A random variable X given in the probability distribution below is the number of persons entering the bank every hour during a span of one week which were recorded by the security guard. a) What is the expected number of persons entering the bank every hour during that week? b) Compute the variance and standard deviation by illustrating its calculation step by step. x 1 2 3 4 P(x) 0.10 0.20 0.45 0.25

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B. Variance and Standard Deviation of a Discrete Random Variable
The variance of a random variable is the expected value of the square of the difference between
the assumed value of random variable and the mean. It is denoted by ² or Var (X).
Var (X)=[(x-µ)² P(x)] or o² = (x-μ)² P(x)]
Where x = outcome, μ = population mean, P(x) =probability of the outcome
The standard deviation of a random variable denoted by o is the square root of the variance.
0 =
Σ[(x − µ)² P(x)]
Note: The larger the value of the variance or standard deviation, the farther are the values of X
from the mean. It means that the larger the value of the variance or standard deviation, the more
spread are the values X from the mean or expected value.
Example:
A researcher surveyed the households in a small town. The random variable X represents
the number of college graduates in the households. The probability distribution of X is shown
below. Determine the variance and standard deviation.
X
0
2
P(X=x)
0.25
0.50
0.25
Solution:
Step 1. Find the expected value.
Based from previous example, E(X) or μ =1.
Step 2. Subtract the expected value from each outcome. Square each difference.
Step 3. Multiply each squared difference by the corresponding probability.
Step 4. Sum up all the figures obtained in step 3.
X
P(x)
x P(x)
x-μ
(x-μ)² P(x)
(x-μ)²
1
0
0.25
0
0-1 -1
1(0.25) = 0.25
1
0.50
0.50
1-1=0
0
0(0.50)=0
2
0.25
0.50
2-1=1
1
1(0.25) = 0.25
[x P(x)] = 1.00
0² =
[(x-μ)² P(x)] = 0.50
Variance, o² = 0.50
(You just need to take the square root of the value of variance to compute the standard deviation)
Standard Deviation, o = √0.50 ≈ 0.71
Based from the standard deviation it is evident that the variance or standard de
shows a low measure of spread which is associated with the random variable X
corresponding probabilities. In other words, the possible values x and their corres
probabilities are close to each other.
Take note that it is easier to interpret the value of standard deviation because it uses
same unit of measure of the random variable X.
Transcribed Image Text:B. Variance and Standard Deviation of a Discrete Random Variable The variance of a random variable is the expected value of the square of the difference between the assumed value of random variable and the mean. It is denoted by ² or Var (X). Var (X)=[(x-µ)² P(x)] or o² = (x-μ)² P(x)] Where x = outcome, μ = population mean, P(x) =probability of the outcome The standard deviation of a random variable denoted by o is the square root of the variance. 0 = Σ[(x − µ)² P(x)] Note: The larger the value of the variance or standard deviation, the farther are the values of X from the mean. It means that the larger the value of the variance or standard deviation, the more spread are the values X from the mean or expected value. Example: A researcher surveyed the households in a small town. The random variable X represents the number of college graduates in the households. The probability distribution of X is shown below. Determine the variance and standard deviation. X 0 2 P(X=x) 0.25 0.50 0.25 Solution: Step 1. Find the expected value. Based from previous example, E(X) or μ =1. Step 2. Subtract the expected value from each outcome. Square each difference. Step 3. Multiply each squared difference by the corresponding probability. Step 4. Sum up all the figures obtained in step 3. X P(x) x P(x) x-μ (x-μ)² P(x) (x-μ)² 1 0 0.25 0 0-1 -1 1(0.25) = 0.25 1 0.50 0.50 1-1=0 0 0(0.50)=0 2 0.25 0.50 2-1=1 1 1(0.25) = 0.25 [x P(x)] = 1.00 0² = [(x-μ)² P(x)] = 0.50 Variance, o² = 0.50 (You just need to take the square root of the value of variance to compute the standard deviation) Standard Deviation, o = √0.50 ≈ 0.71 Based from the standard deviation it is evident that the variance or standard de shows a low measure of spread which is associated with the random variable X corresponding probabilities. In other words, the possible values x and their corres probabilities are close to each other. Take note that it is easier to interpret the value of standard deviation because it uses same unit of measure of the random variable X.
III. Practice Exercises
A random variable X given in the probability distribution below is the number of persons entering
the bank every hour during a span of one week which were recorded by the security guard. a) What is
the expected number of persons entering the bank every hour during that week? b) Compute the variance
and standard deviation by illustrating its calculation step by step.
1
x
2
3
4
P(x)
0.10
0.20
0.45
0.25
IV Evaluation
Transcribed Image Text:III. Practice Exercises A random variable X given in the probability distribution below is the number of persons entering the bank every hour during a span of one week which were recorded by the security guard. a) What is the expected number of persons entering the bank every hour during that week? b) Compute the variance and standard deviation by illustrating its calculation step by step. 1 x 2 3 4 P(x) 0.10 0.20 0.45 0.25 IV Evaluation
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