OCR H240/02 2019 June — Question 13 5 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2019
SessionJune
Marks5
PaperDownload PDF ↗
TopicApproximating Binomial to Normal Distribution
TypeFind minimum/maximum n for probability condition
DifficultyStandard +0.8 This question requires normal approximation to binomial with continuity correction, finding a critical value using inverse normal tables, and careful interpretation of strict inequality P(X<n)<0.025 to determine the largest integer n. The multi-step process and need for precise handling of the inequality boundary makes this moderately challenging, though the underlying techniques are standard A-level material.
Spec2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial

13 It is known that \(26 \%\) of adults in the UK use a certain app. A researcher selects a random sample of 5000 adults in the UK. The random variable \(X\) is defined as the number of adults in the sample who use the app. Given that \(\mathrm { P } ( X < n ) < 0.025\), calculate the largest possible value of \(n\).

Question 13:
AnswerMarks Guidance
\(N(5000 \times 0.26,\ 5000 \times 0.26 \times 0.74)\) may be implied \(= N(1300, 962)\)M1 or, using Binomial not Normal, \(B(5000, 0.26)\) soi; Attempt \(P(X < n)\) for \(1230 \leq n \leq 1250\)
\(1300 - 2 \times \sqrt{962}\) or \(1300 - 1.96 \times \sqrt{962}\); \((= 1238)\) or \((= 1239)\)M1 or \(\phi^{-1}(0.025)\) may be implied; or \(-1.96\) or \(1239(.2)\) seen; or inverse Bin\((0.025)\) \((= 1239)\)
\(P(X < 1239)\) or \(P(X \leq 1238)\) or \(P(X < 1240)\) or \(P(X \leq 1239)\)M1 One of these attempted by Binomial or Normal
\(P(X \leq 1238) = 0.0233\) OR \(P(X \leq 1239) = 0.0251\)A1 BC; Allow 0.0232 instead of 0.0233
\(P(X \leq 1238) = 0.0233\) AND \(P(X \leq 1239) = 0.0251\) AND Largest \(n\) is 1239 or \(n \leq 1239\)A1 [5] Correct \(P(X \leq 1238)\) and \(P(X \leq 1239)\) seen and conclusion; If use normal to find these probs \((0.0265\) & \(0.0246)\) A0A0; NB If two methods used, mark the better one.; NB: SC 1239, no working or incorrect or inadequate working — SC: 4 marks out of 5
# Question 13:
| $N(5000 \times 0.26,\ 5000 \times 0.26 \times 0.74)$ may be implied $= N(1300, 962)$ | M1 | or, using Binomial not Normal, $B(5000, 0.26)$ soi; Attempt $P(X < n)$ for $1230 \leq n \leq 1250$ |
| $1300 - 2 \times \sqrt{962}$ or $1300 - 1.96 \times \sqrt{962}$; $(= 1238)$ or $(= 1239)$ | M1 | or $\phi^{-1}(0.025)$ may be implied; or $-1.96$ or $1239(.2)$ seen; or inverse Bin$(0.025)$ $(= 1239)$ |
| $P(X < 1239)$ or $P(X \leq 1238)$ or $P(X < 1240)$ or $P(X \leq 1239)$ | M1 | One of these attempted by Binomial or Normal |
| $P(X \leq 1238) = 0.0233$ OR $P(X \leq 1239) = 0.0251$ | A1 | **BC**; **Allow 0.0232 instead of 0.0233** |
| $P(X \leq 1238) = 0.0233$ AND $P(X \leq 1239) = 0.0251$ AND Largest $n$ is 1239 or $n \leq 1239$ | A1 [5] | Correct $P(X \leq 1238)$ and $P(X \leq 1239)$ seen and conclusion; If use normal to find these probs $(0.0265$ & $0.0246)$ A0A0; **NB If two methods used, mark the better one.**; NB: SC 1239, no working or incorrect or inadequate working — SC: 4 marks out of 5 |
13 It is known that $26 \%$ of adults in the UK use a certain app. A researcher selects a random sample of 5000 adults in the UK. The random variable $X$ is defined as the number of adults in the sample who use the app.

Given that $\mathrm { P } ( X < n ) < 0.025$, calculate the largest possible value of $n$.

\hfill \mbox{\textit{OCR H240/02 2019 Q13 [5]}}