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Computing the Mean of the Discrete Probability Distribution.pptx
- 2. OBJECTIVES: •ILLUSTRATE AND CALCULATETHE MEAN OF A DISCRETE RANDOM VARIABLE •INTERPRET THE MEAN OF A DISCRETE RANDOM VARIABLE •SOLVE PROBLEMS INVOLVING MEAN OF PROBABILITY DISTRIBUTION
- 3. REVIEW •WHAT ARE THESTEPS IN CONSTRUCTING PROBABILITY DISTRIBUTION? •WHAT ARE THE PROPERTIES OF THE PROBABILITY DISTRIBUTION?
- 4. WHY IS THECONCEPT OF COMPUTING THE MEAN OF THE RANDOM VARIABLE RELEVANT IN A REAL- LIFE SITUATION? • THE CONCEPT OF COMPUTING THE MEAN OF A RANDOM VARIABLE IS RELEVANT IN REAL-LIFE SITUATIONS FOR SEVERAL REASONS: • UNDERSTANDING CENTRAL TENDENCY: THE MEAN PROVIDES A MEASURE OF CENTRAL TENDENCY, INDICATING WHERE THE DATA POINTS TEND TO CLUSTER AROUND. IN REAL-LIFE SITUATIONS, THIS HELPS IN UNDERSTANDING TYPICAL OR AVERAGE VALUES WITHIN A DATASET. FOR EXAMPLE, THE MEAN SALARY OF EMPLOYEES IN A COMPANY GIVES INSIGHT INTO THE TYPICAL EARNINGS.
- 5. • DECISION MAKING:IN MANY DECISION-MAKING PROCESSES, ESPECIALLY IN BUSINESS AND ECONOMICS, THE MEAN OF A RANDOM VARIABLE IS CRUCIAL. FOR INSTANCE, COMPANIES MIGHT USE THE MEAN DEMAND FOR THEIR PRODUCTS TO MAKE PRODUCTION AND INVENTORY DECISIONS. • PREDICTION AND FORECASTING: MEAN VALUES ARE OFTEN USED IN PREDICTION AND FORECASTING MODELS. FOR INSTANCE, IN WEATHER FORECASTING, HISTORICAL MEAN TEMPERATURES OR RAINFALL LEVELS ARE USED AS A BASELINE FOR PREDICTING FUTURE CONDITIONS.
- 6. • RESOURCE ALLOCATION:UNDERSTANDING THE MEAN OF A RANDOM VARIABLE HELPS IN RESOURCE ALLOCATION. FOR INSTANCE, IN HEALTHCARE, KNOWING THE AVERAGE LENGTH OF HOSPITAL STAYS HELPS IN PLANNING RESOURCES SUCH AS BEDS, STAFF, AND MEDICAL SUPPLIES. • RISK ASSESSMENT: MEAN VALUES ARE USED IN RISK ASSESSMENT AND MANAGEMENT. FOR INSTANCE, IN FINANCE, THE MEAN RETURN ON AN INVESTMENT IS USED TO ASSESS ITS POTENTIAL PROFITABILITY AND ASSOCIATED RISKS.
- 7. •OVERALL, COMPUTING THEMEAN OF A RANDOM VARIABLE IS RELEVANT IN REAL-LIFE SITUATIONS BECAUSE IT PROVIDES VALUABLE INSIGHTS INTO THE CHARACTERISTICS OF DATA DISTRIBUTIONS, AIDS IN DECISION-MAKING PROCESSES, AND SUPPORTS VARIOUS ANALYTICAL AND PREDICTIVE TASKS ACROSS DIFFERENT DOMAINS.
- 8. CONSIDER ROLLING ADIE WHAT IS THE AVERAGE NUMBER OF SPOTS THAT WOULD APPEAR Steps Solution Step 1 Construct the probability distribution for the random variable X representing the number of spots that would appear Number of Spots X Probability (X) 1 𝟏 𝟔 2 𝟏 𝟔 3 𝟏 𝟔 4 𝟏 𝟔 5 𝟏 𝟔
- 9. Steps Solution Step 2 Multiplythe value of the random variable X by the corresponding probability Number of Spots X Probability (X) X.P(X) 1 𝟏 𝟔 𝟏 𝟔 2 𝟏 𝟔 𝟐 𝟔 3 𝟏 𝟔 𝟑 𝟔 4 𝟏 𝟔 𝟒 𝟔 5 𝟏 𝟔 𝟓 𝟔
- 10. Steps Solution Step 3 Addthe result obtained in step 2 𝜮𝑿 · 𝑷 𝑿 = 𝟐𝟏 𝟔 = 𝟑 · 𝟓 Number of Spots X Probability (X) X.P(X) 1 𝟏 𝟔 𝟏 𝟔 2 𝟏 𝟔 𝟐 𝟔 3 𝟏 𝟔 𝟑 𝟔 4 𝟏 𝟔 𝟒 𝟔 5 𝟏 𝟔 𝟓 𝟔 The Value obtained in step number 3 is called the mean of the random variable X· The mean tells us the average number of spots that would appear in a roll of a die. So the average number of spots that would appear is 𝟑 · 𝟓
- 11. THE FORMULA FORTHE MEAN OF THE PROBABILITY DISTRIBUTION THE MEAN OF A RANDOM VARIABLE WITH A DISCRETE PROBABILITY DISTRIBUTION IS = 𝑿𝟏 · 𝑷 𝑿𝟏 + 𝑿𝟐 · 𝑷 𝑿𝟐 + 𝑿𝟑 · 𝑷 𝑿𝟑 +. . . , +𝑿𝒏 · 𝑷 𝑿𝒏 = 𝜮𝑿 · 𝑷 𝑿 WHERE 𝑿𝟏, 𝑿𝟐𝑿𝟑,. . . , 𝑿𝒏 ARE THE VALUES OF THE RANDOM VARIABLE X 𝑷 𝑿𝟏 , 𝑷 𝑿𝟐 , 𝑷 𝑿𝟑 . . . , 𝑷 𝑿𝒏 ARE THE CORRESPONDING PROBABILITY
- 12. GROCERY ITEMS The Probabilitiesthat the customer will buy 1, 2, 3, 4, or, 5 items in a grocery store are 3 10 , 3 10 , 3 10 , 3 10 , and, 3 10 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 What is the average number of the items that a customer will buy.
- 13. GROCERY ITEMS Steps Solution Step1 Construct the probability distribution for the random variable X representing the number of spots that would appear Number of Spots X Probability (X) 1 𝟑 𝟏𝟎 2 𝟏 𝟏𝟎 3 𝟏 𝟏𝟎 4 𝟐 𝟏𝟎 5 𝟑 𝟏𝟎
- 14. Steps Solution Step 2 Multiplythe value of the random variable X by the corresponding probability Number of Spots X Probability (X) X.P(X) 1 𝟑 𝟏𝟎 𝟑 𝟏𝟎 2 𝟏 𝟏𝟎 𝟐 𝟏𝟎 3 𝟏 𝟏𝟎 𝟑 𝟏𝟎 4 𝟐 𝟏𝟎 𝟖 𝟏𝟎 5 𝟑 𝟏𝟎 𝟏𝟓 𝟏𝟎
- 15. Steps Solution Step 3 Addthe result obtained in step 2 𝜮𝑿 · 𝑷 𝑿 = 𝟑𝟏 𝟏𝟎 = 𝟑 · 1 Number of Spots X Probability (X) X.P(X) 1 𝟑 𝟏𝟎 𝟑 𝟏𝟎 2 𝟏 𝟏𝟎 𝟐 𝟏𝟎 3 𝟏 𝟏𝟎 𝟑 𝟏𝟎 4 𝟐 𝟏𝟎 𝟖 𝟏𝟎 5 𝟑 𝟏𝟎 𝟏𝟓 𝟏𝟎 So the mean of the probability distribution is 3.1. This implies that the average number of items that the customer will buy is 3.1
- 16. THINK, PAIR, AND,SHARE COMPLETE THE TABLE BELOW AND FIND THE MEAN OF THE FOLLOWING PROBABILITY DISTRIBUTION X P(X) X·P(X) 1 1 7 6 1 7 11 3 7 16 1 7 21 1 7
- 17.
- 18. QUIZ INSTRUCTIONS: COMPUTE THEMEAN OF THE PROBABILITY DISTRIBUTION. 5 POINTS FOR EACH ITEM. X P(X) X·P(X) 1 𝟏 𝟕 6 𝟏 𝟕 11 𝟑 𝟕 16 𝟏 𝟕 21 𝟏 𝟕 X P(X) X·P(X) 1 𝟑 𝟏𝟎 2 𝟏 𝟏𝟎 3 𝟐 𝟏𝟎 4 𝟐 𝟏𝟎 5 𝟐 𝟏𝟎
- 19.