ADDITIONAL MATHEMATICS FORM 5

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ADDITIONAL MATHEMATICS FORM 5 MODULE 13

PROBABILITY DISTRIBUTIONS

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MODULE 13 : PROBABILITY DISTRIBUTIONS

CONTENT

PAGE

13.1. CONCEPT MAP

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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

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13.8. ANSWERS

12


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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


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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


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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


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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


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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.


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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.


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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


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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


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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.


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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?


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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.


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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


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PROGRAM DIDIK CEMERLANG AKADEMIK

SPM

ADDITIONAL MATHEMATICS FORM 5 MODULE 14

PROBABILITY DISTRIBUTIONS


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MODULE 14 : PROBABILITY DISTRIBUTIONS

CONTENT

PAGE 2

14.1. CONCEPT MAP 14.2. PROBABILITY IN NORMAL DISTRIBUTION

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14.3 Score- z

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1 14.4 ACTIVITY 1

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14.5 ACTIVITY 2

10 11

14.6. ACTIVITY 3 14.7 SPM QUESTIONS

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14.8. SELF ASSESSMENT

15 18

14.9. ANSWERS


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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


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14.2

Probability in Normal Distribution

Standardised Normal Distribution


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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


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(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


1.2

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(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


0.13

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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


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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


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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

ADDITIONAL MATHEMATICS FORM 5

ADDITIONAL MATHEMATICS FORM 5

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