CAIE S2 (Statistics 2) 2024 June

1 The random variable \(X\) has the distribution \(\mathrm { B } ( 4000,0.001 )\).

1. Use a suitable approximating distribution to find \(\mathrm { P } ( 2 \leqslant X < 5 )\).
2. Justify your approximating distribution in this case.

2 The widths, \(w \mathrm {~cm}\), of a random sample of 150 leaves of a certain kind were measured. The sample mean of \(w\) was found to be 3.12 cm . Using this sample, an approximate \(95 \%\) confidence interval for the population mean of the widths in centimetres was found to be [3.01, 3.23].

1. Calculate an estimate of the population standard deviation.
2. Explain whether it was necessary to use the Central Limit theorem in your answer to part (a). [1]

3 The masses in kilograms of large and small bags of cement have the independent distributions \(\mathrm { N } ( 50,2.4 )\) and \(\mathrm { N } ( 26,1.8 )\) respectively. Find the probability that the total mass of 5 randomly chosen large bags of cement is greater than the total mass of 10 randomly chosen small bags of cement.
\includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-04_2714_34_143_2012}
\includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-05_2724_35_136_20}

4 In this question you should not use an approximating distribution.
At an election in Menham last year, \(24 \%\) of voters supported the Today Party. A student wishes to test whether support for the Today Party has decreased since last year. He chooses a random sample of 25 voters in Menham and finds that exactly 2 of them say that they support the Today Party. Test at the 5\% significance level whether support for the Today Party has decreased.

5 A random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} a x - x ^ { 3 } & 0 \leqslant x \leqslant \sqrt { 2 }
0 & \text { otherwise } \end{cases}$$ where \(a\) is a constant.

1. Show that \(a = 2\) .
2. Find the median of \(X\) .
3. Find the exact value of \(\mathrm { E } ( X )\).

6 The numbers of green sweets in 200 randomly chosen packets of Frutos are summarised in the table.

1. Calculate an unbiased estimate for the population mean of the number of green sweets in a packet of Frutos, and show that an unbiased estimate of the population variance is 0.783 correct to 3 significant figures.
   The manufacturers of Frutos claim that the mean number of green sweets in a packet is 1.65 .
   Anji believes that the true value of the mean, \(\mu\), is less than 1.65 . She uses the results from the 200 randomly chosen packets to test the manufacturers’ claim.
2. State suitable null and alternative hypotheses for the test.
   \includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-08_2714_37_143_2008}
3. Show that the result of Anji's test is significant at the \(5 \%\) level but not at the \(1 \%\) level.
4. It is given that Anji made a Type I error. Explain how this shows that the significance level that Anji used in her test was not \(1 \%\).

7 The independent random variables \(X\) and \(Y\) have the distributions \(\operatorname { Po } ( 1.9 )\) and \(\operatorname { Po } ( 2.2 )\) respectively.

1. Find \(\mathrm { P } ( X + Y < 4 )\).
   \includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-10_74_1581_406_322}
   \includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-10_75_1581_497_322}
2. Find the probability that \(X = 2\) given that \(X + Y < 4\).
   \includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-10_2715_35_144_2012}
3. A sample of 60 randomly chosen pairs of values of \(X\) and \(Y\) is taken,and the value of \(X + Y\) is calculated for each pair.The sample mean of these 60 values is found. Find the probability that the sample mean of \(X + Y\) is less than 4.0 .
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.