CAIE S1 2009 November — Question 5 8 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2009
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeProbability distribution from formula
DifficultyModerate -0.3 This is a straightforward discrete probability distribution question requiring basic substitution and table construction. Part (i) is simple verification, parts (ii-iii) involve routine calculations and identifying the mode, and part (iv) requires adding probabilities. The arithmetic is slightly tedious but conceptually this is below average difficulty for A-level, requiring only basic probability distribution understanding with no problem-solving insight.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

5 In a particular discrete probability distribution the random variable \(X\) takes the value \(\frac { 120 } { r }\) with probability \(\frac { r } { 45 }\), where \(r\) takes all integer values from 1 to 9 inclusive.
  1. Show that \(\mathrm { P } ( X = 40 ) = \frac { 1 } { 15 }\).
  2. Construct the probability distribution table for \(X\).
  3. Which is the modal value of \(X\) ?
  4. Find the probability that \(X\) lies between 18 and 100 .

5 In a particular discrete probability distribution the random variable $X$ takes the value $\frac { 120 } { r }$ with probability $\frac { r } { 45 }$, where $r$ takes all integer values from 1 to 9 inclusive.\\
(i) Show that $\mathrm { P } ( X = 40 ) = \frac { 1 } { 15 }$.\\
(ii) Construct the probability distribution table for $X$.\\
(iii) Which is the modal value of $X$ ?\\
(iv) Find the probability that $X$ lies between 18 and 100 .

\hfill \mbox{\textit{CAIE S1 2009 Q5 [8]}}