1. The document discusses the mean of discrete random variables. It provides examples to illustrate how to calculate the mean of a discrete random variable by taking the sum of each possible value multiplied by its probability.
2. Calculating the mean involves identifying all possible outcomes, determining the probability of each outcome, multiplying each outcome by its probability, and finding the sum of these products. The mean represents the average value expected from the random variable.
3. Examples are provided to demonstrate calculating the mean for distributions describing number of cakes sold, artist clinic visits, family size, number of times a patient coughs after medicine, and the expected value of a coin tossing game. The mean is interpreted based on the context
1. The document discusses the mean of discrete random variables. It provides examples to illustrate how to calculate the mean of a discrete random variable by taking the sum of each possible value multiplied by its probability.
2. Calculating the mean involves identifying all possible outcomes, determining the probability of each outcome, multiplying each outcome by its probability, and finding the sum of these products. The mean represents the average value expected from the random variable.
3. Examples are provided to demonstrate calculating the mean for distributions describing number of cakes sold, artist clinic visits, family size, number of times a patient coughs after medicine, and the expected value of a coin tossing game. The mean is interpreted based on the context
1. The document discusses the mean of discrete random variables. It provides examples to illustrate how to calculate the mean of a discrete random variable by taking the sum of each possible value multiplied by its probability.
2. Calculating the mean involves identifying all possible outcomes, determining the probability of each outcome, multiplying each outcome by its probability, and finding the sum of these products. The mean represents the average value expected from the random variable.
3. Examples are provided to demonstrate calculating the mean for distributions describing number of cakes sold, artist clinic visits, family size, number of times a patient coughs after medicine, and the expected value of a coin tossing game. The mean is interpreted based on the context
1. The document discusses the mean of discrete random variables. It provides examples to illustrate how to calculate the mean of a discrete random variable by taking the sum of each possible value multiplied by its probability.
2. Calculating the mean involves identifying all possible outcomes, determining the probability of each outcome, multiplying each outcome by its probability, and finding the sum of these products. The mean represents the average value expected from the random variable.
3. Examples are provided to demonstrate calculating the mean for distributions describing number of cakes sold, artist clinic visits, family size, number of times a patient coughs after medicine, and the expected value of a coin tossing game. The mean is interpreted based on the context
Download as PPTX, PDF, TXT or read online from Scribd
Download as pptx, pdf, or txt
You are on page 1/ 19
STATISTICS
AND PROBABILITY Module 2
LESSON 1: Mean of Discrete Random Variables RECALL: • What is a random variable? Random Variables is a set whose elements are the numbers assigned to the outcomes of an experiment. It is a variable whose values are determined by chance. • What is a discrete random variables? Discrete Random Variables is a random variable which takes only finitely many or countably infinitely many different values. It is obtained by counting values for which there are no in-between values, such as the integers 0, 1, 2,3,4,5……
• What is a continuous random variables?
Continuous random variables is a random variable which contains uncountably infinite number of Possible values. It’s values are obtained by measurement And may assume all values In the The interval between any two litle think we are. S fight and here we us right now. EXPECTATIONS: After exploring this supplementary learning material, you
should be able to:
Illustrate the mean of a discrete random variable;
Calculate the mean of a discrete random variable;
Interpret the mean of a discrete random variable; and
Solve problems on expected values. LOOKING BACK
The importance of probability distribution is to help us to calculates various probabilities connected with the different values expected in the random variables, and the summary measures of a random variables, and this summary includes the mean.
This section explains how the mean of probability distributions
is computed and interpreted. ACTIVITY #1 Directions: Give all the possible values of each random variables.
1. A: Number of odd number outcomes in a roll of a die. {1,3,5} 2. B: Scores of a Grade 11 student in a 5 - item quiz. {0,1,2,3,4,5} 3. C: Height (in cm) of a child that does not exceed 98 cm. ≤98 cm; {98, 97,96,95,94....} 4. Find your final grade in Statistics class at the end of the quarter if you obtained the following data: Weight Raw Score Weight*RS Weighted Score
Final Grade:______ 88 BRIEF INTRODUCTION Mean is one of the most important probabilistic concepts in statistics. The mean of the discrete random variable X, denoted by X and µ, it is the weighted average of all possible values of X. It does not have to be a value of discrete random variable can assume. E(X) = µ = ΣxP(x) E(X) is the mean of the outcomes x µ is the mean ΣxP(x) is the sum of each random variable value x multiplied by i ts own probability P(x) Example 1 Let X be the number of cakes sold in a certain store during Valentine’s day, along with its corresponding probabilities is given in the table below. Solve the mean of X. X 10 15 20 25 30 35
P(X) 0.05 0.10 0.24 0.35 0.14 0.12
SOLUTION: Step 1: Identify all Step 2: Determine the Step 3: Multiply each Step 4: Find the sum of the products. possible outcomes. probability of each outcome value with possible outcome. its respective Σ X· P(X) = 23.95 probability. Step 5: Interpret the result. The value obtained Number of Cakes Probability P(X) X· P(X) is called the mean of the random variable X or the Sold (X) Mean of the probability distribution of X. The 10 0.05 0.5 mean tells us that the average number cakes sold 15 0.10 1.5 By a certain store during valentine’s day is 23.95. 20 0.24 4.8 Since we are referring to a number of cakes, thus 25 0.35 8.75 , 30 0.14 4.20 the mean is approximately 24 cakes. 35 0.12 4.20 Note: The mean of random variable X is also referred
to as the expected value, denoted by µx = E[X]. Example 2 Calculate the mean of a discrete random variable. Below is the probability for the artist visits nearby beauty clinic in a one-month period. X 0 1 3 5 7 P(X) 3/10 3/10 2/10 1/10 1/10 Solution: Following the steps above, the answers are as follows:
X P(X) X· P(X) 0 3/10 0
1 3/10 3/10 Therefore, the mean of the
distribution is 2.10. 3 2/10 6/10
5 1/10 5/10
7 1/10 7/10
Σ X· P( X) = 21/10 = 2.10 Example 3
Let X be the number of children in a certain family and their corresponding
probabilities. Compute for the expected value. Number of children 1 2 3 4 5 6 7
Solution: We can find the mean of the given probability distribution by multiplying X to its corresponding P(X) in decimal or fraction and adding the products. We therefore have: X 1 2 3 4 5 6 7 E[X] = P(X) 0.2450 0.3280 0.1020 0.0750 0.0875 0.0725 0.0900 Σ X· P(X) X· P(X) 0.2450 0.6560 0.3060 0.3000 0.4375 0.4350 0.6300 = 3.0095
Interpretation: The computed mean is 3.0095, we only consider an approximated value of 3 since we refer to a person . Hence, the average size is 3 for each family. Example 4 The doctor is interested in the number of times a patient’s coughing wakes them after taking a medicine. For a random sample of 40 patients, the following information was obtained.
Solution: Let X = the number of times a patient’s coughing wakes them after taking a medicine. X P(X) X· P(X) 1 9/40 9/40or 0.225
2 3/40 3/20or 0.15
3 15/40 9/8or 1.125 Therefore, the mean of the distribution
is 2.95. 4 7/40 7/10or 0.7
5 6/40 3/4or 0.75
Σ X· P(X) = 59/20 =2.95 Example 5 In a game of tossing two coins, you will receive Php50 if two tails appear; otherwise, you pay Php15. Is this game fair? Why?
Solution: Let X be the random variable for the gain in this game. The probability distribution is shown below. The table shows that a gain of Php50 (denoted by 50) is a winning X 50 -15 amount when exactly two tails appear in tossing two coins, while a P(X=x) 1/4 3/4 loss of Php15 (denoted by -15) when getting at most 1 tail. The expected value is: E[X] = Σ Xi · P( X i ) = X1 · P(X1) + X2 · P(X2) = 50 (1/4 ) + (—15)(3/4 ) Thus, thisgame is not fair because the expected value = 1.25 of the gain is 1.25 and not zero. (Note that the game is fair if the expected value of the gain or loss equal t o zero). It means that if you play this game several times, the average gain is Php1.25 Activity 2
1. Let X be the number of typographical errors found per page in certain books. The table below shows a probability distribution of X. X 0 1 2 3 4 P(X) 0.65 0.15 0.10 0.09 0.01 Find the mean of X.
2. Calculate the mean of the number of friends who are an anime fans (X) with the probability
distribution given the values 3, 5, 7, 9 and P(3) = 1/8 , P(5) = 1/2 , P(7) = 1/8and P(9) = 1/4 . Solve for µ .
CHECKING YOUR UNDERSTANDING
A. Assume that X is the discrete random variable with the probability shown on table below. The µ. .
B. Solve the following problems. 1.Below is probability distribution of the number of condominiums sold per month. X is the number of units sold. Calculate the µ. X P(x) X 2 3 4 5 P(X) 2/7 3/7 1/7 1/7
2.Find the missing probability and calculate the mean.
X 5 10 15 20 25
P (x) 3/10 1/10 ? 3/10 2/10
Student A has a synchronous 4-days a week. He attends classes 4 days a week 85% of the time, 3 days 7% of the time, 2 days 5% of the time and 1 day 3% of th e time, and a week is randomly chosen. Let X = the number of days Student A attends his asynchronous class per week. Solve the mean. REMEMBER
In computing the mean of the discrete random variable, we should follow the steps such as: (1) identifying all possible outcomes; (2) determining the probability for each possible outcome; (3) computing the product of the random variable and its corresponding probabilities; (4) getting the sum of all the products; and (5) interpreting the results.
The mean of the discrete random variable X, denoted by X and µ, is the
weighted average of expected outcome. It does not have to be a value of discrete random variable can assume. The formula used to compute for the mean of the discrete random variable is: E(X) = µ = ΣxP(x) • E(X) = is the mean of the outcomes x • µ is the mean • ΣxP(x) is the sum of each random variable value x multiplied by its own probability P(x) Word of the Day:
