MEAN-VARIANCE-AND-DISCRETE-RANDOM-VARIABLE.pptx
- 1. MEAN VARIANCE AND DISCRETE RANDOM VARIABLE STATISTICS AND PROBABILITY
- 2. REVIEW • A basket contains 10 red balls and 4 white balls. If three balls are taken from the basket one after the other, determine the possible values of the random variable R representing the number of red balls. • Four coins are tossed. Let T be the random variable representing the number of tails that occur. Find the values of the random variable T.
- 3. REVIEW • Julia is a boutique shop owner in her town. Due to COVID-19 pandemic, wearing a face mask of a person is required for their safety. Since there are limited stocks available, she decided to start another business by aking a face mask. She started selling face masks from day 1 to day 10. The data she collected is shown in the table below.
- 4. DAY NUMBER OF FACEMASK (X) 1 25 2 20 3 15 4 14 5 15 6 10 7 12 8 10 9 15 10 14
- 5. Mean of a Discrete Random Variable • The Mean µ of a discrete random variable is the central value or average of its corresponding probability mass function. It is also called as the Expected Value. It is computed using the formula: • µ = ∑ X.P( ) 𝑥 • Where x is the outcome and P(x) is the probability of the outcome.
- 6. STEPS IN FINDING THE MEAN Step 1: Construct the probability distribution for the random variable X representing the number of items that the customer will buy. x P(x) 10 2/10 12 1/10 14 2/10 15 3/10 20 1/10 25 1/10
- 7. STEPS IN FINDING THE MEAN Step 2: Multiply the value of the random variable X by the corresponding probability. x P(x) x.P(x) 10 2/10=.20 20/10= 2 12 1/10=.10 12/10=1.2 14 2/10=.20 28/10=2.8 15 3/10=.30 45/10=4.5 20 1/10=.10 20/10=2 25 1/10=.10 25/10=2.5
- 8. STEPS IN FINDING THE MEAN Step 3: Add the results obtained in Step 2. Results obtained is the mean of the probability distribution. µ = ∑ X.P( ) 𝑥 = 2+1.2+2.8+4.5+2+2.5 = 15
- 9. ACTIVITY The probabilities that a physician attends to 2, 3, 4, or 5 patients in any day are 0.40, 0.25, 0.15, and 0.20, respectively. Find the average number of patients that a physician attends on a day. Interpret its value in the context of medical treatment.
- 10. ACTIVITY A bakeshop owner determines the boxes of pandesal that are delivered each day. Find the mean of the probability distribution shown. If the manager stated that 35 boxes of pandesal were delivered in one day, do you think that this is believable claim? NUMBER OF BOXES (X) PROBABIL ITY P(X) 35 0.10 36 0.20 37 0.30 38 0.30 39 0.10
- 11. Variance and Standard Deviation of a Random Variable The variance and standard deviation are two values that describe how scattered or spread out the scores are from the mean value of the random variable. The variance, denoted as , is determined using the formula: = ∑.p(x)
- 12. Variance and Standard Deviation of a Random Variable The standard deviation σ is the square root of the variance, thus, σ = √∑( − µ)² ( ) 𝑥 𝑝 𝑥 - variance σ – standard deviation µ - mean p(x) – probability of the outcome
- 13. DAY NUMBER OF FACEMASK (X) 1 25 2 20 3 15 4 14 5 15 6 10 7 12 8 10 9 15 10 14
- 14. 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. =15 x P(x) x.P(x) X- 10 0.20 2 -5 12 0.10 1.2 -3 14 0.20 2.8 -1 15 0.30 4.5 0 20 0.10 2 5 25 0.10 2.5 10 15
- 15. STEPS IN FINDING THE VARIANCE AND STANDARD DEVIATION 3. Square the result obtained in Step 2. x P(x) x.P(x) X- 10 0.20 2 -5 25 12 0.10 1.2 -3 9 14 0.20 2.8 -1 1 15 0.30 4.5 0 0 20 0.10 2 5 25 25 0.10 2.5 10 100 15
- 16. STEPS IN FINDING THE VARIANCE AND STANDARD DEVIATION 4. Multiply the results obtained in Step 3 by the corresponding probability. x P(x) x.P(x) X- .P(x) 10 0.20 2 -5 25 5 12 0.10 1.2 -3 9 0.90 14 0.20 2.8 -1 1 0.20 15 0.30 4.5 0 0 0 20 0.10 2 5 25 2.5 25 0.10 2.5 10 100 10 15
- 17. STEPS IN FINDING THE VARIANCE AND STANDARD DEVIATION 5. Get the sum of the results obtained in Step 4. Results obtained is the value of the variance of probability distribution. x P(x) x.P(x) X- .P(x) 10 0.20 2 -5 25 5 12 0.10 1.2 -3 9 0.90 14 0.20 2.8 -1 1 0.20 15 0.30 4.5 0 0 0 20 0.10 2 5 25 2.5 25 0.10 2.5 10 100 10 15 18.60
- 18. Standard Deviation of a Random Variable σ = √∑( − µ)² ( ) 𝑥 𝑝 𝑥 σ = √18.60 σ = 4.312771730569565 or 4.31
- 19. ACTIVITY The probability that a printer produces 0, 1, 2, 3, and 4 misprints are 15%, 22%, 10%, 38% and 15%, respectively. Construct a probability distribution table and compute the mean value, variance and standard deviation of the random variable and then interpret each value.