Download DISCRETE VARIABLE with KEY

DISCRETE VARIABLE with KEY
1.
A biased die with four faces is used in a game. A player pays 10 counters to roll the die. The table
below shows the possible scores on the die, the probability of each score and the number of
counters the player receives in return for each score.
Score
1
2
3
4
Probability
1
2
1
5
1
5
1
10
Number of counters player receives
4
5
15
n
Find the value of n in order for the player to get an expected return of 9 counters per roll.
[4]
1.
Let X be the number of counters the player receives in return.
E(X) = p(x) × x = 9
1
 1  1
 1

   4     5     15     n  = 9
2
 5  5
  10

1

n=3
10
 n = 30
(M1)
(M1)(A1)
(A1) (C4)
[4]
2.
The quality control department of a company making computer chips knows that 2% of the chips
are defective. Use the normal approximation to the binomial probability distribution, with a
continuity correction, to find the probability that, in a batch containing 1000 chips, between 20
and 30 chips (inclusive) are defective.
[7]
2.
Let n = number of chips = 1000,
p = the probability that a randomly chosen chip is defective = 0.02.
Hence, the mean np = (1000) (0.02) = 20 and the
variance = np(1 – p) = (1000) (0.02) (0.98) = 19.6.
(A1)(A1)
Suppose X is the normal random variable that approximates the
binomial distribution.
The X  N (20, 19.6).
(M1)(M1)
 19.5  20
30.5  20 

Z 
Thus p(19.5  X  30.5) = p 
(M1)(M1)
19.6 
 19.6
= p(–0.11  Z  2.37)
= 0.5349
(A1)
Note: Line before last should be p(–0.113  Z  2.37) = 0.536.
Accept 0.535 or 0.536.
If student’s work is not shown but there is evidence that he/she
used the calculator to find the answer, accept the answer.
[7]
1
3.
A supplier of copper wire looks for flaws before despatching it to customers.
It is known that the number of flaws follow a Poisson probability distribution
with a mean of 2.3 flaws per metre.
(a)
Determine the probability that there are exactly 2 flaws in 1 metre of the wire.
(b)
Determine the probability that there is at least one flaw in 2 metres of the wire.
[3]
3.
Note: Throughout the whole question, students may be using their graphic display calculators
and should not be penalized if they do not show as much work as the marking scheme.
(a)
(b)
Let X denote the number of flaws in one metre of the wire.
(2.3) 2
Then E(X) = 2.3 flaws and P(X = 2) = e–2.3
(M1)(M1)
2!
= 0.265.
(A1)
Note: Award (C3) for a correct answer from a graphic display
calculator.
Let Y denote the number of flaws in two metres of wire.
Then Y has a Poisson distribution with mean E(Y) = 2 × 2.3 = 4.6
flaws for 2 metres.
Hence, P(Y  1) = 1 – P(Y = 0) = 1 – e–4.6
= 0.990 (3 sf)
–4.6
Note: Accept 1 – e
(M1)
(M1)
(A1)
3
3
.
[6]
4.
The continuous random variable X has probability density function f (x) where
 e  ke kx ,
fk (x) = 
 0,
(a)
0  x 1
otherwise
Show that k = 1.
(b) What is the probability that the random variable X has a value that lies between
1 and 1 ? Give your answer in terms of e.
4
2
(c)
Find the mean and variance of the distribution. Give your answers exactly, in terms of e.
The random variable X above represents the lifetime, in years, of a certain type of battery.
(d)
Find the probability that a battery lasts more than six months.
A calculator is fitted with three of these batteries. Each battery fails independently of the other
two. Find the probability that at the end of six months
(e)
(f)
4.
(a)
none of the batteries has failed;
exactly one of the batteries has failed.
For f (x) to be a probability distribution function,


1
0
(e – ke kx )dx = 1

1
0
f ( x ) dx  1 .
(M1)
2

kx
 ex  e

1
0
=1
(M1)
e–e +1
=1
k
e=e 
k
Thus f (x) = e – ex, 0  x  1
k
(b)

=
(c)

1/ 2
1/ 4
(e – e x )dx  ex  e x

1/ 2
1/ 4
(A1)
= 1 (AG)

e
e
 e 4 e
2
4
(M1)
e
 e4 e
4
=

1
0
(A1)
( e – e x ) dx 
1
 (ex  xe
0
1
 ex 2 
= 
 
 2 0
e
= –1
2
Variance =

1
0
3
xe x dx 
x
(M1)
)dx
e
 [ xe x – e x ]10
2
(M1)
(A1)

e

x 2 (e – e x )dx   – 1
0
2


1
2
(M1)
1
2
 ex 3

e

x
2
= 
 e ( x  2 x  2)   – 1
 3
0  2 
2
e2
=2– e–
+e–1
3
4
=1+
(d)
(M1)
e e2

3 4
(A1)

p(battery lasts more than 6 months) = p  x 

=

1
1/ 2
1

2
(e – e x ) dx
(M1)
= [ex  e x ]11 / 2
e
= e
or 0.290(3sf)
2
(e)
(f)
p(no battery failed) = p(all lasted more than 6 months)
3
e

e

 or 0.0243 (3 sf)
= 
2

 3 
e 
e
p(exactly one battery failed) =  1  e   e  
2
2
2


 
 0.179 (3 sf)
(A1)
(M1)
(A1)
2
(M1)
(A1)
[17]
3
5.
In a game a player rolls a biased tetrahedral (four-faced) die. The probability of each possible
score is shown below.
Score
1
2
3
4
Probability
1
5
2
5
1
10
x
Find the probability of a total score of six after two rolls.
[3]
5.
 P( X  x)  1
all x
1 2 1
 
+x=1
5 5 10
3
Therefore,
x=
10
Therefore,
(A1)
1 1
2 3 
P(scoring six after two rolls) =     2    
 10 10 
 5 10 
1
=
4
(M1)
(A1) (C3)
[3]
6.
The probability distribution of a discrete random variable X is given by
x
P(X = x) = k  2  , for x = 0, 1, 2, ......
3
Find the value of k.
6.
[3]
 P( X  x)
Since X is a random variable,
=1
all x
2
Therefore, k +
3
2
2
2
k    k    k + ......... = 1
3
3
3




1

 =1
k
2

1  
3

1
k=
3
(M1)
(M1)
(A1) (C3)
[3]
7. A satellite relies on solar cells for its power and will operate provided that at least one of the cells
is working. Cells fail independently of each other, and the probability that an individual cell fails
within one year is 0.8.
(a)
For a satellite with ten solar cells, find the probability that all ten cells fail within one year.
(b)
For a satellite with ten solar cells, find the probability that the satellite is still operating at
the end of one year.
(c)
For a satellite with n solar cells, write down the probability that the satellite is still
4
operating at the end of one year. Hence, find the smallest number of solar cells required so
that the probability of the satellite still operating at the end of one year is at least 0.95.
[9]
7.
(a)
P(all ten cells fail) = 0.810 = 0.107.
(b)
P(satellite is still operating at the end of one year)
= 1 – P (all ten cells fail within one year)
= 1 – 0.107
= 0.893.
(c)
(M1)(A1)
(M1)
(A1)
P(satellite is still operating at the end of one year)
= 1 – 0.8n.
(C1)
We require the smallest n for which 1 – 0.8n  0.95.
Thus, 0.8n  0.05
5
 
4
(M1)
n
 20
log 20
= 13.4
log 1.25
Therefore, 14 solar cells are needed.
n
(M1)(A1)
(C1)
[9]
8.
In a school,
of the students travel to school by bus. Five students are chosen at random. Find
the probability that exactly 3 of them travel to school by bus.
[3]
8.
Let p be the probability of choosing a student who travels to school by bus.
Let X be the random variable “the number of students who travel to
school by bus.”
Then X ~ B(n, p) with n = 5 and p =
 5  1   2 
Therefore P(X = 3) =     
 3  3   3 
40
=
or 0.165
243
3
(M1)
1
3
2
(using formulae and statistical tables)
(A1)
(A1)
[3]
9.
X is a binomial random variable, where the number of trials is 5 and the probability of success of
each trial is p. Find the values of p if P(X = 4) = 0.12.
[3]
9.
If X ~ Bin (5, p) and P(X = 4) = 0.12 then
5 4
  p (1 – p) = 0.12
 4
5p5 – 5p4 + 0.12 = 0
p = 0.459 (3 s.f ) or 0.973 (3 s.f)
(M1)
(A1)
(G1) (C3)
5
10.
(a)
Patients arrive at random at an emergency room in a hospital at the rate of 15 per hour
throughout the day. Find the probability that 6 patients will arrive at the emergency room
between 08:00 and 08:15.
(b) The emergency room switchboard has two operators. One operator answers calls for
doctors and the other deals with enquiries about patients. The first operator fails to answer
1% of her calls and the second operator fails to answer 3% of his calls. On a typical day, the
first and second telephone operators receive 20 and 40 calls respectively during an
afternoon session. Using the Poisson distribution find the probability that, between them,
the two operators fail to answer two or more calls during an afternoon session.
[8]
10. (a) Let X be the number of patients arriving at the emergency room
15 in
a 15 minute period. Rate of arrival in a 15 minute period =
= 3.75.
4
(3.75) 6 –3.75
P(X = 6) =
e
6!
= 0.0908
(A1)
(M1)
(M1)
OR
P(6 patients) = 0.0908
(b)
(G2)
Let F1, F2 be random variables which represent the number of
failures to answer telephone calls by the first and the second
operator, respectively.
F1 ~ P0 (0.01 × 20) = P0 (0.2).
F2 ~ P0 (0.03 × 40) = P0 (1.2).
Since F1 and F2 are independent
F1 + F2 ~ P0 (0.2 + 1.2) = P0(1.4)
P(F1 + F2  2) = 1 – P(F1 + F2 = 0) – P(F1 + F2 = 1)
= 1 – e–1.4 – (1.4)e–1.4 = 0.408
3
(A1)
(A1)
(M1)
(M1)
(A1)
OR
P(F1 + F2  2) = 0.408
(M0)(G2)
5
[8]
11.
A coin is biased so that when it is tossed the probability of obtaining heads is
2
. The coin is
3
tossed 1800 times. Let X be the number of heads obtained. Find
11.
(a)
the mean of X;
(b)
the standard deviation of X.
n = 1800, p =
[3]
2
3
(a)
E(X) = np = 1200
(b)
SD(X) =
np (1  p)  1200 
(A1) (C1)
1
= 20
3
(M1)(A1) (C2)
[3]
6
12.
12.
When John throws a stone at a target, the probability that he hits the target is 0.4.
He throws a stone 6 times.
(a)
Find the probability that he hits the target exactly 4 times.
(b)
Find the probability that he hits the target for the first time on his third throw.
(a)
Probability =   × (0.4)4 × (0.6)2
4
6
(M1)(A1)
 


= 0.138  accept
432

or 0.13824 
3125



Probability = (0.6)2 × 0.4 = 0.144  or
(b)
[6]
(A2) (C4)
18 

125 
(M1)(A1) (C2)
[6]
13.
Two children, Alan and Belle, each throw two fair cubical dice simultaneously. The score for
each child is the sum of the two numbers shown on their respective dice.
(a)
(i)
Calculate the probability that Alan obtains a score of 9.
(ii)
Calculate the probability that Alan and Belle both obtain a score of 9.
(b) (i)
(ii)
(c)
Calculate the probability that Alan and Belle obtain the same score,
Deduce the probability that Alan’s score exceeds Belle’s score.
Let X denote the largest number shown
on the four dice.
4
 x
(i)
Show that for P(X  x) =   , for x = 1, 2,... 6
6
(ii)
Copy and complete the following probability distribution table.
x
P(X = x)
(iii)
1
1
1296
2
15
1296
3
4
5
6
671
1296
Calculate E(X).
[13]
13.
(a)
1
(= 0.111)
9
(i)
P(Alan scores 9) =
(A1)
(ii)
1  1 
P(Alan scores 9 and Belle scores 9) =     
 9   81 
2
(= 0.0123)
(A1)
2
7
2
(b)
(i)
(ii)
P(A>B) =
=
(c)
(i)
2
2
2
 1 
 2 
 6 
 2 
P(Same score) =   +   + … +   + … +  
 36 
 36 
 36 
 36 
2
1
 
 
+  36 
(M1)
73
= 648 (= 0.113)
(A1)
1
2
73 

1 –

 648 
(M1)
575
(= 0.444)
1296
(A1)
P(One number  x) =
x
(with some explanation)
6
 x
P(X  x) = P(All four numbers  x) =  
6
4
(R1)
4
(M1)(AG)
4
 x
 x – 1
(ii) P(X = x) = P(X  x) – P(X  x – l) =   – 

6
 6 
4
x
1
2
3
4
5
6
P(X = x)
1
1296
15
1296
65
1296
175
1296
369
1296
671
1296
(A1)(A1)(A1)
Note: Award (A3) if table is not completed but calculation of
E(X) in part (iii) is correct.
(iii)
1
15
671
+2×
+…+6×
1296
1296
1296
6797
=
(= 5.24)
1296
E(X) = 1 ×
(M1)
(A1)
7
[13]
14. The random variable X is Poisson distributed with mean  and satisfies
P(X = 3) = P(X = 0) + P(X = 1).
(a)
Find the value of , correct to four decimal places.
(b) For this value of  evaluate P(2  X  4).
[6]
14.
(a)
The given condition implies that
3
6
e–  = e–  +  e– 
(M1)
8
 3 – 6 – 6 = 0
   2.8473
(b)
(A1)
(G1)
P(2  X  4) = P(X = 2) + P(X = 3) + P(X = 4)
2 –
P(X = 2) =
e
3
(M1)
3 –
= 0.235, P(X = 3) =
e
2
4 –
e
P(X = 4) =
= 0.159
24
6
= 0.223,
(G1)
Hence P(2  X  4) = 0.617
OR
P(2  X  4) = P(X  4) – P(X  1)
= 0.8402 – 0.2231
= 0.617
(A1)
(M1)
(G1)
(G1)
3
[6]
15.
Give your answers to four significant figures.
A machine produces cloth with some minor faults. The number of faults per metre is a random
variable following a Poisson distribution with a mean 3. Calculate the probability that a metre of
the cloth contains five or more faults.
[4]
15.
Note: Accept answers to an accuracy of at least 4
significant figures – do not apply AP.
P(X = 0) = 0.04979, P(X = 1) = 0.14936, P(X = 2) = 0.22404,
P(X = 3) = 0.22404, P(X = 4) = 0.16803
Sum = 0.81526
OR
P(X  4) = 0.8153 (accept 0.8152)
Hence, P(X  5) = l – 0.8153 = 0.1847 (accept 0.1848)
OR
P(X  5) = 1 – P(X  4) = 0.1847
(M1)
(A1)
(G2)
(M1)(A1)
(G4)
[4]
16.
16.
When a boy plays a game at a fair, the probability that he wins a prize is 0.25. He plays the game
10 times. Let X denote the total number of prizes that he wins. Assuming that the games are
independent, find
(a)
E(X)
(b)
P(X  2).
(a)
X is B(10, 0.25) (seen or implied)
so E(X) = 10 × 0.25 = 2.5
(b)
P(X  2) = (0.75)10 + 
[6]
10 
 (0.75)9(0.25) +
1
= 0.526
(R1)
(M1)(A1) (C3)
10 
  (0.75)8(0.25)2
2
(M1)(A1)
(A1) (C3)
[6]
9
17.
Two typists were given a series of tests to complete. On average, Mr Brown made 2.7 mistakes per
test while Mr Smith made 2.5 mistakes per test. Assume that the number of mistakes made by any
typist follows a Poisson distribution.
(a)
(b)
Calculate the probability that, in a particular test,
(i)
Mr Brown made two mistakes;
(ii)
Mr Smith made three mistakes;
(iii)
Mr Brown made two mistakes and Mr Smith made three mistakes.
In another test, Mr Brown and Mr Smith made a combined total of five mistakes. Calculate
the probability that Mr Brown made fewer mistakes than Mr Smith.
[11]
17.
B ~ P(2.7), S ~ P(2.5)
(i)
P(B = 2) =
e –2.7 2.7 
2
= 0.245
(M1)(A1)
(ii)
e –2.5 2.5
P(S = 3) =
6
= 0.214
(M1)(A1)
2
(a)
3
(iii)
(b)
The two events are independent.
P((B = 2)  (S = 3)) = P(B = 2) × P(S = 3)
= 0.214 × 0.245
= 0.0524
e –5.2 (5.2) 5
P(B + S = 5) =
 0.175
120
e –2.7 2.7 
P((B = 2)  (S = 3)) =
2
2
(M1)
(A1)
6
(A1)
e –2.5 2.5
×
6
3
 0.245 × 0.214 = 0.0524
e –2.7 2.7 
e -2.5 (2.5) 4
×
 0.181 × 0.133
24
1
1
P((B = 1)  (S = 4)) =
= 0.0242
P((B = 0)  (S = 5)) =
e
–2.7
2.7 
0
1
×
e
–2.5
2.5
120
(A1)
5
 0.067 × 0.067
= 0.0045
(A1)
0.0524  0.0242  0.0045 0.0811

P(B < S) =
0.175
0.175
= 0.464 (or 0.463)
(M1)
(A1)
5
[11]
18. On a television channel the news is shown at the same time each day. The probability that Alice
watches the news on a given day is 0.4. Calculate the probability that on five consecutive days,
she watches the news on at most three days.
[6]
18.
METHOD 1
X is Binomial
n=5
p = 0.4
P(X  3) = 1 – P(X = 4) – P(X = 5)
= 1 – 0.0768 – 0.01024
(A1)(A1)
(M1)
(A1)(A1)
10
= 0.91296... (0.913 to 3 sf)
(A1) (C6)
METHOD 2
P(X  3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.07776 + 0.2592 + 0.3456 + 0.2304
= 0.91296... (0.913 to 3 sf)
(M1)
(A2)
(A1) (C6)
[6]
19. The random variable X has a Poisson distribution with mean λ.
(a) Given that P(X = 4) = P(X = 2) + P(X = 3), find the value of λ.
(b) Given that λ = 3.2, find the value of
(i)
P(X  2);
(ii)
P(X  3  X  2).
[8]
19.
(a)
(b)
e   4 e   2 e   3


4!
2!
3!
2
λ – 4λ – 12 = 0  λ = 6
(i)
P(X ≥ 2) = 1 – e–3.2 – e–3.2 × 3.2 = 0.829
(M1)
(A1)(A1)
3
(M1)(A1)
OR
P(X ≥ 2) = 0.829
(ii)
P(X ≤ 3X  2) =
(G2)
P(2  X  3)
P( X  2)
e 3.2  3.2 2 e 3.2  3.2 3

2
6
=
1  4 . 2 e  3 .2
= 0.520
(M1)
(A1)
(A1)
OR
P(X ≤ 3X ≥ 2) = 0.520
(G3)
5
[8]
20.
The random variable X has a Poisson distribution with mean λ. Let p be the probability that X
takes the value 1 or 2.
(a)
Write down an expression for p.
(b) Sketch the graph of p for 0  λ  4.
(c)
Find the exact value of λ for which p is a maximum.
[7]
20.
(a)
p = λe–λ +
2
2
e–
(A1)
1
11
(b)

0
(c)

4
(A1)
Note: Award (A1) for a maximum point in [0, 4]; sketch need not
be accurate.
dp
2
= e–λ + λ(–e–) + e– × λ +
(–e–)
(M1)(A1)
d
2

2 
= e–λ 1  
(A1)
2 


dp
2 
pmax when
= 0  1   = 0
(M1)
d
2 

 λ2 = 2
  = 2 (do not accept 1.41)
(A1)
Note: If no working shown, award (A2) for an answer of 1.41
obviously obtained from a GDC.
1
5
[7]
21.
Let X be a random variable with a Poisson distribution such that Var(X) = (E(X))2 – 6.
(a)
Show that the mean of the distribution is 3.
(b) Find P(X  3).
Let Y be another random variable, independent of X, with a Poisson distribution such that
E(Y) = 2.
(c)
(d)
Find P(X + Y < 4).
Let U = X + 2Y.
(i)
Find the mean and variance of U.
(ii)
State with a reason whether or not U has a Poisson distribution.
[10]
21. Note: In this question do not penalize answers given to more than three significant figures.
(a)
Let  = E(X) = Var(X). Then λ = λ2 – 6
(M1)
12
Therefore  
(b)
(c)
(d)
1  25
2
(M1)
and since λ must be positive λ = 3.
(A1)
3
Then P(X  3) = 0.647
(A1)
(N1)
1
E(X + Y) = 3 + 2 = 5. Since X and Y are independent X + Y
has a Poisson distribution with mean = 5.
(M1)
Hence P(X + Y  4) = 0.265.
(A1)
Note: Award (N0) if P (X + Y  4) is given instead.
(N1)
(i)
E(U) = E(X) + 2E(Y) = 7
(A1)
Var(U) = Var(X) + 4Var(Y) = 11
(A1)
U does not have a Poisson distribution
(A1)
because Var(U)  E(U).
(R1)
(ii)
2
4
[10]
22. A discrete random variable X has its probability distribution given by
P(X = x) = k(x + 1), where x is 0, 1, 2, 3, 4.
(a)
Show that k =
(b)
Find E(X).
1
15
[6]
22.
(a)
Using
 P( X  x)  1
 k × 1 + k × 2 + k × 3 + k × 4 + k × 5 = 15k = 1
k=
(b)
1
15
Using E(X) =
M1
M1A1
AG
 xP( X  x)
3
5
=0× 1 +l× 2 +2×
+3× 4 +4×
15
15
15
15
15
8
=  2 2 , 2.67 
3 3

N0
3
N2
3
(M1)
A1
A1
[6]
23.
The random variable X has a Poisson distribution with mean 4. Calculate
(a)
P(3  X  5);
(b)
P(X  3);
(c)
P(3  X < 5X  3).
[6]
13
23.
(a)
(b)
P(3  X  5) = P(X  5) – P(X  2)
= 0.547
(M1)
A1
N2
2
P(X  3) = 1 – P(X  2)
= 0.762
(M1)
A1
N2
2
N2
2
P(3  X  5)  0.547 


P(X  3)  0.762 
= 0.718
P(3  X  5|X  3) =
(c)
(M1)
A1
[6]
10
24.
Consider the 10 data items x1, x2, ... x10. Given that
x
2
i
= 1341 and the standard deviation is
i 1
6.9, find the value of x .
24.
[6]
Var( X )  6.92  47.61
(M1)(A1)
10
2
2
2
Using Var( X )  E( X )  (E( X )) or s 
47.61 
x
i 1
10
2
i
 x2
1341
 x2
10
(M1)
(A1)
x 2  86.49
(A1)
x   9.3 (Do not penalize the absence of the minus)
(A1) (C6)
[6]
26.
25
The number of car accidents occurring per day on a highway follows a Poisson distribution with
mean 1.5.
(a)
Find the probability that more than two accidents will occur on a given Monday.
(b)
Given that at least one accident occurs on another day, find the probability that more than
two accidents occur on that day.
[6]
X  Po (m)
EITHER

m2 
P ( X  2)  e  m 1  m 

2 

(M1)
 1 0.404  0.596
 m  2.30
(A1)
P( X  2)  e2.3 (1  2.3)
(M1)
 0.3309  0.331 (3 s.f.)
(A1) (N4)
OR
Solving the equation cdf (m , 2)  1  0.404  0.596
m  2.30
cdf (2.3, 1)  0.3309  0.331(3 s.f.)
(M1)
(A1)
(M1)(A1) (N4)
14
[4]
26. The number of car accidents occurring per day on a highway follows a Poisson distribution with
mean 1.5.
26.
(a)
Find the probability that more than two accidents will occur on a given Monday.
(b)
Given that at least one accident occurs on another day, find the probability that more than
two accidents occur on that day.
[6]
(a)
Let A be the number of accidents which occur in a day
Using a Poisson distribution in an inequality
M1
P(A > 2) = 1  P(A  2)

1 .5 2
= 1  e1.5 1  1.5 
2





= 0.191
(b)
(A1)
A1
N2
P(A  1) = 1  P(A = 0)
= 0.776(869...)
(A1)
Indication of conditional probability as a ratio:
P A  2 
P A  1
or
0.191153 ...
0.776 869 ...
P(A > 2 | A  1) = 0.246
M1
A1
N1
[6]
27. Andrew shoots 20 arrows at a target. He has a probability of 0.3 of hitting the target. All shots are
independent of each other. Let X denote the number of arrows hitting the target.
(a)
Find the mean and standard deviation of X.
(b) Find
(i)
P(X = 5);
(ii)
P(4  X  8).
Bill also shoots arrows at a target, with probability of 0.3 of hitting the target. All shots are
independent of each other.
(c)
(d)
Calculate the probability that Bill hits the target for the first time on his third shot.
Calculate the minimum number of shots required for the probability of at least one shot
hitting the target to exceed 0.99.
[19]
27.
(a)
X ~ B (20, 0.3)
Mean = 20  0.3 = 6
(A1)
A1
Variance = 20  0.3  0.7
= 4.2
Standard deviation = 2.05
(M1)
(A1)
A1
15
(b)
 20 
P X  5     0.35  0.715 (M1)
5
(i)
= 0.179
(ii)
(c)
(d)
A2
N3
P(4  X  8) = P(X  8)  P(X  3)
M1
= 0.780
A2
N2
M1A1
A1
N2
(accept 0.779)
Probability = 0.7  0.7  0.3
= 0.147
METHOD 1
P(at least 1 hit) = 1  0.7n
For solving 1  0.7n = 0.99 (or 1  0.7n > 0.99)
ie
(A1)
M1
n = 12.9...
(A1)
n > 12.9...
(M1)
n = 13
A1
N2
METHOD 2
P (at least 1 hit) =1  0.7n
(A1)
Substituting either n = 12 or n = 13
(M1)
1  0.712 = 0.986...
A1
1  0.713 = 0.990...
A1
n = 13
Note: Award the final R1 only if the preceding
two A marks have been awarded.
R1
N2
[19]
28. The random variable X follows a Poisson distribution. Given that P(X  1) = 0.2, find
(a)
the mean of the distribution;
(b)
P(X  2).
[6]
28.
(a)
(b)
P(X  1) = 0.2
P(X = 0) + P(X = 1) = 0.2
 e + e = 0.2
 e (1 + )=0.2
  = 2.99
P(X  2) = 0.424
(accept 0.425)
(M1)
A1
A2
N2
A2
N2
[6]
16
29.
A bag contains a very large number of ribbons. One quarter of the ribbons are yellow and the rest
are blue. Ten ribbons are selected at random from the bag.
(a)
Find the expected number of yellow ribbons selected.
(b) Find the probability that exactly six of these ribbons are yellow.
(c)
Find the probability that at least two of these ribbons are yellow.
(d)
Find the most likely number of yellow ribbons selected.
(e)
What assumption have you made about the probability of selecting a yellow ribbon?
[12]
29.
(a)
Let X be the number of yellow ribbons in the sample
X ~ B 10 , 


1
4
(M1)
 E(X) = 2.5
A1
10   1   3 
P(X = 6) =      
 6  4   4 
6
(b)
4
(M1)
= 0.0162
30.
N2
A1
N2
In an experiment, a trial is repeated n times. The trials are independent and the probability p of
success in each trial is constant. Let X be the number of successes in the n trials. The mean of X is
0.4 and the standard deviation is 0.6.
(a)
Find p.
(b)
Find n.
(c)
P(X  2) = 1  P(X  1) = 1  (P(X = 0) + P(X = 1)) (M1)
[6]
3
4
10
 1  3 
 10    
 4  4 
= 1  
9
(A1)
= 0.756
A1
10   1   3 
P(X = x) =      
 x  4   4 
x
(d)
N3
10  x
Using GDC (or by substituting values of x into above) it is
possible to calculate relevant probabilities.
x
P(X = x)
1
0.188
2
0.282
3
0.250
(M1)
A2
From these values the most likely number of yellow ribbons is 2.
A1
N0
17
(e)
The probability that a ribbon is yellow remains constant.
R1
[12]
30.
(a)
(b)
X has a binomial distribution
0.4 = np 0.36 = np (1  p)
Attempting to solve.
p = 0.1
n=4
Note:
(M1)
A1A1
(M1)
A1
N3
A1
N1
Allow FT only on integer values of n
for p between 0 and 1.
[6]
31.
31.
A biology test consists of seven multiple choice questions. Each question has five possible
answers, only one of which is correct. At least four correct answers are required to pass the test.
Juan does not know the answer to any of the questions so, for each question, he selects the answer
at random.
(a)
Find the probability that Juan answers exactly four questions correctly.
(b)
Find the probability that Juan passes the biology test.
(a)
X ~ B7, 


1
5
(A1)
7  1   4 
P(X = 4) =        
 4  5   5 
4
= 0.0287
(b)
3
 448 


 15625 
P(pass) = P(X  4) OR 1  P(X  3) (or equivalent)
= 0.0333
Note:
[6]
 521 


 15625 
(M1)
A1
N3
A2
A1
N2
Accept 0.0334
[6]
32.
The time, T minutes, spent each day by students in Amy’s school sending text messages may be
modelled by a normal distribution.
30 of the students spend less than 10 minutes per day.
35 spend more than 15 minutes per day.
(a)
Find the mean and standard deviation of T.
The number of text messages received by Amy during a fixed time interval may be modelled by a
Poisson distribution with a mean of 6 messages per hour.
18
(b)
Find the probability that Amy will receive exactly 8 messages between 16:00 and 18:00 on
a random day.
(c)
Given that Amy has received at least 10 messages between 16:00 and 18:00 on a random
day, find the probability that she received 13 messages during that time.
(d)
During a 5-day week, find the probability that there are exactly 3 days when Amy receives
no messages between 17:45 and 18:00.
[18]
32.
(a)
T ~ N(, 2)
Finding the z-values
(M1)
z1 = 0.5244, z2 = 0.3853
10   =  0.5244, 15   = 0.3853
 = 12.9,  = 5.50
M1
A1A1 N2N2
Note:
(b)
(A1)(A1)
Do not accept any other values that
come from premature rounding.
Let X be the random variable “number of text messages in 2 hours”
e –12 12 
P(X = 8) =
8!
8
M1A1
= 0.0655
(c)
A1
Let X be the random variable “number of text messages in 2 hours”
P(X = 13) = 0.1055...
(A1)
P(X  10) = 0.7576...
(A1)
P(X = 13 | X  10)

M1
P X  13  P X  13 


P X  10   1  P X  9 
= 0.139
Note:
(d)
N3
(A1)
A1
N4
(A1)
(A1)
M1
A1
N1
Do not accept any other value that
comes from premature rounding.
Let Y be the random variable “number of text messages in
15 minutes”
Let D be the random variable “number of days with no
messages received”
Y ~ Po(1.5)
P(Y = 0) = 0.2231...
D ~ B(5, 0.2231...)
P(D = 3) = 0.0670
Note:
0.067 incurs an accuracy penalty.
[18]
19
33.
The times taken for buses travelling between two towns are normally distributed with a mean of
35 minutes and standard deviation of 7 minutes.
(a)
Find the probability that a randomly chosen bus completes the journey in less than 40
minutes.
(b) 90 of buses complete the journey in less than t minutes. Find the value of t.
(c)
A random sample of 10 buses had their travel time between the two towns recorded. Find
the probability that exactly 6 of these buses complete the journey in less than 40 minutes.
[11]
33.
(a)
5

P(T < 40) = P  Z  
7

= 0.762
Note:
(b)
(M1)
A1
Accept 0.761 from tables.
Stating P(T < t) = 0.90 or sketching a labelled diagram
t  35
 1.2815 ...
7
= 44.0 (min)
(M1)
A1
Recognizing binomial distribution with correct parameters
or stating X ~ B (10, 0.762...)
10 
6
4
P(X = 6) =    0.762...  0.237...
6
 
= 0.131
Notes:
A1
(M1)(A1)
t = (1.2815...)(7) + 35
(c)
N2
N4
(A1)(A1)
(M1)
A1
Accept 0.132 or 0.133.
Award FT for their value of p from
(a) but they must have n =10.
[11]
34.
The number of bus accidents that occur in a given period of time has a Poisson distribution with a
mean of 0.6 accidents per day.
(a)
Find the probability that at least two accidents occur on a randomly chosen day.
(b) Find the most likely number of accidents occurring on a randomly chosen day.
Justify your answer.
(c)
Find the probability that no accidents occur during a randomly chosen seven-day week.
[11]
34.
(a)
P(X  2) = 1  [P(X = 0) + P(X = 1)]
= 0.122
(b)
(M1)(A1)
A2
EITHER
Using P(X = x) = e0.6 
0 .6 x
to generate a decreasing
x!
20
sequence of at least three numbers
M1A1
OR
Sketching an appropriate graph of P(X = x) against x
M1A1
OR
Finding P(X = 0) = e0.6 and stating that P(X = 0) > 0.5
M1A1
OR
Using P(X = x) P(X = x  1) 
μ
where  < 1
x
Hence P(X = x) is maximised when x = 0 and so the most
likely number of accidents is zero.
(c)
M1A1
R1
N0
(A1)
M1A1
N1
METHOD 1
Y ~ Po(4.2)
P(Y = 0) = e4.2
(= 0.0150)
METHOD 2
P(X = 0) = e0.6 (= 0.5488...)
(A1)
P(Y = 0) = (e0.6)7 (= (0.5488...)7) (binomial approach)
= e4.2 (= 0.0150)
M1
A1
N1
[10]
35
On a particular road, serious accidents occur at an average rate of two per week and can be
modelled using a Poisson distribution.
(a)
(i)
What is the probability of at least eight serious accidents occurring during a
particular four-week period?
(ii)
Assume that a year consists of thirteen periods of four weeks. Find the probability
that in a particular year, there are more than nine four-week periods in which at least
eight serious accidents occur.
(10)
(b)
Given that the probability of at least one serious accident occurring in a period of n weeks is
greater than 0.99, find the least possible value of n, n +.
[18]
35.
(a)
(i)
X  Po (2) for one week
 X  Po (8) for 4 week
(A1)
 P(X  8) = 1 – P(X = 0) + P(X = 1) + ... + P(X = 7)

8
= 1 e 1  8 

8 2 83 8 4 85 8 6 8 7 
     
2! 3! 4! 5! 6! 7 ! 
= 0.547
(ii)
X  B(13, 0.547)
P(X  9) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13)
=
(M1)(A1)
(A1)
A1
N5
(M1)(A1)
(M1)
13!
13!
13!
0.547 10  0.453 3 
0.547 11  0.453 2 
0.547 12  0.453  0.547 13 (A1)
10!3!
11!2!
12 !1!
21
= 0.0894
(b)
A1
N5
X  Po (2n)
2 n
P (no serious accidents) = e
1 – P (no serious accidents) = P (at least one serious accident)
= 1 –e
2 n
(M1)(A1)
1 – P (no serious accidents)  0.99
2 n
 0.99
A1
2 n
 e
 0.01
(A1)
 –2n  ln 0.01
(A1)
1 –e
n
ln 0.01
2
 n  2.30
(M1)
n=3
A1
N5
[18]
36. The probability distribution of a discrete random variable X is defined by
P(X = x) = cx(5 − x), x = 1, 2, 3, 4.
(a)
Find the value of c.
(b) Find E(X).
[6]
36.
(a)
Using
 P X  x   1
4c + 6c + 6c + 4c = 1 (20c = 1)
c
(b)
1
 0.05 
20
Using E(X) =
 xP  X  x 
= (1  0.2) + (2  0.3) + (3  0.3) + (4  0.2)
= 2.5
Notes: Only one of the first two marks can be
implied.
Award M1A1A1 if the x values are averaged
only if symmetry is explicitly mentioned.
(M1)
A1
A1
N1
(M1)
(A1)
A1
N1
[6]
37. The lifts in the office buildings of a small city have occasional breakdowns. The breakdowns at
any given time are independent of one another and can be modelled using a Poisson distribution
with mean 0.2 per day.
(a)
Determine the probability that there will be exactly four breakdowns during the month of
June (June has 30 days).
22
(b) Determine the probability that there are more than 3 breakdowns during the month of June.
(c)
Determine the probability that there are no breakdowns during the first five days of June.
(d)
Find the probability that the first breakdown in June occurs on June 3 rd.
(e)
It costs 1850 Euros to service the lifts when they have breakdowns. Find the expected cost
of servicing lifts for the month of June.
(f)
Determine the probability that there will be no breakdowns in exactly 4 out of the first 5
days in June.
[13]
37.
(a)
mean for 30 days: 30  0.2 = 6.
P X  4  
6 4 6
e  0.134
4!
(b)
P(X > 3) = 1  P(X  3) = 1  e6(1 + 6 + 18 + 36) = 0.849
(c)
EITHER
mean for five days: 5  0.2 = 1
P(X = 0) = e1 (= 0.368)
(A1)
(M1)A1
N3
(M1)A1
N2
(A1)
A1
N2
OR
mean for one day: 0.2
(A1)
P(X = 0) = (e0.2 )5 = e1 (= 0.368)
(d)
Required probability = e0.2  e0.2  (1  e0.2)
= 0.122
(e)
Expected cost is 1850  6 = 11100 Euros
A1
N2
M1A1
A1
N3
A1
On any one day P(X = 0) = e 0.2
(f)

 5  0.2
Therefore,   e
1
 1  e   0.407
4
 0.2
M1A1
N2
[13]
38.
Over a one month period, Ava and Sven play a total of n games of tennis.
The probability that Ava wins any game is 0.4. The result of each game played is independent of
any other game played.
Let X denote the number of games won by Ava over a one month period.
(a)
Find an expression for P(X = 2) in terms of n.
(b) If the probability that Ava wins two games is 0.121 correct to three decimal places, find the
value of n.
[6]
23
38.
(a)
X ~ B(n, 0.4)
(A1)
n
Using P(X = x) =   0.4  0.6 
r
x
n x
(M1)
 
(b)
 n
n n  1
2
n2 
0.42 0.6n2 

P(X = 2) =   0.4 0.6
2


r
A1
P(X = 2) = 0.121
A1
Using an appropriate method (including trial and error) to solve
their equation.
(M1)
n = 10
A1
Note:
N3
N2
Do not award the last A1 if any other
solution is given in their final answer.
[6]
39.
The distance travelled by students to attend Gauss College is modelled by a normal distribution
with mean 6 km and standard deviation 1.5 km.
(a)
(i)
Find the probability that the distance travelled to Gauss College by a randomly
selected student is between 4.8 km and 7.5 km.
(ii)
15 of students travel less than d km to attend Gauss College. Find the value of d.
At Euler College, the distance travelled by students to attend their school is modelled by a normal
distribution with mean  km and standard deviation  km.
(b)
If 10 of students travel more than 8 km and 5 of students travel less than 2 km, find the
value of  and of .
The number of telephone calls, T, received by Euler College each minute can be modelled by a
Poisson distribution with a mean of 3.5.
(c)
(i)
Find the probability that at least three telephone calls are received by Euler College
in each of two successive one-minute intervals.
(ii)
Find the probability that Euler College receives 15 telephone calls during a
randomly selected five-minute interval.
[21
39.
(a)
(i)
P(4.8 < X < 7.5) = P(0.8 < Z < 1)
Note:
(ii)
= 0.629
Accept P(4.8  X  7.5) = P(0.8  Z  1).
A1
Stating P(X < d) = 0.15 or sketching an appropriately labelled
diagram.
d 6
  1.0364 ...
1 .5
d = (1.0364...)(1.5) + 6
= 4.45 (km)
(b)
(M1)
Stating both P(X > 8) = 0.1 and P(X < 2) = 0.05 or sketching an
appropriately labelled diagram.
Setting up two equations in  and 
N2
A1
(M1)(A1)
(M1)
A1
N4
R1
(M1)
24
(c)
8 =  + (1.281...) and 2 =   (1.644...)
Attempting to solve for  and  (including by graphical means)
 = 2.05 (km) and  = 5.37 (km)
Note: Accept  = 5.36, 5.38.
A1
(M1)
A1A1
N4
(i)
M1
(A1)
A1
M1A1
N3
(ii)
Use of the Poisson distribution in an inequality.
P(T  3) = 1  P(T  2)
= 0.679...
Required probability is (0.679...)2 = 0.461
Note: Allow FT for their value of P(T  3).
 ~ Po(17.5)
Pτ 15  
A1
e 17.5 17 .5
15!
15
(M1)
= 0.0849
A1
N2
[21]
40.
(a)
The independent random variables X and Y have Poisson distributions and Z = X +Y. The
means of X and Y are  and  respectively. By using the identity
n
P Z  n    P  X  k  P Y  n  k 
k 0
show that Z has a Poisson distribution with mean ( +).
(6)
(b)
Given that U1, U2, U3, … are independent Poisson random variables each having mean m,
n
use mathematical induction together with the result in (a) to show that
U
r 1
r
has a
Poisson distribution with mean nm.
[12]
λk
 nk

P(Z = n) =  e   e 
n  k !
k!
k 0
n
40.
(a)
e     
n!
=
=
λ
n
n!
 k !n  k ! λ
k
M1A1
nk
e     
   n
n!
A1
This shows that Z is Poisson distributed with mean ( + ).
(b)
M1A1
k 0
R1
The result is (trivially) true for n = 1. A1
k
Assuming it to be true for n = k, ie
U
r 1
k 1
Consider
k
U  U
r 1
r
r 1
r
r
~ Po km
 U k 1
which, using (a) is Po(km + m) ie Po([k + 1]m)
M1
M1A1
A1
25
Hence proved by induction since true for n = k  true for
n = k + 1 and we have shown true for n = 1.
R1
[12]
41.
The lengths of a particular species of lizard are normally distributed with a mean length of 50 cm
and a standard deviation of 4 cm. A lizard is chosen at random.
(a)
Find the probability that its length is greater than 45 cm.
(b)
Given that its length is greater than 45 cm, find the probability that its length is greater than
55 cm.
[6]
41.
(a)
P(X > 45) = P(Z < 1.25)
(M1)
= 0.894
(b)
A1
Using conditional probability P(X > 55 | X > 45)
N2
(A1)

P X  55  X  45 
P X  45 
(A1)

P X  55   0.1056 ... 


P X  45   0.8943 ... 
(A1)
FT numerator = 1  (their answer to a).
= 0.118
42.
A1
N2
The discrete random variable X has the following probability distribution.
k
 , x  1, 2, 3, 4
P(X = x) =  x
0, otherwise
Calculate
(a)
the value of the constant k;
(b)
E(X).
[6]
42.
(a)
(b)
 1 1 1
k 1     = 1
 2 3 4
12
k=
(= 0.48)
25
E(X) = 1 ×
=
12
6
4
3
 2
 3
 4
25
25
25
25
48
( = 1.92)
25
(M1)(A1)
(A1) (C3)
(M1)(A1)
(A1) (C3)
[6]
26
43.
In a game a player pays an entrance fee of $n. He then selects one number from 1, 2, 3, 4, 5, 6 and
rolls three standard dice.
If his chosen number appears on all three dice he wins four times his entrance fee.
If his number appears on exactly two of the dice he wins three times the entrance fee.
If his number appears on exactly one die he wins twice the entrance fee.
If his number does not appear on any of the dice he wins nothing.
(a)
Copy and complete the probability table below.
−n
Profit ($)
n
2n
3n
75
216
Probability
(4)
(b)
 17 n 
Show that the player’s expected profit is $  
.
 216 
(2)
(c)
What should the entrance fee be so that the player’s expected loss per game is 34 cents?
(2)
(Total 8 marks)
3
43.
(a)
1
1
5
15
 5  125
; P ( 2n)  3 

; P(  n)    
P(3n) = 3 
216 216
216
6 216
6
(M1)
Profit
n
n
2n
3n
Probability
215
216
75
216
15
216
1
216
(A1)(A1)(A1) (N3)
(b)
E(X) =
 n 125  n
216
= 
(c)

75
15
1
 2n
 3n
216
216
216
17n
216
(AG) (N0)
17n
  0.34
216
n = 4.32
(M1)(A1)
(M1)
(accept $ 4.32)
(A1) (N1)
27
1
NORMAL DISTRIBUTION with KEY
1.
A factory has a machine designed to produce 1 kg bags of sugar. It is found that the average weight of sugar
in the bags is 1.02 kg. Assuming that the weights of the bags are normally distributed, find the standard
deviation if 1.7% of the bags weigh below 1 kg.
Give your answer correct to the nearest 0.1 gram.
[4]
1.
Let (z) = 0.017
then (–z) = 1 – 0.017 = 0.983
(M1)
z = –2.12
(A1)
But z =
x


1  1.02

where x = 1 kg
1  1.02
Therefore

= –2.12
(M1)
  = 0.00943 kg = 9.4 g (to the nearest 0.1 g)
2.
2.
3.
(A1)
The random variable X is distributed normally with mean 30 and standard deviation 2.
Find p(27  X  34).
34  30 
 27  30
Z 
p(27  X  34) = p 

2 
 2
= p(–1.5  Z  2)
= 0.433… + 0.477...
= 0.910 (3 sf) (accept 0.911)
(C4)
[4]
(M1)
(A1)
(M1)
(A1) (C4)
A machine is set to produce bags of salt, whose weights are distributed normally, with a mean of 110 g and
standard deviation of 1.142 g. If the weight of a bag of salt is less than 108 g, the bag is rejected. With these
settings, 4% of the bags are rejected.
The settings of the machine are altered and it is found that 7% of the bags are rejected.
(a)
(i)
If the mean has not changed, find the new standard deviation, correct to three decimal
places.
The machine is adjusted to operate with this new value of the standard deviation.
(ii)
Find the value, correct to two decimal places, at which the mean should be set so that only
4% of the bags are rejected.
(b) With the new settings from part (a), it is found that 80% of the bags of salt have a weight which lies
between A g and B g, where A and B are symmetric about the mean. Find the values of A and B,
giving your answers correct to two decimal places.
[4]
3.
Note: In all 3 parts, award (A2) for correct answers with no working.
Award (M2)(A2) for correct answers with written evidence of the
correct use of a GDC (see GDC examples).
(a)
(i)
Let X be the random variable “the weight of a bag of salt”.
2
Then X ~ N(110, σ2), where σ is the new standard deviation.
Given P(X < 108) = 0.07, let Z =
X  110

108  110 

Then P  Z 
 = 0.07



2
Therefore,
= –1.476

(M1)
(M1)(A1)
Therefore, σ = 1.355.
(A1)
GDC Example: Graphing of normal cdf with σ as the variable, and
finding the intersection with p = 0.07.
Let the new mean be µ, then X ~ N(µ, 1.3552).
108   

Then P  Z 
(M1)
 = 0.04
1.355 

108  
Therefore,
= 1.751
(M1)(A1)
1.355
Therefore, µ = 110.37
(A1)
GDC Example: Graphing of normal cdf with µ as the variable and finding
intersection with p = 0.04.
(ii)
(b)
4.
If the mean is 110.37 g then X ~ N(110.37, 1.3552), P(A < X < B) = 0.8
Then P(X < A) = 0.1, and P(X < B) = 0.9.
A  110.37
Therefore,
= –1.282
(M1)
1.355
A = 108.63
(A1)
B  110.37
Therefore,
= 1.282
(M1)
1.355
B = 112.11
(A1)
GDC Example: Graphing of normal cdf with X as the variable and finding
intersection with p = 0.1 and 0.9.
The lifetime of a particular component of a solar cell is Y years, where Y is a continuous random variable
with probability density function
0 when y  0

f ( y)  
- y/2
when y  0.
0.5e
(a)
Find the probability, correct to four significant figures, that a given component fails within six
months.
Each solar cell has three components which work independently and the cell will continue to run if at least
two of the components continue to work.
4.
(b)
Find the probability that a solar cell fails within six months.
(a)
Required probability
= P(Y  12 )
=

1/ 2
0
0.5e – y / 2 dy
= 0.2212.
[7]
(M2)
(G1)
3
OR
Required probability =

1/ 2
0
0.5e – y / 2 dy

= – e–y/2

1/ 2
0
(M1)
(M1)
–1/4
=1–e
= 0.2212 (4 sf)
(b)
Required probability
= P(2 or 3 of the components fail in six months)
 3
=   (0.2212)2(0.7788) + (0.2212)3
 2
= 0.125.
(A1)
(M1)
(M2)
(G1)
5.
The diameters of discs produced by a machine are normally distributed with a mean of 10 cm and standard
deviation of 0.1 cm. Find the probability of the machine producing a disc with a diameter smaller than 9.8
cm.
[3]
5.
Let X be the random variable “the diameter of the disc,” then X ~ N(10, 0.12).
X 
Let Z =
, the standardised normal variable (using formulae and

statistical tables).
Then P(X < 9.8) = P(Z < –2)
= 1 – (2)
= 1 – 0.9773
= 0.0227
(A1)
(A1)
(A1)
(using formulae and statistical tables)
OR
Using a graphic display calculator,
P(X < 9.8) = 0.0228
(A3)
Note: A different answer is obtained if a graphic display calculator is used
6.
Z is the standardized normal random variable with mean 0 and variance 1. Find the value of a such that
P(  Z   a) = 0.75.
[3]
4
6.
0.75
1–(a)
–a
a
From the diagram 1 – 2(1 – (a)) 0.75
2(a) = 1.75
a = 1.15
7.
(M1)
(A1)
(A1) (C3)
A continuous random variable X has probability density function
4

, for 0  x  1 ,

f ( x)    (1  x 2 )

0,
elsewhere

Find E(X).
7.
[3]
METHOD 1
E(X) =

4x
dx
0 π (1  x 2 )
1
(M1)
= 0.441.
(G2) (C3)
METHOD 2
E(X) =
8.

4x
dx
0 π (1  x 2 )
1
=
2
[ln (1  x 2 )]10
π
=
2
(ln 2)
π
(M1)
(M1)
 ln 4 
or π  .


(A1) (C3)
The weights of a certain species of bird are normally distributed with mean 0.8 kg and standard deviation
0.12 kg. Find the probability that the weight of a randomly chosen bird of the species lies between 0.74 kg
and 0.95 kg.
[6]
8.
 0.95 – 0.8 
 0.74 – 0.8 
 – 

 0.12 
 0.12 
P(0.74 < X < 0.95) =  
(M1)(A1)
5
=  (1.25) –  (–0.5)
= 0.8944 – (l – 0.6915)
= 0.586
9.
(A1)
(A1)(A1)
(A1) (C6)
The probability density function f (x), of a continuous random variable X is defined by
1
 x( 4 – x 2 ), 0  x  2
f (x) =  4

0, otherwise.
Calculate the median value of X.
[6]
9.
Let m be the median.
m

Then
0
1
x (4 – x2)dx = 0.5.
4
 4x – x dx
(M1)
m
=>
3
=2
(A1)
0
1 4 m
x ]0 = 2
4
1
=> 2m2 – m4 = 2
4
=> [2x2 –
(M1)
=> m4 – 8m2 + 8 = 0
(A1)
(G2)
m = 1.08
OR
8  64 – 32
8  32
=
=4
2
2
=> m = 4 – 8  4 – 2 2 


m2 =
8 (4  2 2 )
(M1)
(A1) (C6)
Note: Award (C5) if other solutions to the equation
m4 – 8m2 + 8 = 0 appear in the answer box.
10.
The random variable X is normally distributed and
P(X  10) = 0.670
P(X  12) = 0.937.
Find E(X).
[6]
10.
Let E(X) = .
From tables, z1 = 0.44, z2 = 1.53
10 =  + 0.44
12 =  + 1.53
=
1.53  10 – 0.44  12
1.53 – 0.44
 E(X) = 9.19
(A1)(A1)
(A1)
(A1)
(M1)
(A1) (C6)
[6]
6
11.
A random variable X is normally distributed with mean  and standard deviation σ, such that
P(X > 50.32) = 0.119, and P(X < 43.56) = 0.305.
(a)
Find  and .
(b) Hence find P(|X – | < 5).
[7]
11.
(a)
30.5%
11.9%
43.56
50.32
(z) = 0.305  z = –0.51
and (z) = 0.881  z = 1.18
43.56  
50.32  
= 1.18 and
= –0.51


(A1)
(A1)
(M1)
Solving simultaneously
 50.32 =  + 1.18σ and 43.56 = µ – 0.51σ
 1.69σ = 6.76
 σ = 4  µ = 45.6
(b)
12.
P(X – < 5) = P(40.6 < x < 50.6)
= 0.789
(A1)(A1)
5
(M1)
(G1)
2
The following diagram shows the probability density function for the random variable X, which is normally
distributed with mean 250 and standard deviation 50.
f( x )
180
250
280
x
Find the probability represented by the shaded region.
[6]
12.
z1 = 0.6  Φ(z1) = 0.7257
z2 = –1.4  Φ(z2) = 0.0808
Probability = 0.7257 – 0.0808 = 0.6449 = 0.645 (3 sf)
(A1)(A1)
(A1)(A1)
(M1)(A1) (C6)
7
13.
Let f (x) be the probability density function for a random variable X, where
kx 2 , for 0  x  2
f (x) 
0, otherwise
(a)
3
.
8
Show that k =
(b) Calculate
(i)
E(X);
(ii)
the median of X.
[6]
13.
(a)
k

2
0
x 2 dx = 1
(M1)
2
 x3 
k8
k   
=1
3
3
 0
3
k=
8
(b)
(i)
3
E(X) =
8
(AG)
2

2
0
3  x4 
xx dx (   )
8  4 0
2
(M1)
3
2
=
(ii)
(A1)
(A1)
The median m must be a number such that
3 m 2
1 3 2 2 
x dx   or
x dx 
8 0
2 8 m



(M1)(A1)
m
 1
3  x3 
3  m3




0
 
 2
8  3  0 8  3

(A1)
m3 1
  m3 = 4.
8
2
 m = 3 4 (= 1.59 to 3 sf) .
14.
(A1)
A continuous random variable X has probability density function given by
f (x) = k (2x – x2),
f (x) = 0,
(a)
Find the value of k.
(b)
Find P(0.25  x  0.5).
for 0  x  2
elsewhere.
[6]
8
14.
(a)
2
Since X is a continuous r.v.   k (2 x – x 2 )dx  1
0
(M1)
2

x3 
 k  x 2 –  1
3 0

(A1)
 8 

 4 –  – 0  1
 3 

k 
3
4
(A1) (C3)
0 .5
(b)
 f ( x ) dx
P(0.25  x  0.5) =
(M1)
0.25

15.
29
 0.113
256
(A2) (C3)
Ian and Karl have been chosen to represent their countries in the Olympic discus throw. Assume that the
distance thrown by each athlete is normally distributed. The mean distance thrown by Ian in the past year
was 60.33 m with a standard deviation of 1.95 m.
(a)
In the past year, 80% of Ian’s throws have been longer than x metres.
Find x, correct to two decimal places.
(b)
In the past year, 80% of Karl’s throws have been longer than 56.52 m. If the mean distance of his
throws was 59.39 m, find the standard deviation of his throws, correct to two decimal places.
(c)
This year, Karl’s throws have a mean of 59.50 m and a standard deviation of 3.00 m. Ian’s throws
still have a mean of 60.33 m and standard deviation 1.95 m. In a competition an athlete must have at
least one throw of 65 m or more in the first round to qualify for the final round. Each athlete is
allowed three throws in the first round.
(i)
(ii)
Determine which of these two athletes is more likely to qualify for the final on their first
throw.
Find the probability that both athletes qualify for the final. (11)
[17]
15.
(a)
X = length of Ian’s throw.
X  N (60.33, 1.952)
P(X > x) = 0.80  z = –0.8416
(A1)
x – 60.33
1.95
(M1)
– 0.8416 
x = 58.69 m
(A1)
(b)
(N3)
Y = length of Karl’s throw.
Y  N (59.39, 2)
P(Y > 56.52) = 0.80  z = –0.8416
– 0.8416 
56.52 – 59.39

(A1)
(M1)
9
 = 3.41 (accept 3.42)
(c)
(i)
(A1)
Y  N (59.50, 3.002)
X  N (60.33, 1.952)
EITHER
P(Y  65) = 0.0334 P(X  65) = 0.00831 (no (AP) here)(A2)(A2)
OR
65 – 59.50 

P(Y  65)  P  Z 

3.00 

(M1)
= P(Z  1.833)
= 0.0334 (accept 0.0336)
(A1)
65 – 60.33 

P(X  65)  P  Z 

1.95 

(M1)
= P(Z  2.395)
= 0.0083 (accept 0.0084)
(A1)
THEN
Karl is more likely to qualify since P(Y  65)  P(X  65)
(R1)
Note: Award full marks if probabilities are not calculated but the correct
conclusion is reached with the reason 1.833 < 2.395.
(ii)
If p represents the probability that an athlete throws 65 metres
or more then with 3 throws the probability of qualifying
for the final is
1 – (1 – p)3, or p + (1 – p)p + (1 – p)2p,
or p3 + 3(1 – p)p2 + 3(1 – p)2p
(M1)
Therefore P(Ian qualifies) = 1 – (1 – 0.00831)3
= 0.0247
(A1)
P (Karl qualifies) = 1 – (1 – 0.0334)3
= 0.0969
(A1)
Assuming independence
(R1)
P(both qualify) = (0.0247)(0.0969)
(M1)
= 0.00239
(A1)
11
Note: Depending on use of tables or gdc answers may vary from 0.00239
to 0.00244.
[17]
16.
The speeds of cars at a certain point on a straight road are normally distributed with mean μ and standard
deviation σ. 15% of the cars travelled at speeds greater than 90 km h–1 and 12% of them at speeds less than
40 km h–1. Find μ and σ.
[6]
16.
P(X > 90) = 0.15 and P(X < 40) = 0.12
Finding standardized values 1.036, –1.175
(M1)
A1A1
10
90  
Setting up the equations 1.036 =

µ = 66.6,  =22.6
, –1.175 =
40  

(M1)
A1A1 N2N2
[6]
17.
The continuous random variable X has probability density function
1
x(1 + x2) for 0  x  2,
6
f (x) = 0 otherwise.
f (x) =
(a)
Sketch the graph of f for 0  x  2.
(b) Write down the mode of X.
(c)
(d)
Find the mean of X.
Find the median of X.
[12]
17.
(a)
y
2
f( x )
1
0
1
x
2
A2
(b)
Mode = 2
A1
(c)
Using E(X) =

b
a
(M1)
x f ( x ) dx
2

2
4
Mean= 1 ( x  x ) dx
6
A1
0
2
5
 3

= 1 x  x 
6 3
5 0
(A1)
= 68 (1.51)
45
A1
N2
m
(d)

3
The median m satisfies 1 ( x  x ) dx  1
6
2
M1A1
0
m2  m4  3
2
2
4
 m + 2m2 – 12 = 0
 2  4  48
m2 =
= 2.60555...
2
(A1)
(A1)
11
m = 1.61
A1
18.
N3
The probability density function f (x) of the continuous random variable X is defined on the interval [0, a]
by
1x

f ( x)   8
27
 2
 8x
for 0  x  3.
for 3  x  a.
Find the value of a.
18.

Using
3

0

3
a

a
0
f ( x) dx  1
[6]
(M1)
1
9
x dx 
(  0.5625)
8
16
(M1)(A1)
27 9
27
27  1 1 
dx        
2
8a 8
8x
8  a 3
(M1)(A1)
27 9 7
 
8a 8 16
a
54
( 4.91)
11
(A1) (C6)
[6]
19.
A company buys 44% of its stock of bolts from manufacturer A and the rest from manufacturer B. The
diameters of the bolts produced by each manufacturer follow a normal distribution with a standard
deviation of 0.16 mm.
The mean diameter of the bolts produced by manufacturer A is 1.56 mm.
24.2% of the bolts produced by manufacturer B have a diameter less than 1.52 mm.
(a)
Find the mean diameter of the bolts produced by manufacturer B.
A bolt is chosen at random from the company’s stock.
(b)
Show that the probability that the diameter is less than 1.52 mm is 0.312, to three significant figures.
(c)
The diameter of the bolt is found to be less than 1.52 mm. Find the probability that the bolt was
produced by manufacturer B.
(d)
Manufacturer B makes 8000 bolts in one day. It makes a profit of $1.50 on each bolt sold, on
condition that its diameter measures between 1.52 mm and 1.83 mm. Bolts whose diameters measure
less than 1.52 mm must be discarded at a loss of $0.85 per bolt.
Bolts whose diameters measure over 1.83 mm are sold at a reduced profit of $0.50 per bolt.
Find the expected profit for manufacturer B.
[16]
12
19.
(a)
Let B be the random variable “diameter of the bolts produced by
manufacturer B”.
 P( B  1.52)  0.242
1.52   

 P Z 
  0.242
0.16 

(M1)
1.52  
  0.69988
0.16
(A1)
   1.63
(b)
(A1)
Let A be the random variable “diameter of the bolts produced by
manufacturer A”.
1.52  1.56 

 P( A  1.52)  P  Z 

0.16 

(M1)
 P(Z  0.25)  0.40129 (0.4013)
P(diameter less than 1.52 mm)  0.44  0.40129  0.56  0.242
(A1)
(M1)(A1)
 0.312 (3 sf)
(c)
P  bolt produced by B d  1.52 
0.242  0.56
0.31209
(AG)
1.83 1.63 

P( B  1.83)  P  Z 
  0.10564
0.16 

(A1)
(M1)
 0.65236
(A1)
Expected gain  $ 8000  0.242  (0.85)  0.65236 1.50  0.10564  0.50 
20.
N2
(M1)(A1)
P(1.52  B  1.83)  1  0.242  0.10564
 $ 6605.28
N0
(M1)(A1)
 0.434
(d)
N2
(M1)
(A1)
N2
A random variable X is normally distributed with mean  and variance 2. If P (X  6.2) = 0.9474 and P (X
 9.8) = 0.6368, calculate the value of  and of .
[6]
20.
P(X > 6.2) = 0.9474 gives z =  1.62
(A1)
P(X < 9.8) = 0.6368 gives z = 0.35
(A1)
6.2  
9.8  
  1.62 and
 0.35
σ
σ
(M1)(A1)
6.2   =  1.62 
9.8   = 0.35 
13
Solving gives  = 1.83 and  = 9.16
21.
(A1)(A1) (C6)
Use mathematical induction to prove that 5n + 9n + 2 is divisible by 4, for n 
+
.
[9]
21.
Let f (n) = 5n + 9n + 2 and let Pn be the proposition that f(n) is divisible by 4.
Then f (1) = 16
So P1 is true
Let Pn be true for n = k ie f (k) is divisible by 4
Consider f (k + l) = 5k+1 + 9k+l + 2
= 5k(4 + 1) + 9k(8 + 1) + 2
= f (k) + 4(5k + 2 × 9k)
Both terms are divisible by 4 so f (k +1) is divisible by 4.
Pk true  Pk+l true
Since P1 is true, Pn is proved true by mathematical induction for n 
+
.
A1
A1
M1
M1
A1
A1
R1
R1
R1
N0