IB DP Maths Topic 5.8 Mean and variance of the binomial distribution SL Paper 2

Question

A factory makes switches. The probability that a switch is defective is 0.04. The factory tests a random sample of 100 switches.

Find the mean number of defective switches in the sample.

[2]
a.

Find the probability that there are exactly six defective switches in the sample.

[2]
b.

Find the probability that there is at least one defective switch in the sample.

[3]
c.
Answer/Explanation

Markscheme

evidence of binomial distribution (may be seen in parts (b) or (c))     (M1)

e.g. np, \(100 \times 0.04\)

\({\text{mean}} = 4\)     A1     N2

[2 marks]

a.

\({\rm{P}}(X = 6) = \left( {\begin{array}{*{20}{c}}
{100}\\
6
\end{array}} \right){(0.04)^6}{(0.96)^{94}}\)     (A1)

\( = 0.105\)     A1     N2

[2 marks]

b.

for evidence of appropriate approach     (M1)

e.g. complement, \(1 – {\rm{P}}(X = 0)\)

\({\rm{P}}(X = 0) = {(0.96)^{100}} = 0.01687 \ldots \)     (A1)

\({\rm{P}}(X \ge 1) = 0.983\)     A1     N2

[3 marks]

c.

Question

Paula goes to work three days a week. On any day, the probability that she goes on a red bus is \(\frac{1}{4}\) .

Write down the expected number of times that Paula goes to work on a red bus in one week.

[2]
a.

In one week, find the probability that she goes to work on a red bus on exactly two days.

[2]
b.

In one week, find the probability that she goes to work on a red bus on at least one day.

[3]
c.
Answer/Explanation

Markscheme

evidence of binomial distribution (seen anywhere)     (M1)

e.g. \(X \sim {\text{B}}\left( {3{\text{, }}\frac{1}{4}} \right)\)

\({\rm{mean}} = \frac{3}{4}\) (\(= 0.75\))     A1     N2

[2 marks]

a.

\({\rm{P}}(X = 2) = \left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right){\left( {\frac{1}{4}} \right)^2}\left( {\frac{3}{4}} \right)\)   
 (A1)

\({\rm{P}}(X = 2) = 0.141\) \(\left( { = \frac{9}{{64}}} \right)\)     A1     N2

[2 marks]

b.

evidence of appropriate approach     M1

e.g. complement, \(1 – {\rm{P}}(X = 0)\) , adding probabilities

\({\rm{P}}(X = 0) = {(0.75)^3}\) \(\left( { = 0.422,\frac{{27}}{{64}}} \right)\)     (A1)

\({\rm{P}}(X \ge 1) = 0.578\) \(\left( { = \frac{{37}}{{64}}} \right)\)     A1    N2

[3 marks]

c.

Question

A test has five questions. To pass the test, at least three of the questions must be answered correctly.

The probability that Mark answers a question correctly is \(\frac{1}{5}\) . Let X be the number of questions that Mark answers correctly.

Bill also takes the test. Let Y be the number of questions that Bill answers correctly.

The following table is the probability distribution for Y .


(i)     Find E(X ) .

(ii)    Find the probability that Mark passes the test.

[6]
a(i) and (ii).

(i)     Show that \(4a + 2b = 0.24\) .

(ii)    Given that \({\rm{E}}(Y) = 1\) , find a and b .

[8]
b(i) and (ii).

Find which student is more likely to pass the test.

[3]
c.
Answer/Explanation

Markscheme

(i) valid approach     (M1)

e.g. \(np\) , \(5 \times \frac{1}{5}\)

\({\rm{E}}(X) = 1\)     A1     N2

(ii) evidence of appropriate approach involving binomial     (M1)

e.g. \(X \sim B\left( {5,\frac{1}{5}} \right)\)

recognizing that Mark needs to answer 3 or more questions correctly     (A1)

e.g. \({\rm{P}}(X \ge 3)\)

valid approach     M1

e.g. \(1 – {\rm{P}}(X \le 2)\) , \({\rm{P}}(X = 3) + {\rm{P}}(X = 4) + {\rm{P}}(X = 5)\)

\({\text{P(pass)}} = 0.0579\)     A1     N3

[6 marks]

a(i) and (ii).

(i) evidence of summing probabilities to 1     (M1)

e.g. \(0.67 + 0.05 + (a + 2b) + \ldots + 0.04 = 1\)

some simplification that clearly leads to required answer

e.g. \(0.76 + 4a + 2b = 1\)     A1

\(4a + 2b = 0.24\)     AG      N0

(ii) correct substitution into the formula for expected value     (A1)

e.g. \(0(0.67) + 1(0.05) + \ldots + 5(0.04)\)

some simplification     (A1)

e.g. \(0.05 + 2a + 4b + \ldots + 5(0.04) = 1\)

correct equation     A1

e.g. \(13a + 5b = 0.75\)

evidence of solving     (M1)

\(a = 0.05\) , \(b = 0.02\)     A1A1     N4

[8 marks]

b(i) and (ii).

attempt to find probability Bill passes     (M1)

e.g. \({\rm{P}}(Y \ge 3)\)

correct value 0.19     A1

Bill (is more likely to pass)     A1     N0

[3 marks]

c.

Question

A box holds 240 eggs. The probability that an egg is brown is 0.05.

Find the expected number of brown eggs in the box.

[2]
a.

Find the probability that there are 15 brown eggs in the box.

[2]
b.

Find the probability that there are at least 10 brown eggs in the box.

[3]
c.
Answer/Explanation

Markscheme

correct substitution into formula for \({\rm{E}}(X)\)     (A1)

e.g. \(0.05 \times 240\)

\({\rm{E}}(X) = 12\)     A1     N2

[2 marks]

a.

evidence of recognizing binomial probability (may be seen in part (a))     (M1)

e.g. \(\left( {\begin{array}{*{20}{c}}
{240}\\
{15}
\end{array}} \right){(0.05)^{15}}{(0.95)^{225}}\) , \(X \sim {\rm{B}}(240,0.05)\)

\({\rm{P}}(X = 15) = 0.0733\)     A1     N2

[2 marks]

b.

\({\rm{P}}(X \le 9) = 0.236\)     (A1)

evidence of valid approach     (M1)

e.g. using complement, summing probabilities

\({\rm{P}}(X \ge 10) = 0.764\)     A1     N3

[3 marks]

c.

Question

A forest has a large number of tall trees. The heights of the trees are normally distributed with a mean of \(53\) metres and a standard deviation of \(8\) metres. Trees are classified as giant trees if they are more than \(60\) metres tall.

A tree is selected at random from the forest.

Find the probability that this tree is a giant.

[3]
a(i).

A tree is selected at random from the forest.

Given that this tree is a giant, find the probability that it is taller than \(70\) metres.

[3]
a(ii).

Two trees are selected at random. Find the probability that they are both giants.

[2]
b.

\(100\) trees are selected at random.

Find the expected number of these trees that are giants.

[3]
c(i).

\(100\) trees are selected at random.

Find the probability that at least \(25\) of these trees are giants.

[3]
c(ii).
Answer/Explanation

Markscheme

valid approach     (M1)

eg     \({\text{P}}(G) = {\text{P}}(H > 60,{\text{ }}z = 0.875,{\text{ P}}(H > 60) = 1 – 0.809,{\text{ N}}\left( {53, {8^2}} \right)\)

\(0.190786\)

\({\text{P}}(G) = 0.191\)     A1     N2

[3 marks]

a(i).

finding \({\text{P}}(H > 70) = 0.01679\)  (seen anywhere)     (A1)

recognizing conditional probability     (R1)

eg     \({\text{P}}(A\left| {B),{\text{ P}}(H > 70\left| {H > 60)} \right.} \right.\)

correct working     (A1)

eg     \(\frac{{0.01679}}{{0.191}}\)

\(0.0880209\)

\({\text{P}}(X > 70\left| {G) = 0.0880} \right.\)     A1     N3

[6 marks]

a(ii).

attempt to square their \({\text{P}}(G)\)     (M1)

eg     \({0.191^2}\)

\(0.0363996\)

\({\text{P}}({\text{both }}G) = 0.0364\)     A1     N2

[2 marks]

b.

correct substitution into formula for \({\text{E}}(X)\)     (A1)

eg     \(100(0.191)\)

\({\text{E}}(G) = 19.1{\text{ }}[19.0,{\text{ }}19.1]\)     A1     N2

[3 marks]

c(i).

recognizing binomial probability (may be seen in part (c)(i))     (R1)

eg     \(X \sim {\text{B}}(n,{\text{ }}p)\)

valid approach (seen anywhere)     (M1)

eg     \({\text{P}}(X \geqslant 25) = 1 – {\text{P}}(X \leqslant 24),{\text{ }}1 – {\text{P}}(X < a)\)

correct working     (A1)

eg     \({\text{P}}(X \leqslant 24) = 0.913 \ldots ,{\text{ }}1 – 0.913 \ldots \)

\(0.0869002\)

\({\text{P}}(X \geqslant 25) = 0.0869\)     A1     N2

[3 marks]

c(ii).

Question

A factory has two machines, A and B. The number of breakdowns of each machine is independent from day to day.

Let \(A\) be the number of breakdowns of Machine A on any given day. The probability distribution for \(A\) can be modelled by the following table.

M16/5/MATME/SP2/ENG/TZ1/08

Let \(B\) be the number of breakdowns of Machine B on any given day. The probability distribution for \(B\) can be modelled by the following table.

M16/5/MATME/SP2/ENG/TZ1/08.c+d

On Tuesday, the factory uses both Machine A and Machine B. The variables \(A\) and \(B\) are independent.

Find \(k\).

[2]
a.

(i)     A day is chosen at random. Write down the probability that Machine A has no breakdowns.

(ii)     Five days are chosen at random. Find the probability that Machine A has no breakdowns on exactly four of these days.

[3]
b.

Find \({\text{E}}(B)\).

[2]
c.

(i)     Find the probability that there are exactly two breakdowns on Tuesday.

(ii)     Given that there are exactly two breakdowns on Tuesday, find the probability that both breakdowns are of Machine A.

[8]
d.
Answer/Explanation

Markscheme

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(0.55 + 0.3 + 0.1 + k = 1\)

\(k = 0.05{\text{ (exact)}}\)    A1     N2

[2 marks]

a.

(i)     0.55     A1     N1

(ii)     recognizing binomial probability     (M1)

eg\(\,\,\,\,\,\)\(X:{\text{ }}B(n,{\text{ }}p),{\text{ }}\left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right),{\text{ }}{(0.55)^4}(1 – 0.55),{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n – r}}\)

\(P(X = 4) = 0.205889\)

\(P(X = 4) = 0.206\)    A1     N2

[3 marks]

b.

correct substitution into formula for \({\text{E}}(X)\)     (A1)

eg\(\,\,\,\,\,\)\(0.2 + (2 \times 0.08) + (3 \times 0.02)\)

\({\text{E}}(B) = 0.42{\text{ (exact)}}\)    A1     N2

[2 marks]

c.

(i)     valid attempt to find one possible way of having 2 breakdowns     (M1)

eg\(\,\,\,\,\,\)\(2A,{\text{ }}2B,{\text{ }}1A\) and \(1B\), tree diagram

one correct calculation for 1 way (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(0.1 \times 0.7,{\text{ }}0.55 \times 0.08,{\text{ }}0.3 \times 0.2\)

recognizing there are 3 ways of having 2 breakdowns     (M1)

eg\(\,\,\,\,\,\)A twice or B twice or one breakdown each

correct working     (A1)

eg\(\,\,\,\,\,\)\((0.1 \times 0.7) + (0.55 \times 0.08) + (0.3 \times 0.2)\)

\({\text{P(2 breakdowns)}} = 0.174{\text{ (exact)}}\)     A1     N3

(ii)     recognizing conditional probability     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A|B),{\text{ P}}(2A|{\text{2breakdowns}})\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\frac{{0.1 \times 0.7}}{{0.174}}\)

\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402298\)

\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402\)     A1     N2

[8 marks]

d.

Question

The probability of obtaining heads on a biased coin is 0.4. The coin is tossed 600 times.

(i)     Write down the mean number of heads.

(ii)    Find the standard deviation of the number of heads.

[4]
a(i) and (ii).

Find the probability that the number of heads obtained is less than one standard deviation away from the mean.

[3]
b.
Answer/Explanation

Markscheme

(i) recognizing binomial with \(n = 600\) , \(p = 0.4\)     M1

\({\rm{E}}(X) = 240\)     A1     N2

(ii) correct substitution into formula for variance or standard deviation     A1

e.g. 144, \(\sqrt {600 \times 0.4 \times 0.6} \)

sd = 12     A1     N1

[4 marks]

a(i) and (ii).

attempt to find range of values     M1

e.g. \(240 \pm 12\) \(228 < X < 252\)

evidence of correct approach     A1

e.g. \({\rm{P}}(X \le 251) – {\rm{P}}(X \le 228)\)

\({\rm{P}}(228 < X < 252) = 0.662\)     A1     N2

[3 marks]

b.
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