prob distribution

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PROBABILITY DISTRIBUTION

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PROBABILITY DISTRIBUTION

8.1 BINOMIAL DISTRIBUTION

Probability of an event in a binomial distribution:

Where n = total number of independent trials

r = number of successful trials = 0, 1, 2, 3, …, n

p = probability of success

q = probability of failure = 1- p

1. Find the following probabilities using the formula ( ) n r n r

rP X r C p q .

a) 1

, 5, 32

p n r

Solution: 3 2

5

3

1 1( 3)

2 2

= 0.3125

P X C

b) 1

, 6, 23

p n r

[0.3292]

c) 1

, 7, 54

p n r

[0.0115]

d) 0.8, 7, 3p n r

[0.0287]

e) 0.4, 5, 2p n r

[0.3456]

f) 0.6, 8, 5p n r

[0.2787]

( ) n r n r

rP X r C p q

2. Solve each of the following problems:

Example: The probability that Ali can solve a given problem is 1

3. Find the probability

that Ali can solve 3 problems out of 4 given problems.

Solution : 1 1 2

, 1 , 3, 43 3 3

p q r n

3 1

4

3

1 2( 3) 0.0988

3 3P X C

a) The probability that all the oranges in a

packet are good is 2

3. May has bought 6

packets of oranges. Find the probability

that 3 of the packets contain good oranges.

[0.2195]

b) A die is tossed 8 times. The probability

of getting the number ‘1’ is 1

6. Find the

probability of getting the number ‘1’ for 5

times.

[0.0042]

c) A coin is tossed four times and the

results are recorded. Find the probability of

getting exactly 3 ‘heads’.

[0.25]

d) In a survey, 60% of the teenagers owned

a handphone. Find the probability that 4 of

6 teenagers own a handphone.

[0.3110]

e) Given that 2 out of four people are left-

handed. Find the probability that in a

family of 8, exactly 5 are left-handed.

[0.2188]

f) In a group of students, the probability

that a students cycle to school is 0.8. Find

the probability that among a group of 10

students, exactly 6 students cycle to school.

[0.0881]

3. Solve each of the following problems.

Example : A box contains 5 identical beads of which 3 are blue and the rest are red. 3

beads were drawn at random one after another with replacement. Find the

probability of getting

a) 2 blue beads,

b) at least 2 blue beads.

Solution : 3 2

, , 35 5

p q n

a) P( 2 blue beads) =

2 35

2

3 2( ) ( ) 0.23045 5

C

b) P(at least 2 blue beads)

= ( 2) ( 3)P X P X

=

2 3 3 25 5

2 3

3 2 3 2( ) ( ) ( ) ( ) 0.5765 5 5 5

C C

a) An unbiased die is flipped 4 times. Find

the probability of getting

a) an odd number for 3 times

b) a prime number which is less than 4

for 2 times

[a)0.25 b) 0.2963]

b) A doctor knows that 10% of the

patients will have undesirable side effects

from a certain drug. 8 patients are chosen at

random. Find the probability of getting

a) 2 patients will have undesirable side

effects

b) at least 2 patients will have

undesirable side effects

[a) 0.1488 b) 0.1869]

c) The probability that Ali wins a match is

0.8. If he plays 6 consecutive match, find

the probability that he wins

a) 2 matches

b) at least 5 matches

[a) 0.0154 b) 0.6553]

d) The pass rate in an examination is 90%.

If a group of 10 students were chosen at

random, find the probability that

a) 8 of them pass the

examination

b) at least 9of the pass the

examination

[a) 0.1937 b) 0.7361]

When

4. A random variable, X, has a binomial distribution with n trials and p as the probability

of success. Find the mean, variance and standard deviation of each of the following.

Example:

p = 0.89. n = 250

Solution:

0.89, 0.11

250(0.89) 223

var 250(0.89)(0.11) 24.475

tan 24.475 4.947

p q

mean np

iance npq

s dard deviation npq

a) p = 0.47, n = 100

[mean= 47, variance=24.91,

standard deviation=4.99]

b) p = 0.6, n = 50

[mean= 30, variance=12,


c) p = 0.15, n = 24

[mean= 3.6, variance=3.06,


d) p = 0.9, n = 150



e) p = 0.52, n = 200



2

( , )

,

var ,

tan deviation,

X B n p

mean np

iance npq

s dard npq

8.2 Normal Distribution

1. Find the following probability of Z-values by using the standard normal distribution

table.

Example:

P( Z > 1.5) = 0.0668

Example:

P( Z < -0.3) = P( Z > 0.3) = 0.3821

a) P( Z > 0.89) =

[0.1867]

b) P( Z > 2.24) =

[0.0123]

c) P(Z < -1.72) =

[0.0427]

d) P(Z < -0.644) =

[0.2598]

-0.3 0.3 1.5

2. Find the following probability of Z-values

Example:

P( 0.5 < Z < 1)

= P( Z > 0.5)- P( Z > 1)

= 0.3085 – 0.1587

= 0.1498

Example:

P( -2.18 < Z < -2)

= P( Z > 2)- P( Z > 2.18)

= 0.0228 – 0.0146

= 0.0082

a) P( 0.5 < Z < 1) =

[0.1498]

b) P( 1.6 < Z < 2) =

[0.032]

c) P( -0.62< Z < -0.54)

[0.027]

d) P( -1.74 < Z < -1.3)

[0.0559]

0.5 1 -2.18 -2 2.18 2

3. Find the following probability of Z-values

Example:

P( Z < 2.2) = 1- P( Z > 2.2)

= 1- 0.0139

= 0.9861

Example:

P( Z > -1.83) = 1 – P( Z > 1.83)

= 1 – 0.0336

= 0.9664

a) P( Z < 1.45)

[0.9265]

b) P( Z < 0.714)

[0.7624]

c) P( Z > -1.067)

[0.8570]

d) P( Z > -0.269)

[0.6060]

-1.83 2.2

4. Find the following probability of Z-values.

Example:

P( -1 < Z < 2) = 1 – P(Z > 2) –P(Z < -1)

= 1 – P(Z > 2) – P(Z >1)

= 1 – 0.228 – 0.1587

= 0.6133

a) P( -0.14 < Z < 1.52)

[0.4914]

b) P( -2.02 < Z < 0.689)

[0.7329]

c) P( -0.837 < Z < 1.97)

[0.7743]

d) P( -0.568 < Z < 0.724)

[0.4805]

e) P( -0.36< Z < 1.58)

[0.5835]

2 -1

5. Using the standard normal distribution X

Z

, complete the following table.

X Z

Example:

10

6

2 10 6

22

Z

a) 26

10 5

[3.2]

b) -5

25 20

[-3]

c)

[1]

3 8 -0.25

d)

[18]

9 3 3

e)100

80 40

[0.5]

f) 152

125 12

[2.25]

g) 104

116 10

[-1.2]

6. A random variable X is distributed normally with mean 20 and variance 25. Find the

following probabilities.

Example:

P( X > 30 ) = 30 20

( )5

P Z

= ( 2)P Z

= 0.0228

a) P( X > 26.8)

[0.0869]

c) P( X < 10)

[0.0228]

c) P( 28 < X < 32)

[0.032]

d) P( 15 < X < 28)

[0.7865]

e) P( 35 < X < 40)

[0.1554]

1.36 2

7. A random variable X is distributed normally with the following mean and standard

deviation. Find each of the following probability.

Example:

Mean = 3.7

Standard deviation = 1.2

P( X < 6) = 6 3.7

( )1.2

P Z

= ( 1.9167)P Z

= 1 ( 1.9167)P Z

= 1 – 0.02764 = 0.9724

a) Mean = 180

Standard deviation = 20

P(X < 190)

[0.6915]

b) Mean = 750


P( X < 800)

[0.9773]

c) Mean = 46


P( X > 60)

[0.0913]

1.9167

d) Mean = 190

Standard deviation =30

P(X>240)

[0.0478]

e) Mean = 30


P(28 < X < 40)

[0.3002]

f) Mean = 200


P(300 < X < 460)

[0.0619]

g) Mean = 200


P(300 < X < 460)

[0.0657]

prob distribution

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