CABT SHS Statistics & Probability - Expected Value and Variance of Discrete Probability Distributions
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This document discusses key concepts about finding the mean and variance of discrete probability distributions. It defines that the mean of a random variable X is equal to the sum of the products of each possible value of X and its corresponding probability. The variance is the weighted average of the squared differences between each value and the mean. Several examples are provided to demonstrate calculating the mean, variance, and standard deviation of distributions.
CABT SHS Statistics & Probability - Expected Value and Variance of Discrete Probability Distributions
- 3. Mean and Variance of Discrete Probability Distributions Statistics & Probability Grade 11
- 4. Mean and Variance of Random Variables For a discrete random variable X with probability distribution function P(X), the EXPECTED VALUE or MEAN of X is given by E X x P x Mean of a Random Variable The mean is equal to the SUM of the PRODUCTS of the values of X and their corresponding probabilities P(X)
- 5. Mean and Variance of Random Variables Remarks: 1. “The mean of the random variable” can also mean “the mean of the probability distribution.” Mean of a Random Variable
- 6. Mean and Variance of Random Variables Remarks: 2. The mean of a random variable is the weighted AVERAGE value of all the possible values of X, with the weights being the probabilities of the respective values of X. Mean of a Random Variable
- 7. Find the mean of the given probability distribution. Mean and Variance of Random Variables Mean of a Random Variable X P(X) 0 0.05 1 0.15 2 0.35 3 0.45
- 8. number of head X 0 1 2 3 P(X) 1/8 3/8 3/8 1/8 If three coins are tossed, find the mean of the number of heads X that occur. Mean and Variance of Random Variables Mean of a Random Variable Solution Recall that the pdf for the number of heads is as follows:
- 9. Mean and Variance of Random Variables For a discrete random variable X with probability distribution function P(X), the VARIANCE of X is given by 2 2 2 Var X x P x Variance of a Random Variable Note: so 2 2 x P x E X 2 2 2 E X
- 10. Mean and Variance of Random Variables The STANDARD DEVIATION of a discrete random variable X is the SQUARE ROOT of its variance: 2 2 x P x The Standard Deviation
- 11. X P(X) 0 0.05 1 0.15 2 0.35 3 0.45 Find the variance and standard deviation of the given probability distribution. Mean and Variance of Random Variables Variance of a Random Variable
- 12. A study conducted by a TV station showed the number of televisions per household and the corresponding probabilities for each. Find the mean, variance, and standard deviation. Mean and Variance of Random Variables Mean of a Random Variable
- 13. a.An investment in Project A will result in a loss of P 26,000 with probability 0.30, break even with probability 0.50, or result in a profit of P 68,000 with probability 0.20. What is the expected value of this investment? Mean and Variance of Random Variables Mean of a Random Variable
- 14. b. Meanwhile, an investment in Project B will result in a loss of P 71,000 with probability 0.20, break even with probability 0.65, or result in a profit of P 143,000 with probability 0.15. Which investment is better: A or B? Mean and Variance of Random Variables Mean of a Random Variable
- 15. In the Mr. and Ms. CABT Raffle Draw, there is one winner of P 10,000, one winner of P 5000, and five winners of P1000. One thousand tickets are sold at P10 each. Find the expectation if a person buys one ticket. Mean and Variance of Random Variables Mean of a Random Variable
- 16. In the Mr. and Ms. CABT Raffle Draw, there is one winner of P 10,000, one winner of P 5000, and five winners of P1000. One thousand tickets are sold at P10 each. Find the expectation if a person buys one ticket. Mean and Variance of Random Variables Mean of a Random Variable Prize 0 1,000 5,000 10,000 Gain (X) 10 990 4,990 9,990 No. of tickets 993 5 1 1 Probability 0.993 0.005 0.001 0.001
- 17. In the Mr. and Ms. CABT Raffle Draw, there is one winner of P 10,000, one winner of P 5000, and five winners of P1000. One thousand tickets are sold at P10 each. Find the expectation if a person buys one ticket. Mean and Variance of Random Variables Mean of a Random Variable Gain X 10 990 4,990 9,990 P(X) 0.993 0.005 0.001 0.001 The following is the pdf:
- 18. JM the financial adviser suggests that his client Manay Ehljie select one of two types of bonds in which to invest her P 5,000. Bond X pays a return of 4% and has a default rate of 2%. Bond Y has a return of 2.5% and a default rate of 1%. Find the expected rate of return and decide which bond would be a better investment. When the bond defaults, the investor loses all the investment. Mean and Variance of Random Variables Mean of a Random Variable
- 19. JM the financial adviser suggests that his client Manay Ehljie select one of two types of bonds in which to invest her P 5,000. Bond X pays a return of 4% and has a default rate of 2%. Bond Y has a return of 2.5% and a default rate of 1%. Find the expected rate of return and decide which bond would be a better investment. When the bond defaults, the investor loses all the investment. Mean and Variance of Random Variables Mean of a Random Variable Amount of return for X: 5,000 0.04 200 Amount of return for Y: 5,000 0.025 125 The probability of obtaining a return (that is, NOT defaulting): For X: 100% 2% = 98% or 0.98 For Y: 100% 2.5% = 97% or 0.975
- 20. JM the financial adviser suggests that his client Manay Ehljie select one of two types of bonds in which to invest her P 5,000. Bond X pays a return of 4% and has a default rate of 2%. Bond Y has a return of 2.5% and a default rate of 1%. Find the expected rate of return and decide which bond would be a better investment. When the bond defaults, the investor loses all the investment. Mean and Variance of Random Variables Mean of a Random Variable Return X 5,000 200 P(X) 0.02 0.98 Return Y 5,000 125 P(Y) 0.01 0.975
- 21. Do you have any QUESTIONs?
- 24. Thank you!