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Republic of the Philippines
DEPARTMENT OF EDUCATION
Region I
SCHOOLS DIVISION OF PANGASINAN II
LUCIANO MILLAN NATIONAL HIGH SCHOOL
Asingan, Pangasinan
STATISTICS
AND
PROBABILITY
QUARTER 3- Week 1
Module 1
Most Essential Learning Competencies
Illustrates a random variable (discrete and continuous) (M11/12SP-IIIa-1) ;
Distinguished between a discrete and a continuous random variable (M11/12SP-IIIa-2) ;
Finds the possible values of a random variable (M11/12SP-IIIa-3) ;
Illustrates a probability distribution for a discrete random variable and its properties (M11/12SP-
IIIa-4) ; and
Computes probabilities corresponding to a given random variable (M11/12SP-IIIa-6).
For you to understand the lesson well, do the following activities. Have fun and good luck! Jumpstart Figure 1 shows how human beings are fascinated with the different events that involve probability. Humans are trying to make decisions based on how likely a particular event or experiment will happen. Since probability is all about chance, human-beings are trying to search for the ways to represent these events. Humans created random variable and use the concept to make it easy for an individual to verify the different results or probabilities of a given events or situation. Figure 1. Different Events that use Probability Lesson 1 Random Variable
Tossing the Coins Outcomes First Tossed Second Tossed Third Tossed
- After recording, you will notice that there are only 8 possible outcomes and no matter how you will repeat the tossing, the result will always be one of the 8 outcomes you already had.
- After writing all the possible outcomes, try to fill the given table below: OUTCOMES Number of Heads (H) Number of Tails (T) Note : Count the number of heads or tails in the given outcomes. Example: HHH , how many heads are there? Correct! There are 3 heads. How about tails? Correct! There are no or 0 tail.
- Based on the given table above, what can you say about the number of head? How about the number of tails? Is there a possibility that the number of heads or tails exceed the highest value or number recorded? Why do you say so? Discover The activities on the Jumpstart part are all essential for you to understand the content of this module which on Random Variable. Before proceeding with the discussion, let us have a short recall on the different concepts that we are going to use in tackling the different parts of this module.
In your basic probability class before, you already discussed the following terms: experiments/events, outcomes, sample, and sample spaces. Experiments are any movement or activity which can be dome repeatedly under similar or comparative condition. The outcomes are the result of a given experiment while the sample space is the set of all possible outcomes of an experiment. Let us have an example to illustrate this: rolling a die (dice) is an example of experiment. Upon rolling the die, it is expected that it will show a dot on the top which are either 1, 2, 3, 4, 5 or 6 , these are what we call the sample space of the given experiment. If you are only interested on a given value of dots, let say, three dots , then “3” is what we call the outcome or the sample. In this module, we are looking for a number that will be assigned for the result of an experiment. What you did on Activity 2 is a preparation for the discussion. If there are 4 coins instead of three coins tossed, what number or value can be assigned for the frequency of heads that will occur? If three cards are drawn from a deck of card, what number can be assigned for the frequency of face cards that will occur? The answer to these questions requires a knowledge of random variable. Recall also that a variable is a characteristic or attribute that can assume different values (e.g. in algebra, variables like x can be any number). We are going to use capital letter (X, Y, Z, and others) to denote or represent a variable. Let us examine the examples below to know more about Random Variable. Example 1: TOSSING THREE COINS Suppose you have three coins. We want to find the number of heads that will occur after tossing the coins. To determine the number of sample space, we are going to use the basic probability rule. Since there are three coins, and each coin will give us 2 outcomes, then () = 2 2 2 = 23 = 8 possible outcomes or sample space. Let us check! We are going to use tree diagram to show us the possible outcomes of the said experiment. Tree diagram is used in probability to show possibilities of an event. First Coin Second Coin Third Coin Sample Space H H HHH T HHT H T H HTH T HTT H H THH T T THT T H TTH T TT
G B GB
G GG
Sample Space / Possible Outcomes Random Variable X (number of boys in the family) BB 2 BG 1 GB 1 GG 0 Based on the sample space, there are 4 possible outcomes. Since we are only interested with the number of boys in the family, then the random variable = {0,1,2}. Observed that you cannot have 3 as the value of the random variable since it is impossible to have 3 boys in the family with two children as what the problem given as a condition. Example 3: DEFECTIVE or NON-DEFECTIVE Suppose 4 laptops are tested at random. Let N represent the non-defective laptops and let D represent the defective laptops. If we let Z be the random variable for the number of non-defective laptops , determine the value of the random variable Z. Solution: Based on the given problem, there are 4 laptops that we will be tested at random and each laptop can either be defective or non-defective. Let us first determine the number of sample space. Since there are 4 laptops and each laptop can be defective or non-defective, then () = 2 2 2 2 = 2^4 = 16 sample space / possible outcomes. First Second Third Fourth Sample Space N NNNN N D NNND N NNDN N D D NNDD N NDNN N D NDND D N NDDN N D D NDDD N DNNN N D DNND N N DNDN
D D D DNDD
N DDNN
N D DDND
D N DDDN
D D DDDD
Here are some enrichment activities for you to work on to master and strengthen the basic concepts you have learned from this lesson. Example 3: Am I DISCRETE or CONTINUOUS! Experiment Random Variable X Types of Random Variable Determine the defective cell phones in the given shipment The number of defective phones. Discrete (Reason: You can count the number of defective phones) Buying two trays of eggs in the market The weight of eggs in kilograms. Continuous (Reason: Since we are talking about the weight of the eggs, and weight is measurable) Rolling a pair of dice. The sum of the number of dots on the top faces. Discrete (Reason: Since the number of dots is countable, it takes a finite number: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12) Vehicular Accidents happened in La Union The number of accidents happened at the intersection Discrete (Reason: Accidents are countable) Learners will prepare for a quiz in Mathematics The time spent by the learners studying for a quiz in Mathematics Continuous (Reason: Time is a measurable unit) Just keep in mind that discrete random variables are countable, while continuous random variable are those measured or uncountable. Explore Activity 1: CLASSIFY ME! Directions : Classify whether the following random variables are DISCRETE or CONTINUOUS. Write D or C only. Write your answer on the space provided.
- The number of deaths per year attributed to lung cancer.
- The average amount of electricity consumed per household per month.
- The number of patient arrivals per hour at a medical clinic.
- The number of bushels of mangoes per hectare this year.
- The number of voters favoring a candidate.
- The number of people who are playing LOTTO each day.
- The amount of sugar in a cup of coffee.
- The time needed to finish the test.
- The number of female athletes in R1AA.
- The speed of a car.
- The number of dropouts in a school district for a period of 10 years.
- The amount of paint utilized in a building project.
- The number of siblings in a family of a region.
- The weight of newborns each year in a hospital.
- The number of COVID-19 cases each day. Activity 2: JUSTIFY ME! Directions: Given the following experiments, give the random variable X that fits the type of random variable given. The first entry is already answered as your guide. Experiment Random Variable X Type of Random Variable Example: The car travelling in La Union tourist spots. The number of La Union tourist spots visited. Discrete Playing mobile games Continuous The learners reading a module. Discrete Eating your breakfast every day. Continuous Learners going to school. Discrete Visiting the social media sites. Discrete Activity 3: FINDING MY VALUE! Directions: Find the possible values of the given random variable of the following experiments below. Write your answer on the space provided.
- Supposed two coins are tossed, let P be the random variable representing the number of heads that occur. Find the values of the random variable P. Possible Outcomes Random Variable P Answer : The values of the random variable P are.
- Inside the box are 2 balls – one white and one yellow. Two balls are picked one at a time with replacement (meaning the ball is replaced once picked). Let X be the random variable representing the number of white balls. Find the values of the random variable X.
Problem: Five coins are tossed. Let G be the random variable representing the number of heads (H) that occur. Find the values of the random variable G. What to find:
1. The number of possible outcomes or sample space (S). 2. The sample spaces (place in the table). 3. The correct values of random variable G. 4. Interpretation or description about the value of the random variable G.
Lesson 2 Probability Distribution and Its Properties Jumpstart Activity 1: FACT or BLUFF? Directions: Determine whether the statement is FACT or BLUFF. If the answer is bluff, you can modify the statement to make it true. Statement Fact Bluff
- There are 4 outcomes if you tossed two coins.
- If you tossed three coins where X be the random variable representing the number of tails that occur. The possible values of the random variable X are 0, 1, and 2.
- The sum of 3 + 1 = 5. 4 2 8
- The sum of 0.25 + 0.06 + 0.36 + 0.28 = 0.
- If = +1^ , then the value of (1) is 1. 6 2
- If = 3 , then the value of (4) is 3. −2 2 Activity 2: WHAT’S RANDOM? Directions: Determine the values of the random variable by answering the given problem below. Find the values of the random variable Y representing the number of green balls when 2 balls are drawn in succession without replacement from a jar containing 4 red balls and 5 green balls.
a. Let D represent the defective keyboard and N for non-defective computer. The number of sample space of the given problem is 8. This is because there are three keyboards subjected to inspection whether defective or non- defective, () = 2 2 2 = 23 = 8 sample space. Testing the First Keyboard Testing the Second Keyboard Testing the Third Keyboard Sample Space N N NNN D NND N D N NDN D NDD N N DNN D D DND D N DDN D DDD The sample space is: = {, , , , , , , }. b. Count the number of defective keyboards in each outcome in the sample space and assign this number to the outcome. For instance, if you list NND, then the number of defective keyboards is 1. Possible Outcome Random Variable X (number of defective keyboards) NNN 0 NND 1 NDN 1 NDD 2 DNN 1 DND 2 DDN 2 DDD 3 There are four possible values of the random variable X representing the number of defective keyboards. The possible values of X are 0, 1, 2, and 3. c. Each of these numbers corresponds to an event in the sample space S of equally likely outcomes for this experiment. Since the value of the random variable X represents the number of defective keyboards. If each of the outcomes is equally likely to occur, then the probability is: =
In assigning probability values for P(X) to each value of the random variable, since we know that the number of the sample space is 8, then the probability that: 1 defective keyboard will come out is 1 or 0 = 1 ; 8 8
2 defective keyboard will come out is 3 or 1 = 3 ; 8 3 3 defective keyboards will come out is 8 8 or 2 = 3 ; 8 and 4 defective keyboards will come out is 1 or 3 = 1 . 8 8 Random Variable X 0 1 2 3 Probability P(X) (0) = 1 8 (1) = 3 8 (2) = 3 8 (3) = 1 8 Note: These are the number of random variable X found in the table above. How many 0, 1, 2, and 3 occurred in the table of random variable X? The number 8 is the number of sample space. (() = ) You can also construct the table in this form: X 0 1 2 3 P(X) (^1) or 0. 8 (^3) or 0. 8 (^3) or 0. 8 (^1) or 0. 8 d. By adding all the probabilities or the values of (): 1 3 3 1 1 + 3 + 3 + 18 8
8
8
8
8
8 = 1 You can also use the decimals in determining the sum: 0.125 + 0.375 + 0.375 + 0.125 = 1 If you add all the probabilities, the sum is equal to 1. e. From the given activity, you can observe that the values of the probability ranges from 0 to 1 only. No values of the probability will be lesser than zero and no values will be greater than 1. The table given below is what we call the probability distribution or also known as the probability mass function. X 0 1 2 3 P(X) (^1) or 0. 8 (^3) or 0. 8 (^3) or 0. 8 (^1) or 0. 8 The probability distribution of a discrete random variable X is a list of the possible values of X and the corresponding probabilities of the values. It specifies the probability associated with each possible value of the random variable. The distribution is generally known as Probability Mass function .’
2
2 7 Example 3: Determine whether the given values can serve as the values of a probability distribution. a. (1) = 0.05, (2) = 1.01, (3) = 0. b. (4) = 3 , (5) = 7 , (3) = 1 20 20 2 Solution: a. The probability of each value of the random variable DOES NOT lie between 0 and 1 because (2) = 1.01. Therefore, this is not a probability distribution. b. The probability of each values of the random variable lies between 0 and 1 because ( 3
20 =^ 0.35^ ,^
(^1) = 0.50). The sum of its probabilities is equal to
3
20
Therefore, this is a probability distribution. Example 4: Determine whether the given values can serve as the values of a probability distribution of a random variable X. a. () = 1 for = 1, 2, 3, … , 9 This means that the value of (1) up to (9) is 1. 7 X 1 2 3 4 5 6 7 8 9 P(X) 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 The probability value () lies between 0 and 1 because^1 (^7) is equal to 0.. But the sum of its probabilities is not equal to 1. 1 1 1 1 1 1 1 1 1 9 7
Therefore, this is NOT a probability distribution. b. () = 12 (^25) for = 1, 2, 3, 4 Evaluating (), given the value of : Note: 1 becomes 10 because we need^2 to make the denominators the same or equal. Note: To convert fraction to decimal,dividethe numeratorbythe denominator.
12 25 3 P 12 12 1 = = 25 =. 3 = = 75 =. 𝟏 2 = =^ 12 =. 50 4 =^ = 12 100 =. Illustrating this in a table would give as: Z 1 2 3 4 P(Z) (^) 0.48 0.24 0.16 0. The probability of each value of the random variable lies between 0 and 1 and the sum of its probabilities is equal to 1. Therefore, this is a probability distribution. Explore Activity 1: AM I A PROBABILITY DISTRIBUTION? Directions: Determine whether the following represents a probability distribution or not. Explain your answer.
1 =
6 for^ =^ 1,2, 12 25 1 12 25 2 12 25 4 X 1 5 7 8 P(X) 1 4 1 8 1 4 1 8 X 1 3 5 7 P(X) 0.35 0.25 1.22 0.
