http://sahatmozac.blogspot.com
ADDITIONAL MATHEMATICS
FORM 5
MODULE 13
PROBABILITY
DISTRIBUTIONS
http://mathsmozac.blogspot.com
1
http://sahatmozac.blogspot.com
MODULE 13 : PROBABILITY DISTRIBUTIONS
CONTENT
PAGE
13.1. CONCEPT MAP
1
13.2. PROBABILITY IN BINOMIAL DISTRIBUTION
2
13.3. ACTIVITY 1
5
1 13.4 BINOMIAL DISTRIBUTION GRAPH
6
13.5. MEAN, VARIANCE, STANDARD DEVIATION
OF BINOMIAL DISTRIBUTION
7
13.6. ACTIVITY 2
8
13.7. ACTIVITY 3 : SPM FOCUS PRACTICE
10
13.8. ANSWERS
12
http://mathsmozac.blogspot.com
2
http://sahatmozac.blogspot.com
13.1
CHAPTER 8
CONCEPT MAP
PROBABILITY DISTRIBUTIONS
BINOMIAL DISTRIBUTION
X ~B( n, p)
Probability
in Binomial Distribution
Mean, Variance, Standard
Deviation of Binomial
Distribution
Binomial Distribution
Graph
http://mathsmozac.blogspot.com
3
http://sahatmozac.blogspot.com
13.2 Probability in Binomial Distribution
P(X = r) = n C r p r q n r
Where
P = probability
X = binomial discrete random variable
r = number of successes (r = 0, 1, 2, …., n)
n = number of trials
p = probability of success (0 < p < 1)
q = probability of failure (q = 1 - p)
Example 1 :
In a town, the probability that rain will fall on any day is 0.3. Calculate the probability
that rain will fall on exactly 2 days in a certain week.
Solution :
Step 1 : Identify the parameters:
p = 0.3
q = 1 – 0.3 = 0.7
r=2
n=7
Step 2 : Substitute into the formula
P(X = r) = n C r p r q n r
P(X = 2) = 7 C 2 (0.3) 2 (0.7) 7 2
= (21)x(0.09)x(0.1681)
= 0.3177
Example 2 :
1
. Calculate
20
the probability that exactly 3 durians are rotten if a sample of 10 durians is chosen.
The probability that a durian chosen at random from a basket is rotten is
http://mathsmozac.blogspot.com
4
http://sahatmozac.blogspot.com
Step 1 : Identify the parameters:
1
p=
20
1
19
q=1–
=
20
20
r=3
n = 10
Step 2 : Substitute into the formula
P(X = r) = n C r p r q n r
P(X = 3) = 10 C 3 (
1 3 19 103
) (
)
20
20
= (120)x(0.000125)x(0.6983)
= 0.0105
Example 3 :
65% of Form Five students of a school pass the SPM Additional Mathematics paper. If a
sample of 5 student is chosen at random, calculate the probability that all of them pass the
SPM Additional Mathematics paper.
Solution :
Step 1 : Identify the parameters:
p = 65% = 0.65
q = 1 – 0.65 = 0.35
r=5
n=5
Step 2 : Substitute into the formula
P(X = r) = n C r p r q n r
P(X = 5) = 5 C 5 (0.65) 5 (0.35) 55
= (1)x(0.116)x(1)
= 0.116
http://mathsmozac.blogspot.com
5
http://sahatmozac.blogspot.com
Example 4 :
The probability that Danial will win a tennis competition is 0.8. If a total of 5 games are played,
find the probability that Danial will win
(a) exactly 3 games,
(b) at least three games,
(c) not more than 3 games.
Solution :
(a) P(X = r) = n C r p r q n r
P(X = 3) = 5 C 3 (0.8) 3 (0.2) 2
= (10)x(0.512)x(0.04)
= 0.2048
(b) P(X  3) = P(X = 3) + P(X = 4) + P(X = 5)
= [ 5 C 3 (0.8) 3 (0.2) 2 ] + [ 5 C 4 (0.8) 4 (0.2) 1 ] + [ 5 C 5 (0.8) 5 (0.2) 0 ]
= 0.2048 + 0.4096 + 0.3277
= 0.9421
(c) P(X  3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= [ 5 C 0 (0.8) 0 (0.2) 5 ] + [ 5 C 1 (0.8) 1 (0.2) 4 ] + [ 5 C 2 (0.8) 2 (0.2) 3 ] + [ 5 C 3 (0.8) 3 (0.2) 2 ]
= 0.00032 + 0.0064 + 0.0512 + 0.2048
= 0.2627
ALTERNATIVE METHOD
(d) P(X  3) = 1 - P(X = 4) - P(X = 5)
= 1 - 0.4096 - 0.3277
= 0.2627
http://mathsmozac.blogspot.com
6
http://sahatmozac.blogspot.com
13.3 ACTIVITY 1
1. During a shooting competition, the probability that Rasyidi will strike the target
is 0.8. If Rasyidi fires 8 shoots, calculate the probability that exactly 7 shots
strike the target.
2. In a certain school, 4 out 10 students have a computer at home. Calculate the
probability that from a sample of 5 students, none of them have a computer at
home.
3. The probability that Rafieq will win a badminton competition is 60%. If a total
of 7 games are played, find the probability that Rafieq will win
(a) exactly 4 games,
(b) at least 5 games,
(c) not more than 4 games.
http://mathsmozac.blogspot.com
7
http://sahatmozac.blogspot.com
13.4 Binomial Distribution Graph
Example :
A fair coin is tossed 4 times continuously. X represents the number of times a head appears.
(a) List the possible elements of X.
(b) Calculate the probability for the occurrence of each element of X.
(c) Hence, plot a graph to represent the binomial probability distribution of X.
Solution :
(a) Since the coin is tossed 4 times continuously, X = { 0, 1, 2, 3, 4 }
(b) P(X = r) = n C r p r q n r
1
1
1
n = 4, p = , q = 1 =
2
2
2
1 0 1 4
) ( ) = 0.0625
2
2
1 1 1 3
( ) ( ) = 0.25
2
2
1
1
( ) 2 ( ) 2 = 0.375
2
2
1
1
( ) 3 ( ) 1 = 0.25
2
2
1 4 1 0
( ) ( ) = 0.0625
2
2
P(X = 0) = 4 C 0 (
P(X = 1) = 4 C 1
P(X = 2) = 4 C 2
P(X = 3) = 4 C 3
P(X = 4) = 4 C 4
(c)
r
P(X = r)
0
0.0625
1
0.25
2
0.375
3
0.25
4
0.0625
The graph that represents the binomial probability distribution of X is as follows.
http://mathsmozac.blogspot.com
8
http://sahatmozac.blogspot.com
P(X= r)
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
r
13.5 Mean, Variance, Standard Deviation of Binomial Distribution
 = np
 2 = npq
=
npq
Example 1 :
40% of the students in a school wear spectacles. From a sample of 10 students, calculate the
mean, variance and standard deviation of the number of students who wear spectacles.
Solution :
p = 40% = 0.4, q = 1 – 0.4 = 0.6, n = 10
Mean,  = np
= 10 x 0.4
= 4
Variance,  2 = npq
= 10 x 0.4 x 0.6
= 2.4
Standard deviation,  =
npq
= 2.4
= 1.549
http://mathsmozac.blogspot.com
9
http://sahatmozac.blogspot.com
Example 2 :
In a group of teachers, the mean and variance of the number of teachers who own a Proton car
are 6 and 2.4 respectively. Find the probability that a teacher chosen at random owns a Proton
car.
Solution :
Mean,  = np = 6
np = 6 --------- (1)
Variance,  2 = npq = 2.4
npq = 2.4
(2)
,
(1)
npq
=
np
q=
p =
--------(2)
2 .4
6
0.4
1 – 0.4 = 0.6
Hence, the probability that a teacher chosen at random owns a Proton car is 0.6.
13.6 ACTIVITY 2
1
. If there are 20
16
papayas in the basket, calculate the mean and standard deviation of the rotten papayas in the
basket.
1. The probability that a papaya chosen at random from a basket is rotten is
http://mathsmozac.blogspot.com
10
http://sahatmozac.blogspot.com
2. For a binomial distribution, it is given that mean = 10 and variance = 4,
success and q = probability of failure. Find
(a) the value of p and q
(b) the probability of obtaining 2 successes out of 10 experiments.
p = probability of
3. In a farm, 45% of the chicks hatched from eggs are males.
(a) If 7 eggs are chosen at random, calculate the probability that 2 or more male chicks are
hatched.
(b) If there are 1000 eggs in the farm, calculate the mean and standard deviation of the number
of male chicks are hatched.
http://mathsmozac.blogspot.com
11
http://sahatmozac.blogspot.com
13.7 ACTIVITY 3 : SPM FOCUS PRACTICE
1. SPM 2003, PAPER 1. QUESTION 25.
In an examination, 70% of the students passed. If a sample of 8 students is randomly selected,
find the probability that 6 students from the sample passed the examination.
[3 marks]
2. SPM 2003, PAPER 2. QUESTION 10.
(a) Senior citizens make up 20% of the population of a settlement.
(i)
If 7 people are randomly selected from the settlement, find the probability that at
least two of them are senior citizens.
(ii)
If the variance of the senior citizens is 128, what is the population of the
settlement?
http://mathsmozac.blogspot.com
12
http://sahatmozac.blogspot.com
3. SPM 2004, PAPER 2. QUESTION 11.
(a) A club organizes a practice session for trainees on scoring goals from penalty kicks. Each
trainee takes 8 penalty kicks. The probability that a trainee scores a goal from a penalty kick
is p. After the session, it is found that the mean number of goals for a trainee is 4.8.
(i) Find the value of p.
(ii) If a trainee is chosen at random, find the probability that he scores at least one goal.
4. SPM 2005, PAPER 2. QUESTION 11.
For this question, give your answer correct to three significant figures.
(a) The result of a study shows that 20% of the pupils in a city cycle to school. If 8 pupils from
the city are chosen at random, calculate the probability that
(i) exactly 2 of them cycle to school,
(ii) less than 3 of them cycle to school.
http://mathsmozac.blogspot.com
13
http://sahatmozac.blogspot.com
13.8 ANSWERS
ACTIVITY 1:
1. 0.3355
2. 0.07776
3. (a) 0.2903
(b) 0.4199
(c) 0.5801
ACTIVITY 2:
1.  = 1.25
σ = 1.083
2. (a) p = 0.6, q = 0.4
(b) 0.9983
3. (a) 0.8976
(b) 15.73
ACTIVITY 3:
1. 0.2965
2. (a) (i) 0.4233
(ii) 800
3. (a) (i) p = 0.6
(ii) 0.993
4. (a) (i) 0.294
(ii) 0.797
http://mathsmozac.blogspot.com
14
http://sahatmozac.blogspot.com
PROGRAM DIDIK CEMERLANG AKADEMIK
SPM
ADDITIONAL MATHEMATICS
FORM 5
MODULE 14
PROBABILITY
DISTRIBUTIONS
http://mathsmozac.blogspot.com
15
http://sahatmozac.blogspot.com
MODULE 14 : PROBABILITY DISTRIBUTIONS
CONTENT
PAGE
2
14.1. CONCEPT MAP
14.2. PROBABILITY IN NORMAL DISTRIBUTION
3
14.3 Score- z
7
1 14.4 ACTIVITY 1
9
14.5 ACTIVITY 2
10
11
14.6. ACTIVITY 3
14.7 SPM QUESTIONS
13
14.8. SELF ASSESSMENT
15
18
14.9. ANSWERS
http://mathsmozac.blogspot.com
16
http://sahatmozac.blogspot.com
14.1
CHAPTER 8
CONCEPT MAP
PROBABILITY DISTRIBUTIONS
NORMAL DISTRIBUTION
2
X ~N (  ,  )
Probability
in Normal Distribution
Standardised Normal
Distribution
Z ~N (0,1)
Score- z
http://mathsmozac.blogspot.com
17
http://sahatmozac.blogspot.com
14.2
Probability in Normal Distribution
Standardised Normal Distribution
http://mathsmozac.blogspot.com
18
http://sahatmozac.blogspot.com
Example 1 :
Find the value of each of the following probabilities by reading the standardised normal
distribution table.
(a) P(Z > 0.934)
(b) P(Z  1.25)
Solution
(b) P(Z  1.25) = 1 – P(Z > 1.25)
= 1 – 0.1057
= 0.8944
1.25
1.25
http://mathsmozac.blogspot.com
19
http://sahatmozac.blogspot.com
(c) P(Z  - 0.23)
Solution
(c) P(Z  - 0.23) =
=
=
=
1 – P(Z < - 0.23)
1 – P(Z > 0.23)
1 – 0.40905
0.59095
-0.23
0.23
(d) P(Z > - 1.512)
Solution
(d) P(Z < - 1.512) = P(Z > 1.512)
= 0.06527
-1.512
1.512
(e) P(0.4 < Z < 1.2)
Solution
(e) P(0.4 < Z < 1.2) = P(Z > 0.4) – P(Z > 1.2)
= 0.3446 – 0.1151
= 0.2295
0.4
1.2
0.4
http://mathsmozac.blogspot.com
1.2
20
http://sahatmozac.blogspot.com
(f) P(- 0.828 < Z  - 0. 555)
Solution
(f) P(- 0.828 < Z  - 0. 555) = P(Z > 0.555) – P(Z > 0.828)
= 0.28945 – 0.20384
= 0.08561
-0.828
-0.555
0.828
0.555
(g) P(- 0.255  Z < 0.13)
Solution
(g) P(- 0.255  Z < 0.13) =
=
=
=
-0.255
0.13
1 – P(Z < - 0.255) – P(Z > 0.13)
1 – P(Z > 0.255) – P(Z > 0.13)
1 – 0.39936 – 0.44828
0.15236
-0.255
http://mathsmozac.blogspot.com
0.13
21
http://sahatmozac.blogspot.com
14.3
Score- z
Example 2 :
Find the value of each of the following :
(a)
(b)
(c)
(d)
P(Z  z) = 0.2546
P(Z < z) = 0.0329
P(Z < z) = 0.6623
P(z < Z < z 0.548) = 0.4723
Solution
(a) P(Z  z) = 0.2546
Score-z = 0.66
0.2546
z
(b) P(Z < z) = 0.0329
Score-z = -1.84
(c)
P(Z < z) = 0.6623
1 - P(Z > z) = 0.6623
P(Z > z) = 1 – 0.6623
= 0.3377
Score-z = 0.419
(d)
P(z < Z < z 0.548) = 0.4723
1 – P(Z < z) – P(Z > 0.548) = 0.4723
1 – P(Z < z) – 0.2919 = 0.4723
P(Z < z) = 1 – 0.2919 – 0.4723
= 0.2358
Score-z = -0.72
http://mathsmozac.blogspot.com
22
http://sahatmozac.blogspot.com
Example 3 :
The masses of the loaves of bread produced by a bakery are normally distributed with a
mean of 400 g and a standard deviation of 15 g. Calculate
(a) the standardised score for the mass of 405 g,
(b) the probability that a loaf of bread chosen at random will have a mass of more than
405 g,
(c) the percentage of loaves of bread that have masses of less than 403 g,
(d) the number of the loaves of bread that have masses between 394 g and 409 g if 1000
loaves of bread are produced in a day.
Solution
(a) X = 405,  = 400,  = 15
The standardised score,
Z=
X 
405  400 1
=
=

15
3
(b) P(X > 405)
405  400 

= PZ 

15


= P (Z > 0.3333)
= 0.3696
(c) P(X < 403)
403  400 

= PZ 

15


= P (Z < 0.2)
= 1 – P(Z > 0.2)
= 1 – 0.4207
= 0.5793
= 0.5793 x 100%
= 57.93%
(d) P(394 < X < 409)
409  400 
 394  400
= P
Z

15
15


= P(-0.4 < Z < 0.6)
= 1 - P(Z > 0.4) – P(Z > 0.6)
= 1 – 0.3446 – 0.2743
= 0.3811
http://mathsmozac.blogspot.com
23
http://sahatmozac.blogspot.com
Hence, if 1000 loaves of bread are produced, the number of loaves that masses between
394 g and 409 g is 0.3811 x 1000 = 381.1 = 381
14.4 Activity 8.1
If Z is the variable for standard normal distribution, find the value for each of the
following:
1.P( Z> 0.637)
2. P( Z> 0.1)
3. P( Z  2.018)
4 P( Z<-0.5)
.
5 P( Z  -0.34)
6. P( Z  -3.47)
7. P( -0.225  Z< 0.135)
8. P(-0.25< Z< 0)
http://mathsmozac.blogspot.com
24
http://sahatmozac.blogspot.com
14.5 Activity 8.2
Find the z-score for each of the following:
1. P (Z< z) = 0.5987
2. P (Z< z) = 0.1515
3.P (Z< z) = 0.8599
4. P (Z< z) = 0.4247
5.P ( z< Z< 0.683) = 0.4143
6. P (Z  z) = 0.99865
7. P (0.5<Z< z) = 0.0342
8. P (Z  z) = 0.9898
http://mathsmozac.blogspot.com
25
http://sahatmozac.blogspot.com
14.6 Activity 8.3
1. Determine the probability for each of the following if
Mean =55, standard deviation = 5
(i)
P(X < 68)
(ii)
P(X > 56)
(iii)
P(58<X<63)
2. In a normal distribution of a variable X,the standard scores of
1
and 1 have x
2
scores .11 and 12 respectively. Find the mean and standard deviation of the normal
distribution.
http://mathsmozac.blogspot.com
26
http://sahatmozac.blogspot.com
3. The additional mathematics mark for a group of students has a normal distribution
and the mean is 48 while the standard deviation is 5. Determine the probability for
a student to have the mark of
(a) greater than 55
(b) between 40 and 52
4. Find the probability if:
Mean = 43; standard deviation = 8
(i)
P (X>35)
(ii)
P(X<28)
(iii)
P( X > 33)
http://mathsmozac.blogspot.com
27
http://sahatmozac.blogspot.com
14.7
1.The mass of the workers in a factory is normally distributed with a mean of 67.86 kg
and a variance of 42.25kg².200 of the workers in the factory weigh between 50 kg and 70
kg. Find the total number of worker in the factory.
2.A survey on body-mass is done on a group of students has a normal distribution with a
mean of 50 kg and a standard deviation of 15kg.
(i) If a student is chosen at random,calculate the probability that his mass is less than 41
kg.
(ii) Given that 12% of the students have a mass of more than m kg, find the value of m.
http://mathsmozac.blogspot.com
28
http://sahatmozac.blogspot.com
3.The mass of water- melon chosen randomly from the orchard follows a normal
distribution with a mean of 3.2 kg and a standard deviation of 0.5kg. Find
(i) the probability that a water-melon chosen randomly from the orchard has a mass of
not more than 4.0kg
(ii) the value of m if 60% of the water-melons from the orchard have a mass of more
than m kg.
http://mathsmozac.blogspot.com
29
http://sahatmozac.blogspot.com
14.8 SELF ASSESSMENT
1. The life span of a certain machine is normally distributed with mean 1500 days
and standard deviation 30 days.
(a) what is than probability that a machine chosen at random has a life spam of
(i)
more than 1532 days
(ii)
between 1480 days and 1530 days
(b) Given that 5% of the machines have than spans of more than n days, find the
value of n
http://mathsmozac.blogspot.com
30
http://sahatmozac.blogspot.com
2.The weight of fish reared by a farmer are normally distributed with mean 1.2 kg and
standard deviation 0.1 kg. Calculate
(a) the probability that a fish chosen at random has a weight that is between 0.9kg
and 1.3kg
(b) the percentage of fish with weights less than 1.25 kg
Answer;
http://mathsmozac.blogspot.com
31
http://sahatmozac.blogspot.com
3.The result of a test is normally distributed with mean 55 and standard deviation
10. If the passing mark is at least 40, find the probability of the chosen student
passed the test.
4.The probability for Mazlan whom threw a stone to hit the target is 0.65.Find the
quantity of stones for Mazlan to hit the target at least once so that the probability would
be greater than 0.9
http://mathsmozac.blogspot.com
32
http://sahatmozac.blogspot.com
ANSWERS
14.9 Activity 8.1
1.0.2620
2. 0.4602
3. 0.9782
4. 0.3085
.
5. 0.6331
6. 0.99974
7. 0.1427
8. 0.0987
Activity 8.2
1. 0.25
2. -1.03
3. -1.08
4. -0.19
5. -0.417
6. -3.00
7. 0.600
8. 2.32
Activity 8.3
1. (i) 0.99534 (ii) 0.4207 (iii) 0.2195
2.   10,   2
3. (a) 0.0808
(b) 0.7333
4. (i) 0.8413
(ii) 0.0303
(iii) 0.7888
http://mathsmozac.blogspot.com
33
http://sahatmozac.blogspot.com
SELF ASSESSMENT
1. (a) (i) 0.1430,
(ii) 0.589,
(b) n=1549
2. (a) 0.84,
(b) 69.15%
3. 9332
4. 3
SPM QUESTIONS
1. 319 workers
2. (i) 0.2743,
(ii) m=67.625kg
3. (i) 0.9452,
(ii) m=3.0735
http://mathsmozac.blogspot.com
34