2-computing The Mean, Variance And Standard Deviation.pptx

Computing the Mean, Variance and Standard Deviation of a Discrete Probability Distribution

Review: Given the values of the variables X and Y, evaluate the following summation.

𝑋1 = 4 𝑋2 = 2 𝑋3 = 5 𝑋4 = 1 𝑌1 = 2 𝑌2 = 1 𝑌3 = 0 𝑌4 = 2 1. ) σ 𝑋

2. ) σ 𝑌

3. ) σ 𝑋𝑌

4. ) σ 𝑋 + 𝑌

5. ) σ 4𝑋𝑌

Mean –

Is used to describe a set of data as to what point the item values or scores tend to cluster or concentrate.

Standard Deviation is

the measure of the variation of a set of data in terms of the amounts by which individual values differ from their mean.

The following are the scores of students in a test. Compute the mean

Score 42 50 53 38 46

Number of students 8 12 9 7 4

Activity: Number Spots: Consider rolling a die. What is the average number of spot that would appear?

Steps in Finding the Mean of Discrete probability Distribution  Step

1. Construct the probability distribution for the random variable X representing the number of spots that would appear.  Step 2. Multiply the value of the random variable X by the corresponding probability.  Step 3. Add the results obtained in step 2.

Formula for the Mean of the Probability Distribution The mean of random variable with a discrete probability distribution is: 𝜇 = 𝑋1 ∙ 𝑃 𝑋1 + 𝑋2 ∙ 𝑃 𝑋2 + 𝑋3 ∙ 𝑃 𝑋3 + . . . , + 𝑋𝑛 ∙ 𝑃 𝑋𝑛 𝑜𝑟 𝜇 = ෍ 𝑋 ∙ 𝑃 𝑋 Where: 𝑋1 , 𝑋2 , 𝑋3 … . 𝑋𝑛 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑋; 𝑎𝑛𝑑 𝑃 𝑋1 , 𝑃 𝑋2 , 𝑃 𝑋3 , … , 𝑃 𝑋𝑛 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠.

Group 1: Grocery Items: 𝑇ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 𝑎 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟 𝑤𝑖𝑙𝑙 𝑏𝑢𝑦 1, 2, 3, 4, 𝑎𝑛𝑑 5 𝑖𝑡𝑒𝑚𝑠 𝑖𝑛 3 1 1 2 3 𝑔𝑟𝑜𝑐𝑒𝑟𝑦 𝑠𝑡𝑜𝑟𝑒 𝑎𝑟𝑒 , , , , 𝑎𝑛𝑑 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦. 10 10 10 10 10 What is the average number of items that a costumer will buy?

Group 2. Surgery Patients The probabilities that a surgeon operates on 3,4,5,6 or 7 patients in any day are 0.15, 0.10, 0.20, 0.25 and 0.30 respectively. Find the average number of patients that a surgeon operates on a day.

Mathematical Journal  What

does the mean of probability distribution tell us? How do you interpret the mean of a probability distribution?  Why is it that values of a discrete random variable expressed in whole numbers but the mean of its probability distribution are most of the time expressed in decimal numbers?

Review: Compute the variance of this frequency distribution. Complete the table X 5 8 10 12 15

f 3 5 4 5 3 N = 20

𝑋 − 𝑋ത

ത 2 (𝑋 − 𝑋)

ത 2 𝑓(𝑋 − 𝑋)

ത 2= ෍ 𝑓(𝑋 − 𝑋)

The variance and standard deviation describe the amount of spread, dispersion, or variability of the items in a distribution. How do you describe the spread or dispersion in a probability distribution?

Formula for the Variance and Standard Deviation of a Discrete Probability Distribution The variance of a discrete probability distribution is given by formula: 𝝈𝟐 = ෍(𝑿 − 𝝁)𝟐 ∙ 𝑷(𝑿) 𝒐𝒓 𝝈𝟐 = ෍ 𝑿𝟐 ∙ 𝑷 𝑿 − 𝝁𝟐 The standard deviation of a discrete probability distribution is given by the formula 𝝈=

෍(𝑿 − 𝝁)𝟐 ∙ 𝑷(𝑿)

𝒐𝒓 𝝈 =

෍ 𝑿𝟐 ∙ 𝑷 𝑿 − 𝝁𝟐

Where: X = Value of the random variable

P(X) = Probability of the random variable X 𝝁 = mean probability distribution

STEPS IN FINDING THE VARIANCE AND STANDARD DEVIATION  1.

Find the mean of the probability distribution.

 2.

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

 3.

Square the results obtained in Step 2.

 4.

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

 5.

Get the sum of the results obtained in Step 4.

Halimbawa: Number of Cars Sold The number of cars sold per day at a local car dealership, along with its corresponding probabilities, is shown in the succeeding table. Compute the variance and the standard deviation of the probability distribution by the following the given steps. Complete the table.

Solution 1 No. of cars Probabili 𝑋 ∙ 𝑃(𝑋) sold (X) ty P(X) 0 1/10 1

2/10

2

3/10

3

2/10

4

2/10

𝑋− 𝜇

(𝑋 − 𝜇)2

𝜎 2 = σ(𝑋 − 𝜇)2 ∙ 𝑃(𝑋) =

(𝑋 − 𝜇)2 ∙ 𝑃(𝑋)

Solution 2 X 0 1 2 3 4

P(X) 1/10 2/10 3/10 2/10 2/10 2

𝑿 ∙ 𝑷(𝑿)

2

2

𝜎 = ෍𝑋 ∙ 𝑃 𝑋 − 𝜇 =

𝟐

𝑿 ∙ 𝑷(𝑿)

Group Activity

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