Edexcel S2 — Question 3 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeNormal approximation to binomial
DifficultyModerate -0.3 This is a standard S2 binomial distribution question with straightforward application of formulas. Parts (a) and (b) require direct use of binomial probability calculations with n=8, p=0.3. Part (c) involves a routine normal approximation to binomial (n=150, p=0.3), which is a core S2 technique. All parts follow textbook procedures with no problem-solving insight required, making it slightly easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial

The sales staff at an insurance company make house calls to prospective clients. Past records show that 30% of the people visited will take out a new policy with the company. On a particular day, one salesperson visits 8 people. Find the probability that, of these,
  1. exactly 2 take out new policies, [3 marks]
  2. more than 4 take out new policies. [2 marks]
The company awards a bonus to any salesperson who sells more than 50 policies in a month.
  1. Using a suitable approximation, find the probability that a salesperson gets a bonus in a month in which he visits 150 prospective clients. [5 marks]

AnswerMarks Guidance
(a)let \(X =\) no. out of 8 who take out policies \(\therefore X \sim B(8, 0.3)\) M1
\(P(X = 2) = 0.5518 - 0.2553 = 0.2965\)M1 A1
(b)\(P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9420 = 0.0580\) M1 A1
(c)let \(Y =\) no. out of 150 who take out policies \(\therefore Y \sim B(150, 0.3)\) M1
\(N \text{ approx. } S \sim N(45, 31.5)\)M1
\(P(Y > 50) = P(S > 50.5)\)M1
\(= P\left(Z > \frac{50.5 - 45}{\sqrt{31.5}}\right) = P(Z > 0.98)\)A1
\(= 1 - 0.8365 = 0.1635\)A1 (10)
(a) | let $X =$ no. out of 8 who take out policies $\therefore X \sim B(8, 0.3)$ | M1 |
| $P(X = 2) = 0.5518 - 0.2553 = 0.2965$ | M1 A1 |
(b) | $P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9420 = 0.0580$ | M1 A1 |
(c) | let $Y =$ no. out of 150 who take out policies $\therefore Y \sim B(150, 0.3)$ | M1 |
| $N \text{ approx. } S \sim N(45, 31.5)$ | M1 |
| $P(Y > 50) = P(S > 50.5)$ | M1 |
| $= P\left(Z > \frac{50.5 - 45}{\sqrt{31.5}}\right) = P(Z > 0.98)$ | A1 |
| $= 1 - 0.8365 = 0.1635$ | A1 | (10)
The sales staff at an insurance company make house calls to prospective clients. Past records show that 30% of the people visited will take out a new policy with the company.

On a particular day, one salesperson visits 8 people. Find the probability that, of these,

\begin{enumerate}[label=(\alph*)]
\item exactly 2 take out new policies, [3 marks]
\item more than 4 take out new policies. [2 marks]
\end{enumerate}

The company awards a bonus to any salesperson who sells more than 50 policies in a month.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Using a suitable approximation, find the probability that a salesperson gets a bonus in a month in which he visits 150 prospective clients. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q3 [10]}}