| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Single binomial probability calculation |
| Difficulty | Moderate -0.3 This question involves straightforward probability calculations from frequency tables and basic binomial distribution computations (finding P(X≥6) and expected values). While it requires multiple steps and interpretation, all techniques are standard A-level statistics procedures with no novel problem-solving required. The context is slightly more complex than typical textbook exercises, but the mathematical demands are routine. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Daily mean total cloud cover (oktas) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency (number of days) | 0 | 1 | 4 | 7 | 10 | 30 | 52 | 52 | 28 |
| Daily mean total cloud cover (oktas) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency (number of days) | 0 | 0 | 1 | 1 | 2 | 1 | 5 | 9 | 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{132}{184} = 0.71739...\) awrt 0.717 | B1 | Allow equivalent fraction e.g. \(\frac{33}{46}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X \geq 6) = 1 - P(X \leq 5)\) or \(P([X=]6)+P([X=]7)+P([X=]8)\) | M1 | Writing or using \(1-P(X\leq 5)\) or sum of three terms |
| \(= 1 - 0.296722...\) awrt 0.703 | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(184 \times P(X=7)\) \([= 184 \times 0.2811...]\) | M1 | e.g. \(184 \times [P(X \leq 7) - P(X \leq 6)]\) |
| \(= 51.7385...\) awrt 51.7 | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Parts (a) and (b)(i) similar and expected number of 7s (51.7 or 0.281) matches number of 7s in data (52 or 0.283) so Magali's model is supported | B1ft | Comparing prob. with prob. and days with days; allow not supported or mixed conclusions if consistent with f.t. answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{23}{28} = 0.82142...\) awrt 0.821 | B1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Part (d)/0.821 differs from part (a)/(b)(i)/0.7... or greater/different probability of high cloud cover/more likely to have high cloud if previous day had high cloud cover; independence does not hold | B1 | Any one bullet point |
| ...therefore Magali's (binomial) model may not be suitable | dB1 | Dependent on previous B1; condone not accurate for "not suitable" |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{132}{184} = 0.71739...$ awrt **0.717** | B1 | Allow equivalent fraction e.g. $\frac{33}{46}$ |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \geq 6) = 1 - P(X \leq 5)$ or $P([X=]6)+P([X=]7)+P([X=]8)$ | M1 | Writing or using $1-P(X\leq 5)$ or sum of three terms |
| $= 1 - 0.296722...$ awrt **0.703** | A1 | — |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $184 \times P(X=7)$ $[= 184 \times 0.2811...]$ | M1 | e.g. $184 \times [P(X \leq 7) - P(X \leq 6)]$ |
| $= 51.7385...$ awrt **51.7** | A1 | — |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Parts (a) and (b)(i) similar **and** expected number of 7s (51.7 or 0.281) matches number of 7s in data (52 or 0.283) so Magali's model is supported | B1ft | Comparing prob. with prob. **and** days with days; allow not supported or mixed conclusions if consistent with f.t. answers |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{23}{28} = 0.82142...$ awrt **0.821** | B1 | — |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Part (d)/0.821 differs from part (a)/(b)(i)/0.7... **or** greater/different probability of high cloud cover/more likely to have high cloud if previous day had high cloud cover; independence does not hold | B1 | Any one bullet point |
| ...therefore Magali's (binomial) model may not be suitable | dB1 | Dependent on previous B1; condone not accurate for "not suitable" |
---\begin{enumerate}
\item Magali is studying the mean total cloud cover, in oktas, for Leuchars in 1987 using data from the large data set. The daily mean total cloud cover for all 184 days from the large data set is summarised in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | c | c | c | }
\hline
Daily mean total cloud cover (oktas) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Frequency (number of days) & 0 & 1 & 4 & 7 & 10 & 30 & 52 & 52 & 28 \\
\hline
\end{tabular}
\end{center}
One of the 184 days is selected at random.\\
(a) Find the probability that it has a daily mean total cloud cover of 6 or greater.
Magali is investigating whether the daily mean total cloud cover can be modelled using a binomial distribution.
She uses the random variable $X$ to denote the daily mean total cloud cover and believes that $X \sim \mathrm {~B} ( 8,0.76 )$
Using Magali's model,\\
(b) (i) find $\mathrm { P } ( X \geqslant 6 )$\\
(ii) find, to 1 decimal place, the expected number of days in a sample of 184 days with a daily mean total cloud cover of 7\\
(c) Explain whether or not your answers to part (b) support the use of Magali's model.
There were 28 days that had a daily mean total cloud cover of 8 For these 28 days the daily mean total cloud cover for the following day is shown in the table below.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | l | l | }
\hline
Daily mean total cloud cover (oktas) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Frequency (number of days) & 0 & 0 & 1 & 1 & 2 & 1 & 5 & 9 & 9 \\
\hline
\end{tabular}
\end{center}
(d) Find the proportion of these days when the daily mean total cloud cover was 6 or greater.\\
(e) Comment on Magali's model in light of your answer to part (d).
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q4 [9]}}