IB DP Maths Topic 7.2 Mean of linear combinations of n random variables HL Paper 3

 

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Question

(a)     A random variable, X , has probability density function defined by

\[f(x) = \left\{ {\begin{array}{*{20}{l}}
  {100,}&{{\text{for }} – 0.005 \leqslant x < 0.005} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Determine E(X) and Var(X) .

(b)     When a real number is rounded to two decimal places, an error is made.

Show that this error can be modelled by the random variable X .

(c)     A list contains 20 real numbers, each of which has been given to two decimal places. The numbers are then added together.

(i)     Write down bounds for the resulting error in this sum.

(ii)     Using the central limit theorem, estimate to two decimal places the probability that the absolute value of the error exceeds 0.01.

(iii)     State clearly any assumptions you have made in your calculation.

▶️Answer/Explanation

Markscheme

(a)     f(x)is even (symmetrical about the origin)     (M1)

\({\text{E}}(X) = 0\)     A1

\({\text{Var}}(X) = {\text{E}}({X^2}) = \int_{ – 0.005}^{0.005} {100{x^2}{\text{d}}x} \)     (M1)(A1)

\( = 8.33 \times {10^{ – 6}}\left( {{\text{accept }}0.83 \times {{10}^{ – 5}}{\text{ or }}\frac{1}{{120\,000}}} \right)\)     A1

[5 marks]

 

(b)     rounding errors to 2 decimal places are uniformly distributed     R1

and lie within the interval \( – 0.005 \leqslant x < 0.005.\)     R1

this defines X     AG

[2 marks]

 

(c)     (i)     using the symbol y to denote the error in the sum of 20 real numbers each rounded to 2 decimal places

\( – 0.1 \leqslant y( = 20 \times x) < 0.1\)     A1

 

(ii)     \(Y \approx {\text{N}}(20 \times 0,{\text{ }}20 \times 8.3 \times {10^{ – 6}}) = {\text{N}}(0,{\text{ }}0.00016)\)     (M1)(A1)

\({\text{P}}\left( {\left| Y \right| > 0.01} \right) = 2\left( {1 – {\text{P}}(Y < 0.01)} \right)\)     (M1)(A1)

\( = 2\left( {1 – {\text{P}}\left( {Z < \frac{{0.01}}{{0.0129}}} \right)} \right)\)

\( = 0.44\) to 2 decimal places     A1     N4

 

(iii)     it is assumed that the errors in rounding the 20 numbers are independent     R1

and, by the central limit theorem, the sum of the errors can be modelled approximately by a normal distribution     R1

[8 marks]

Total [15 marks]

Examiners report

This was the only question on the paper with a conceptually ‘hard’ final part. Part(a) was generally well done, either by integration or by use of the standard formulae for a uniform distribution. Many candidates were not able to provide convincing reasoning in parts (b) and (c)(iii). Part(c)(ii), the application of the Central Limit Theorem was only very rarely tackled competently.

Question

Two students are selected at random from a large school with equal numbers of boys and girls. The boys’ heights are normally distributed with mean \(178\) cm and standard deviation \(5.2\) cm, and the girls’ heights are normally distributed with mean \(169\) cm and standard deviation \(5.4\) cm.

Calculate the probability that the taller of the two students selected is a boy.

▶️Answer/Explanation

Markscheme

let \(X\) denote boys’ height and \(Y\) denote girls’ height

if \(BB,{\text{ P(taller is boy)}} = 1\)     (A1)

if \(GG,{\text{ P(taller is boy)}} = 0\)     (A1)

if \(BG\) or \(GB\):

consider \(X – Y\)     (M1)

\(E(X – Y) = 178 – 169 = 9\)     A1

\({\text{Var}}(X – Y) = {5.2^2} + {5.4^2}\;\;\;( = 56.2)\)     (M1)A1

\({\text{P}}(X – Y > 0) = 0.885\)     A1

answer is \(\frac{1}{4} \times 1 + \frac{1}{2} \times 0.885 = 0.693\)     (M1)A1

[9 marks]

Examiners report

[N/A]

Question

Adam does the crossword in the local newspaper every day. The time taken by Adam, \(X\) minutes, to complete the crossword is modelled by the normal distribution \({\text{N}}(22,{\text{ }}{5^2})\).

Beatrice also does the crossword in the local newspaper every day. The time taken by Beatrice, \(Y\) minutes, to complete the crossword is modelled by the normal distribution \({\text{N}}(40,{\text{ }}{6^2})\).

a.Given that, on a randomly chosen day, the probability that he completes the crossword in less than \(a\) minutes is equal to 0.8, find the value of \(a\).[3]

b.Find the probability that the total time taken for him to complete five randomly chosen crosswords exceeds 120 minutes.[3]

c.Find the probability that, on a randomly chosen day, the time taken by Beatrice to complete the crossword is more than twice the time taken by Adam to complete the crossword. Assume that these two times are independent.[6]

▶️Answer/Explanation

Markscheme

\(z = 0.841 \ldots \)    (A1)

\(a = \mu  + z\sigma \)    (M1)

\( = 26.2\)    A1

[3 marks]

a.

let \(T\) denote the total time taken to complete 5 crosswords.

\(T\) is \({\text{N}}(110,{\text{ }}125)\)     (A1)(A1)

Note:     A1 for the mean and A1 for the variance.

\({\text{P}}(T > 120) = 0.186\)    A1

[3 marks]

b.

consider the random variable \(U = Y – 2X\)     (M1)

\({\text{E}}(U) =  – 4\)    A1

\({\text{Var}}(U) = {\text{Var}}(Y) + 4{\text{Var}}(X)\)    (M1)

\( = 136\)    A1

\({\text{P}}(Y > 2X) = {\text{P}}(U > 0)\)    (M1)

\( = 0.366\)    A1

[6 marks]

c.

Examiners report

Part (a) was very well answered with only a very few weak candidates using 0.8 instead of 0.841…

a.

Part (b) was well answered with only a few candidates calculating the variance incorrectly.

b.

Part (c) was again well answered. The most common errors, not often seen, were writing the variance of \(Y – 2X\) as either \({\text{Var}}(Y) + 2{\text{Var}}(X)\) or \({\text{Var}}(Y) – 2(or{\text{ }}4){\text{Var}}(X)\).

c.

Question

Alun answers mathematics questions and checks his answer after doing each one.

The probability that he answers any question correctly is always \(\frac{6}{7}\), independently of all other questions. He will stop for coffee immediately following a second incorrect answer. Let \(X\) be the number of questions Alun answers before he stops for coffee.

Nic answers mathematics questions and checks his answer after doing each one.

The probability that he answers any question correctly is initially \(\frac{6}{7}\). After his first incorrect answer, Nic loses confidence in his own ability and from this point onwards, the probability that he answers any question correctly is now only \(\frac{4}{7}\).

Both before and after his first incorrect answer, the result of each question is independent of the result of any other question. Nic will also stop for coffee immediately following a second incorrect answer. Let \(Y\) be the number of questions Nic answers before he stops for coffee.

a.(i)     State the distribution of \(X\), including its parameters.

(ii)     Calculate \({\text{E}}(X)\).

(iii)     Calculate \({\text{P}}(X = 5)\).[6]

b.(i)     Calculate \({\text{E}}(Y)\).

(ii)     Calculate \({\text{P}}(Y = 5)\).[9]

▶️Answer/Explanation

Markscheme

(i)     \({\text{NB}}\left( {2,{\text{ }}\frac{1}{7}} \right)\)     A1A1A1

Note: The final A1 mark can be awarded for knowing that \(p = \frac{1}{7}\) independent of the other two marks.

(ii)     \({\text{E}}(X) = \frac{r}{p} = 14\)     A1

(iii)     \(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right){\left( {\frac{6}{7}} \right)^3}{\left( {\frac{1}{7}} \right)^2} = 0.0514\)     (M1)A1

Note: Accept any number that rounds to this 3sf number.

[6 marks]

a.

(i)     \(Y = {Y_1} + {Y_2}\) (number up to1st + number up to 2nd)     (M1)

\({Y_1} \sim Geo\left( {\frac{1}{7}} \right),{\text{ }}{Y_2} \sim Geo\left( {\frac{3}{7}} \right)\)    (A1)

Notes: The above (A1) is independent of the (M1).

Could have \({\text{NB }}(1,{\text{ }}p)\), instead of \(Geo(p)\).

\({\text{E}}(Y) = \frac{1}{{\left( {\frac{1}{7}} \right)}} + \frac{1}{{\left( {\frac{3}{7}} \right)}} = 7 + \frac{7}{3} = 9\frac{1}{3}{\text{ (9.33)}}\)    M1A1

(ii)     \(Y = {Y_1} + {Y_2} = 5\) happens when     (M1)

\({Y_1} = 1,{\text{ }}{Y_2} = 4\) or \({Y_1} = 2,{\text{ }}{Y_2} = 3\) or \({Y_1} = 3,{\text{ }}{Y_2} = 2\) or \({Y_1} = 4,{\text{ }}{Y_2} = 1\)     (A1)

so probability is \(\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{3}{7}\)     (M1)(A1)

\( = 0.0928{\text{ }}\left( {\frac{{1560}}{{16807}}} \right)\)    A1

Note: Accept any answer that rounds to 0.093.

[9 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

The weights, X kg, of the males of a species of bird may be assumed to be normally distributed with mean 4.8 kg and standard deviation 0.2 kg.

The weights, Y kg, of female birds of the same species may be assumed to be normally distributed with mean 2.7 kg and standard deviation 0.15 kg.

a.Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.[1]

b.Find the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.[6]

c.Two randomly chosen male birds and three randomly chosen female birds are placed on a weighing machine that has a weight limit of 18 kg. Find the probability that the total weight of these five birds is greater than the weight limit.[4]

▶️Answer/Explanation

Markscheme

Note: In question 1, accept answers that round correctly to 2 significant figures.

P(4.75 < X < 4.85) = 0.197      A1

[1 mark]

a.

Note: In question 1, accept answers that round correctly to 2 significant figures.

consider the random variable X − 2Y     (M1)

E(X − 2Y) =  − 0.6     (A1)

Var(X − 2Y) = Var(X) + 4Var(Y)     (M1)

= 0.13     (A1)

X − 2Y ∼ N(−0.6, 0.13)

P(X − 2Y > 0)     (M1)

= 0.0480     A1

[6 marks]

b.

Note: In question 1, accept answers that round correctly to 2 significant figures.

let W = X1 + X2 + Y1 + Y2 + Y3 be the total weight

E(W) = 17.7     (A1)

Var(W) = 2Var(X) + 3Var(Y) = 0.1475     (M1)(A1)

W ∼ N(17.7, 0.1475)

P(W > 18) = 0.217     A1

[4 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.
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