| Exam Board | OCR |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Topic | Geometric Distribution |
| Type | Mean/expectation of geometric distribution |
| Difficulty | Moderate -0.3 This is a straightforward application of standard geometric distribution formulas. Part (a) requires recalling E(X)=1/p, part (b) uses the CDF formula P(X≤k)=(1-q^k), and part (c) recognizes X~B(30,P(Y≥10)) then applies variance formula. All parts are direct formula application with minimal problem-solving, making it slightly easier than average but not trivial due to the multi-part structure and binomial-geometric combination in part (c). |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
1 Every time a spinner is spun, the probability that it shows the number 4 is 0.2 , independently of all other spins.
\begin{enumerate}[label=(\alph*)]
\item A pupil spins the spinner repeatedly until it shows the number 4 .
Find the mean of the number of spins required.
\item Calculate the probability that the number of spins required is between 3 and 10 inclusive.
\item Each pupil in a class of 30 spins the spinner until it shows the number 4 . Out of the 30 pupils, the number of pupils who require at least 10 spins is denoted by $X$.
Determine the variance of $X$.
\end{enumerate}
\hfill \mbox{\textit{OCR FS1 AS 2021 Q1 [8]}}