Standard +0.3 This is a straightforward application of the Central Limit Theorem to find a probability for a sample mean. Students need to recognize that the sampling distribution of the mean is approximately normal, calculate the standard error (√(1.25/50)), standardize to get a z-score, and look up the normal probability. While it requires understanding of the CLT and careful arithmetic, it's a standard textbook exercise with no novel insight required, making it slightly easier than average.
The discrete random variable \(H\) takes values 1, 2, 3 and 4. It is given that E(\(H\)) = 2.5 and Var(\(H\)) = 1.25. The mean of a random sample of 50 observations of \(H\) is denoted by \(\bar{H}\).
Use a suitable approximation to find P(\(\bar{H} < 2.6\)). [5]
Product of two "\(C_r\)" divided by "\(C_6\)" or "\(^rC_s(\frac{12}{20} × \frac{11}{19} × \frac{10}{18} × \frac{9}{17} × \frac{8}{16} × \frac{7}{15})\)"
\(= 0.3576\)
A1
Final answer, exact fraction or a.r.t. 0.358
[2]
\([= 231/646]\)
Question 5(ii):
Answer
Marks
Guidance
\(B(60, 0.6) \approx N(36, 14.4)\)
B1 B1
\(N(36, \ldots)\) or \(N(24, \ldots)\); 14.4 or \(\sqrt{14.4}\), both from \(B(60, 0.6)\)
Standardise with their \(np\) and \(\sqrt{np q}\) (or \(npq\))
A1
Both their \(\sqrt{npq}\) and cc correct
[30.5 if using 24]
A1
Answer in range [0.956, 0.957]
\(\text{No effect as population is large or yes but not by much}\)
B1
Need all of one of these [not "sample"], or equiv, nothing wrong
[6]
$N(2.5, 0.025)$ | M1 | Normal (any – can be implied by standardisation)
$\Phi\left(\frac{2.59-2.5}{\sqrt{0.025}}\right) = \Phi(0.5692)$ | A1 | Mean 2.5
| A1 | Variance or SD 1.25 ÷ 50 stated or used
| A1 | Standardise 2.59 or 2.61, with $\sqrt{(1.25/50)}$
$= 0.7154$ | A1 | Answer in range [0.715, 0.716] or [0.736, 0.737] from 0.632
| [5] |
## Question 5(i)(a):
$^6C_4 0.6^4 0.4^2 = 0.311[04]$ | M1 | This formula, allow 0.6 ↔ 0.4, or tables used correctly
| A1 | Final answer, exact fraction or a.r.t. 0.311
| [2] | $[= 972/3125]$
## Question 5(i)(b):
$^{12}C_4 \times ^8C_2 \times ^{30}C_6 = [495×28 + 38760]$ | M1 | Product of two "$C_r$" divided by "$C_6$" or "$^rC_s(\frac{12}{20} × \frac{11}{19} × \frac{10}{18} × \frac{9}{17} × \frac{8}{16} × \frac{7}{15})$"
$= 0.3576$ | A1 | Final answer, exact fraction or a.r.t. 0.358
| [2] | $[= 231/646]$
## Question 5(ii):
$B(60, 0.6) \approx N(36, 14.4)$ | B1 B1 | $N(36, \ldots)$ or $N(24, \ldots)$; 14.4 or $\sqrt{14.4}$, both from $B(60, 0.6)$
$1 - \Phi\left(\frac{29.5-36}{\sqrt{14.4}}\right) = 1 - \Phi(-1.713)$ | M1 | Standardise with their $np$ and $\sqrt{np q}$ (or $npq$)
| A1 | Both their $\sqrt{npq}$ and cc correct | [30.5 if using 24]
| A1 | Answer in range [0.956, 0.957]
$\text{No effect as population is large or yes but not by much}$ | B1 | Need all of one of these [not "sample"], or equiv, nothing wrong
| [6] |
The discrete random variable $H$ takes values 1, 2, 3 and 4. It is given that E($H$) = 2.5 and Var($H$) = 1.25. The mean of a random sample of 50 observations of $H$ is denoted by $\bar{H}$.
Use a suitable approximation to find P($\bar{H} < 2.6$). [5]
\hfill \mbox{\textit{OCR S2 2012 Q4 [5]}}