Name: MEAN, VARIANCE AND STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE
Uploaded: 2022-03-09
Duration: 20 min 53 s
Description: MEAN, VARIANCE AND STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE.

MEAN, VARIANCE AND STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE. 

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MEAN, VARIANCE AND STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE

STATISTICS AND PROBABILITY PREPARED BY: LIANA P. BALICAO

MEAN OF A DISCRETE RANDOM VARIABLE. MEAN- is the value that we would expect to observe on average if the experiment is repeated many times.. 

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MEAN- is the value that we would expect to observe on average if the experiment is repeated many times.

MEAN OF A DISCRETE RANDOM VARIABLE. The mean (also called the expected value ) of a discrete random variable X is the number µ = ?[X·P(X)] where: µ = mean; X = values of the random variable X; and P(X) = the corresponding probabilities.. 

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The   mean   (also called the   expected value ) of a discrete random variable   X   is the number µ  = ?[X·P(X)] where:  µ = mean;  X = values of the random variable X; and  P(X) = the corresponding probabilities.

. Example. Consider rolling a die. What is the average number of spots that would appear?. 

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Consider rolling a die. What is the average number of spots that would appear?

. SOLUTION. Step 1 . Construct the probability distribution for the random variable X representing the number of spots that would appear.. 

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Step 1 .  Construct the probability distribution for the random variable X representing the number of spots that would appear.

X FREQUENCY P(X) 1 1 1/6 2 1 1/6 3 1 1/6 4 1 1/6 5 1 1/6 6 1 1/6

. SOLUTION. Step 2 . Multiply the value of the random variable X by the corresponding probability.. 

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Step 2 . Multiply the value of the random variable X by the corresponding probability.

X P(X) X*P(X) 1 1/6 1/6 2 1/6 1/3 3 1/6 ½ 4 1/6 2/3 5 1/6 5/6 6 1/6 1

. SOLUTION. Step 3 . Add the results obtained in Step 2.. 

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Step 3 . Add the results obtained in Step 2.

X P(X) X*P(X) 1 1/6 1/6 2 1/6 1/3 3 1/6 ½ 4 1/6 2/3 5 1/6 5/6 6 1/6 1 Σ X P(X)  = 7/2

. SOLUTION. Step 4 . Substitute the values into the formula. µ = ?[X·P(X)] µ = 7/2 or 3.5 Therefore, the mean or the expected value of the probability distribution is 7/2 or 3.5. 

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Step 4 . Substitute the values into the formula.  µ  = ?[X·P(X)]  µ  = 7/2 or 3.5 Therefore, the mean or the expected value of the probability distribution is  7/2  or  3.5

. Example. The probabilities that a surgeon operates on 2, 3, 4, 5 or 6 patients in any day are 0.20, 0.10, 0.20, 0.20 and 0.30, respectively. Find the mean of patients that a surgeon operates on a day.. 

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The probabilities that a surgeon operates on 2, 3, 4, 5 or 6 patients in any day are 0.20, 0.10, 0.20, 0.20 and 0.30, respectively. Find the mean of patients that a surgeon operates on a day.

. SOLUTION. Step 1 . Construct the probability distribution for the random variable Y representing the number of patients that a surgeon operates on a day.. 

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Step 1 .  Construct the probability distribution for the random variable Y representing the number of patients that a surgeon operates on a day.

Y P(Y) 2 0.20 3 0.10 4 0.20 5 0.20 6 0.30

. SOLUTION. Step 2 . Multiply the value of the random variable Y by the corresponding probability.. 

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Step 2 . Multiply the value of the random variable Y by the corresponding probability.

Y P(Y) Y*P(Y) 2 0.20 0.40 3 0.10 0.30 4 0.20 0.80 5 0.20 1.00 6 0.30 1.80

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Y P(Y) Y*P(Y) 2 0.20 0.40 3 0.10 0.30 4 0.20 0.80 5 0.20 1.00 6 0.30 1.80 Σ Y P(Y) = 4.30

SOLUTION. Step 4 . Substitute the values into the formula. µ = ?[X·P(X)] µ = 4.30 Therefore, the mean or the expected value of the probability distribution is 4.30. 

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Step 4 . Substitute the values into the formula.  µ  = ?[X·P(X)]  µ  = 4.30 Therefore, the mean or the expected value of the probability distribution is  4.30

. Variance and Standard Deviation of a Discrete Random Variable. 

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Variance and Standard Deviation of a Discrete Random Variable

The Variance and The Standard Deviation  of a discrete random variable describe the spread or variability of the random variable using the “mean” or the “expected value”

. The variance, 02, ofa discrete random variable X is the number. 

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The variance, 02, ofa discrete random variable X is the number

. Steps in Finding the Variance and Standard Deviation. 

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Steps in Finding the Variance and Standard Deviation

Find the mean of the probability distribution.  1 Subtract the mean from each value of the random variable X.  2 Square the results obtained in Step 2.  3 Multiply the results obtained in Step 3 by the corresponding probability  4 Get the sum of the results obtained in Step 4 to get the variance.  5 Get the square root of the result in Step 5 to get the standard deviation. 6

EXAMPLE. Find the variance and standard deviation of the probability distribution of the random variable X, which take only the values 1, 2, and 3, given that P(1)=10/33, P(2)=1/3, and P(3)=12/33.. 

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Find the variance and standard deviation of the probability distribution of the random variable X, which take only the values 1, 2, and 3, given that P(1)=10/33, P(2)=1/3, and P(3)=12/33.

. SOLUTION. Step 1. Find the mean of the probability distribution.. 

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Step 1. Find the mean of the probability distribution.

. SOLUTION. Step 2. Subtract the mean from each value of the random variable X. 

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Step 2. Subtract the mean from each value of the random variable X

10
37
37
xtP(X)
36
1-20691.06
2-2.062006
3-2.06=0.94

. SOLUTION. Step 3. Square the results obtained in Step 2. 

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Step 3. Square the results obtained in Step 2

1
2
3
10
33
1
3
12
10
33
2
3
36
-1.06
006
0.94
(-106)2
: 1.1236
= 0.0036

. SOLUTION. Step 4. Multiply the results obtained in Step 3 by the corresponding probability.. 

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Step 4. Multiply the results obtained in Step 3 by the corresponding probability.

3
x tP(X)
37
37
094
(X-g)2
11236
0.0036
0.8836
03405
0.0012
03213

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Step 5. Get the sum of the results obtained in Step 4. The result is the value of the variance.

10
37
12
37
X'P(X)
10
2
36
006
0.94
11236
0.0036
0.8836
0.3405
0.0012
0.3213
0,663

. SOLUTION. Step 6. Get the square root of the variance to get the standard deviation. 

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Step 6. Get the square root of the variance to get the standard deviation