(0.6826)^8(0.3174) + (0.6826)^9$ | M1 | Binomial term $p^r(1-p)^{9-r}{}^9C_r$, ${}^9C_r$ must be seen |
| | M1 | Binomial expression for $P(8)+P(9)$, any $p$ |
| $= 0.167$ | A1 | Correct answer — **Total: 5** |
---5 When Moses makes a phone call, the amount of time that the call takes has a normal distribution with mean 6.5 minutes and standard deviation 1.76 minutes.\\
(i) $90 \%$ of Moses's phone calls take longer than $t$ minutes. Find the value of $t$.\\
(ii) Find the probability that, in a random sample of 9 phone calls made by Moses, more than 7 take a time which is within 1 standard deviation of the mean.
\hfill \mbox{\textit{CAIE S1 2014 Q5 [8]}}