Introduction to Business Analytics Course Part 7
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This document discusses statistical concepts and applications in business, focusing on probability theory and its methods including classical, relative frequency, and subjective approaches. It explains probability distributions and includes a case study on binomial distribution with practical examples such as Amir's chocolate bar purchases. Additionally, it covers the conditions for discrete probability distributions and provides insights into normal distribution concepts.
Introduction to Business Analytics Course Part 7
- 2. STATISTICAL CONCEPTS AND THEIR APPLICATIONS IN BUSINESS
- 3. PROBABILITY THEORY HEAD TAIL • Probability is a branch of mathematics that deals with the uncertainty of an event happening in the future. • Probability value always occurs within a range of 0 to 1. • Probability of an event, P(E) = No. of favorable occurrences No. of possibleoccurrences
- 4. ASSIGNING PROBABILITIES • Classical method – based on equally likely outcomes. E.g.: Rolling a dice. • Relative frequency method – based on experimentation or historical data. E.g.: A car agency has 5 cars. His past record as shown in the table shows his cars used in past 60 days. • Subjective method – based on judgment. E.g.: 75% chance that England will adopt to Euro currency by 2020. No. of carsused No. of days Probability 0 3 (3/60) = 0.05 1 10 (10/60) = 0.17 2 16 (16/60) = 0.27 3 15 (15/60) = 0.25 4 9 (9/60) = 0.15 5 7 (7/60) = 0.11
- 5. PROBABILITY DISTRIBUTION 0.2 0 1 2 3 4 5 6 • Probability distribution for a random variable gives information about how the probabilities are distributed over the values of that random variable. • Its defined by f(x) which gives probability of each value. • E.g. Suppose we have sales data for AC sale in last 300 days. Probability of units sold, f(x) 0.6 0.4 Probability of units sold, f(x) Units sold No. ofdays Probability of units sold,f(x) 0 10 0.03 1 55 0.18 2 150 0.5 3 55 0.18 4 25 0.08 5 5 0.02
- 6. • Discrete probability distribution • Following conditions should be satisfied – • A fixed number of trials • Each trial is independent of the others • The probability of each outcome remains constant from trial to trial. • Examples • Tossing a coin 10 times for occurrences of head • Surveying a population of 100 people to know if they watch television or not • Rolling a die to check for occurrence of a 2 BINOMIAL DISTRIBUTION
- 7. Example of binomial distribution: Amir buys a chocolate bar every day during a promotion that says one out of six chocolate bars has a gift coupon within. Answer the following questions : • What is the distribution of the number of chocolates with gift coupons in seven days? • What is the probability that Amir gets no chocolates with gift coupons in seven days? • Amir gets no gift coupons for the first six days of the week. What is the chance that he will get a one on the seventh day? • Amir buys a bar every day for six weeks. What is the probability that he gets at least three gift coupons? • How many days of purchase are required so that Amir’s chance of getting at least one gift coupon is 0.95 or greater? (Assume that the conditions of binomial distribution apply: the outcomes for Amir’s purchases are independent, and the population of chocolate bars is effectively infinite.) CASE STUDY—BINOMIAL DISTRIBUTION
- 8. Steps: Formula = nCr pr q n-r Where n is the no. of trials , r is the number of successful outcomes , p is the probability of success, and q is the probability of failure. Other important formulae include p + q =1 Hence, q = 1 – p CASE STUDY—BINOMIAL DISTRIBUTION (CONTD.) Thus, p =1/6 q =5/6
- 9. 1. Distribution of number of chocolates with gift coupons in 7 days: 7C r (1/6)r (5/6)7-r 2. Probability of failing 7 days : P(X=0) =(5/6)7 3. Probability of winning a coupon on the 7th day : 1/6 4. The number of winning at least 3 wrappers in six weeks: P(X ≥3)=1 – P(X≤2) =1 – (P(X=0)+P(X=1)+P(X=2) =1 – (0.0005+0.0040+0.0163) = 0.979 5. Number of purchase days required so that probability of success is greater than 0.95: CASE STUDY—BINOMIAL DISTRIBUTION (CONTD.) P(X ≥1) ≥0.95 = 1 – P(X ≤0) ≥ 0.95 = 1 – P(X=0) ≤0.05 = n ≥ 16.43 (applying log function) =17days.
- 10. • Theoretical model of the whole population • Centered around the mean and symmetrical on both sides • Standard normal distribution – mean 0 and standard deviation 1 NORMAL DISTRIBUTION
- 11. Thank You If you are looking for business analytics courses in Bangalore then visit: http://beamsync.com/business-analytics-training-bangalore/ Next Part We will Publish Soon.