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i.
| X | 3 | 2 | 1 | 0 | -1 |
| P(X) | 0.3 | 0.2 | 0.4 | 0.1 | 0.05 |
Solution:
We know that the sum of probability distribution is always 1.
Sum of probabilities (P(X))=P(X=3)+P(X=2)+P(X=1)+P(X=0)+P(X=-1)
=0.3+0.2+0.4+0.1+0.05=1.05>1
The sum of probability distribution is not equal to 1. Hence, it is not the probability distribution of the given random variables.
ii.
| X | 0 | 1 | 2 |
| P(X) | 0.6 | 0.4 | 0.2 |
Solution:
Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)
=0.6+0.4+0.2=1.2>1
The sum of probability distribution is not equal to 1. Hence, it is not the probability distribution of the given random variables.
iii.
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | 0.5 | 0.2 | 0.1 | 0.1 |
Solution:
Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)
=0.1+0.5+0.2+0.1+0.1=1
The sum of probability distribution is equal to 1. Hence, it is the probability distribution of the given random variables.
iv.
| X | 0 | 1 | 2 | 3 |
| P(X) | 0.3 | 0.2 | 0.4 | 0.1 |
Solution:
Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)
=0.3+0.2+0.4+0.1=1
The sum of probability distribution is equal to 1. Hence, it is the probability distribution of the given random variables.
| X | -2 | -1 | 0 | 1 | 2 | 3 |
| P(X) | 0.1 | k | 0.2 | 2k | 0.3 | k |
Find the value of k.
Solution:
We know that the sum of probability distribution is always 1.
Sum of probability distribution (P(X))=P(X=-2)+P(X=-1)+P(X=0)+P(X=1)+P(X=2)+P(X=3)=1
=>0.1+k+0.2+2k+0.3+k=1
=>0.6+4k=1
=>4k=1-0.6
=>k=0.1
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X) | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
i. Find the value of a.
Solution:
We know that the sum of probability distribution is always 1.
Sum of probability distribution (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)=1
=>a+3a+5a+7a+9a+11a+13a+15a+17a=1
=>81a=1
=>a=1/81
ii. Find P(X<3).
Solution:
P(X<3)=P(X=0)+P(X=1)+P(X=2)
=1/81+3/81+5/81
=9/81=1/9
iii. Find P(X>=3).
Solution:
P(X>=3)=P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)
=7/81+9/81+11/81+13/81+15/81+17/81
=72/81=8/9
iv. Find P(0<X<5).
P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)
=3/81+5/81+7/81+9/81
=24/81=8/27=0.296
| X | 0 | 1 | 2 |
| P(X) | 3c3 | 4c-10c2 | 5c-1 |
where c>0. Find c.
Solution:
We know that the sum of probability distributions of a random variable is always 1.
=>3c3 +4c-10c2+5c-1=1
=>3c3-10c2+9c-2=0
Let c=1
3(1)-10(1)+9(1)-2=12-12=0
Therefore c=1.
By horn's method :
👁 ImageWe get a quadratic equation : 3c2-7c+2=0
From this quadratic equation we get,
=>3c2-6c-c+2=0
=>3c(c-2)-1(c-2)=0
=>(3c-1)(c-2)=0
=>3c-1=0; c-2=0
=>3c=1; c=2
=>c=1/3; c=2; c=1
We know that a single probability distribution cannot be 1 or more than one. So we take c=1/3.
Therefore, c=1/3.
Solution:
P(X<2)=P(X=0)+P(X=1)
=3(1/3)3+4(1/3)-10(1/3)2
=1/9+4/3-10/9
=1/3=0.33
Solution:
P(1<X<=2)=P(X=2)
=5(1/3)-1
=5/3-1
=2/3=0.66
Solution:
Sum of probability distributions= P(X=x1)+P(X=x2)+P(X=x3)+P(X=x4)=1
Given,
2P(X=x1)=3P(X=x2)=P(X=x3)=5P(X=x4)
=>P(X=x2)=2/3P(X=x1) ; P(X=x3)=2/1(P(X=x1) ; P(X=x4)=2/5(P(X=x1)
=>P(X=x1)+2/3(P(X=x1)+2/1(P(X=x1)+2/5(P(X=x1)=1
=>61/15(P(X=x1)=1
=>P(X=x1)=15/61=0.24
=>P(X=x2)=2/3(P(X=x1)
=2/3(15/61)
=10/61=0.16
=>P(X=x3)=2/1(P(X=x1)
=2(15/61)
=30/61=0.49
=>P(X=x4)=2/5(P(X=x1))
=2/5(15/61)
=6/61=0.09
Solution:
We know that the sum of probability distributions is equal to 1.
=>P(X=0)+P(X>0)+P(X<0)=1
Given,
P(X=0)=P(X>0)=P(X<0)
=>P(X=0)+P(X=0)+P(X=0)=1
=>3P(X=0)=1
=>P(X=0)=1/3
=>P(X>0)=1/3
=>P(X=1)+P(X=2)+P(X=3)=1/3
Given,
P(X=1)=P(X=2)=P(X=3)
=>3P(X=1)=1/3
=>P(X=1)=1/9 ; P(X=2)=1/9 ; P(X=3)=1/9
=>P(X<0)=1/3
=>P(X=-1)+P(X=-2)+P(X=-3)=1/3
Given,
P(X=-3)=P(X=-2)=P(X=-1)
=>3P(X=-1)=1/3
=>P(X=-1)=1/9 ; P(X=-2)=1/9 ; P(X=-3)=1/9
Solution:
Given that two cards are drawn from a well shuffled deck of 52 cards.
Then the random variables for the probability distribution of the number of aces could be
i. No ace is drawn
ii. One ace is drawn
iii. Two aces are drawn
i. No ace is drawn:
P(X=0)=52-4C2/52C2
=48C2/52C2
=48!/2!x46!/52!/2!x50!
=48x47/52x51
=188/221
=0.85
ii. One ace is drawn:
P(X=1)=4C1x48C1/52C2
=4x48x2/52x51
=32/221
=0.14
iii. Two aces are drawn:
P(X=2)=4C2/52C2
=6x2/52x51
=1/221
=0.004
Solution:
Given that three coins are tossed simultaneously.
Then the random variables for the probability distribution of the number of heads could be,
i. No heads
ii. One head
iii. Two heads
iv. Three heads
i. No heads:
P(X=0)=1C1x1C1x1C1/ 2C1x 2C1x 2C1
=1x1x1/2x2x2
=1/8
=0.125
ii. One head:
P(X=1)=1C1+1C1+1C1/8
=1+1+1/8
=3/8
=0.37
iii. Two heads:
P(X=2)=1C1+1C1+1C1/8
=3/8
=0.37
iv. Three heads:
P(X=3)=1/8
=0.125
Solution:
Given that four cards are drawn simultaneously from a well shuffled pack of 52 cards.
Then the random variables for the probability distribution of the number of aces drawn could be,
i. No aces
ii. One ace
iii. Two aces
iv. Three aces
v. Four aces
i. No aces
P(X=0)=48C4/ 52C4
=48x47x46x45/49x50x51x52
=0.71
ii. One ace
P(X=1)=4C1x48C3/52C4
=4x48x47x46x4/49x50x51x52
=0.25
iii. Two aces
P(X=2)= 4C2x48C2/ 52C4
=6x48x47x12/49x50x51x52
=0.024
iv. Three aces
P(X=3)= 4C3x48C1/52C4
= 4x48x24/49x50x51x52
= 0.0007
v. Four aces
P(X=4)=4C4/ 52C4
=1/ 270725
=0.000003694
Solution:
Given that three balls are drawn at random from a bag.
Then the value of random variable for the probability distribution of number of red balls could be,
i. No red ball
ii. One red ball
iii. Two red balls
iv. Three red balls
i. No red balls:
P(X=0)=6C3/10C3
=6x5x4/10x9x8
=1/6=0.16
ii. One red ball:
P(X=1)=6C2x4C1/10C3
=6x5x4x3/10x9x8
=1/2=0.5
iii. Two red balls:
P(X=2)=6C1x4C2/10C3
=6x4x3x3/10x9x8
=3/10=0.3
iv. Three red balls:
P(X=3)=4C3/10C3
=4x3x2/10x9x8
=1/30=0.03
Solution:
Given that five defective mangoes are mixed with 15 good ones.
Then the values of random variable for the probability distribution could be,
i. No defective
ii. One defective
iii. Two defective
iv. Three defective
v. Four defective
i. No defective:
P(X=0)=15C4/20C4
=15x14x13x12/20x19x18x17
=91/323=0.28
ii. One defective:
P(X=1)=15C3x5C1/20C4
=15x14x13x5x4/20x19x18x17
=455/969=0.469
iii. Two defective:
P(X=2)=15C2x5C2/20C4
=15x14x5x4x6/20x19x18x17
=70/323=0.21
iv. Three defective:
P(X=3)=15C1x5C3/20C4
=15x5x4x3x4/20x19x18x17
=10/323=0.03
v. Four defective:
P(X=4)=5C4/20C4
=5x4x3x2/20x19x18x17
=1/969=0.001
Solution:
Given that two dice are thrown simultaneously.
Then the outcomes would be as follows:
(1,1) ; (1,2) ; (1,3) ; (1,4) ; (1,5) ; (1,6) ;
(2,1) ; (2,2) ; (2,3) ; (2,4) ; (2,5) ; (2,6) ;
(3,1) ; (3,2) ; (3,3) ; (3,4) ; (3,5) ; (3,6) ;
(4,1) ; (4,2) ; (4,3) ; (4,4) ; (4,5) ; (4,6) ;
(5,1) ; (5,2) ; (5,3) ; (5,4) ; (5,5) ; (5,6) ;
(6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (6,6)
The values of the random variable could be: 2,3,4,5,6,7,8,9,10,11,12
P(X=2)=1/36=0.02
P(X=3)=2/36=1/18=0.05
P(X=4)=3/36=1/12=0.08
P(X=5)=4/36=1/9=0.11
P(X=6)=5/36=0.13
P(X=7)=6/36=1/6=0.16
P(X=8)=5/36=0.13
P(X=9)=4/36=1/9=0.11
P(X=10)=3/36=1/12=0.08
P(X=11)=2/36=1/18=0.05
P(X=12)=1/36=0.02
Solution:
Given that the students are selected without any bias.
Then the values of the random variable X could be: 14,15,16,17,18,19,20,21
P(X=14)=2/15=0.13
P(X=15)=1/15=0.06
P(X=16)=2/15=0.13
P(X=17)=3/15=0.2
P(X=18)=1/15=0.06
P(X=19)=2/15=0.13
P(X=20)=3/15=0.2
P(X=21)=1/15=0.06
Solution:
Given that five defective bolts are mixed with 20 good ones.
Then the values of random variable for the probability distribution would be,
i. No defective
ii. One defective
iii. Two defective
iv. Three defective
v. Four defective
i. No defective:
P(X=0)=20C4/25C4
=20x19x18x17/25x24x23x22
=969/2530=0.38
ii. One defective:
P(X=1)=20C3x5C1/25C4
=20x19x18x5x4/25x24x23x22
=114/253=0.45
iii. Two defective:
P(X=2)=20C2x5C2/25C4
=20x19x5x4x6/25x24x23x22
=38/253=0.15
iv. Three defective:
P(x=3)=20C1x5C2/25C4
=20x5x4x4x3/25x24x23x22
=4/253=0.015
v. Four defective:
P(X=4)=5C4/25C4
=5x4x3x2/25x24x23x22
=1/2530=0.0004
Solution:
Given that two cards are drawn with replacement from well shuffled pack of 52 cards.
Then the values of random variable for the probability distribution could be,
i. No ace
ii. One ace
iii. Two aces
i. No ace:
P(X=0)=(48/52 )x(48/52)
=144/169=0.85
ii. One ace:
P(X=1)=(48/52)x(4/52)+(4/52)x(48/52)
=24/169=0.14
iii. Two aces:
P(X=2)=(4/52)x(4/52)
=1/169=0.005
This chapter covers the concepts of mean, variance, and standard deviation of a random variable. It explains how to calculate the expected value (mean) and variance of a discrete random variable, both theoretically and using the formulas. The chapter also discusses the properties of variance, the relationship between mean and variance, and the importance of these measures in probability and statistics. Several examples are provided to illustrate the application of these concepts in solving real-world problems.
1. A random variable X has the following probability distribution:
X: 1 2 3 4
P(X): 0.1 0.2 0.3 0.4
Find the mean and variance of X.
2. The mean and variance of a binomial random variable X are 4 and 2, respectively. Find the parameters n and p.
3. The mean and variance of a Poisson random variable X are 3 and 3, respectively. Find the parameter λ.
4. In a normal distribution, the mean is 50 and the standard deviation is 10. Find the probability that a randomly selected observation will be between 40 and 60.
5. The weights of packages in a shipment are normally distributed with a mean of 25 kg and a standard deviation of 2 kg. What is the probability that a randomly selected package weighs more than 27 kg?
6. The heights of students in a class are normally distributed with a mean of 165 cm and a standard deviation of 8 cm. Find the percentage of students whose heights are between 155 cm and 175 cm.
7. The lifetime of a certain type of light bulb is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours. Find the probability that a randomly selected light bulb will last between 900 and 1100 hours.
8. In a certain population, the IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the percentage of the population that has an IQ score above 120.
9. The scores of a test are normally distributed with a mean of 60 and a standard deviation of 10. If 25% of the students scored below a certain score, find that score.
10. The number of accidents per month in a factory is a Poisson random variable with a mean of 2. Find the probability that in a given month, the number of accidents is less than 3.
