chapter 8 : probability distribution · 0.8. if he plays 6 consecutive match, find the probability...
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PROBABILITY DISTRIBUTION
Name
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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,
standard deviation=3.464]
c) p = 0.15, n = 24
[mean= 3.6, variance=3.06,
standard deviation=1.749]
d) p = 0.9, n = 150
[mean= 135, variance=13.5,
standard deviation=3.674]
e) p = 0.52, n = 200
[mean= 104, variance=49.92,
standard deviation=7.065]
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
Standard deviation = 25
P( X < 800)
[0.9773]
c) Mean = 46
Standard deviation = 3
P( X > 60)
[0.0913]
1.9167
d) Mean = 190
Standard deviation =30
P(X>240)
[0.0478]
e) Mean = 30
Standard deviation = 15
P(28 < X < 40)
[0.3002]
f) Mean = 200
Standard deviation =40
P(300 < X < 460)
[0.0619]
g) Mean = 200
Standard deviation =40
P(300 < X < 460)
[0.0657]