Moderate -0.5 This is a straightforward application of the normal approximation to the binomial distribution with continuity correction. Students need to identify n=90, p=1/6, calculate mean and variance, apply continuity correction (P(X≥14) = P(Z≥13.5)), and use normal tables. It's slightly easier than average because it's a single probability calculation with clear parameters and a standard method, though the continuity correction and table lookup require care.
Standardise, \(np\) and \(npq\), allow errors in \(\sqrt{\text{or cc}}\) or both \(\sqrt{\text{and cc}}\) both right
\(= -0.424\)
A1
Final answer, a.r.t. 0.664. \([\text{Po}(15): 1/6]\)
0.6643
A1
6
Let $R$ be the number of 1s | A1 | 2 | Non-invalidating reason
$R \sim B(90, 1/6)$ | B1 | $B(90, 1/6)$ stated or implied
| B1 | Normal, $\mu = 15$ stated or implied
| B1 | 12.5 or $\sqrt{12.5}$ or $12.5^2$ seen
$\frac{13.5-15}{\sqrt{12.5}}$ | M1 | Standardise, $np$ and $npq$, allow errors in $\sqrt{\text{or cc}}$ or both $\sqrt{\text{and cc}}$ both right
$= -0.424$ | A1 | Final answer, a.r.t. 0.664. $[\text{Po}(15): 1/6]$
**0.6643** | A1 | 6 |
3 A fair dice is thrown 90 times. Use an appropriate approximation to find the probability that the number 1 is obtained 14 or more times.
\hfill \mbox{\textit{OCR S2 2007 Q3 [6]}}