Discrete and Normal Random Variables: Mean, Variance, and Applications, Study notes of Family and Consumer Science

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STATISTICS

and PROBABILITY

Quarter 3 – Module 3

Mean and Variance of a Discrete Random

Variable and Normal Random Variable

SENIOR HIGH SCHOOL

Statistics and Probability – Grade 11 Alternative Delivery Mode Quarter 3 – Module 3 : Mean and Variance of a Discrete Random Variable and Normal Random Variable First Edition, 2020 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio Printed in the Philippines by ________________________ Department of Education – Region VII Schools Division of Negros Oriental Office Address: Kagawasan, Ave., Daro, Dumaguete City, Negros Oriental Telephone #: (035) 225 2376 / 541 1117 E-mail Address: [email protected] Development Team of the Module Writer: Didith T. Yap Editor: Mercyditha D. Enolpe Reviewer: Rickleoben V. Bayking Layout Artist: Jerry Mar B. Vadil Management Team: Senen Priscillo P. Paulin, CESO V Rosela R. Abiera Fay C. Luarez, TM, Ed.D., Ph.D. Maricel S. Rasid Nilita L. Ragay, EdD Elmar L. Cabrera Elisa L. Baguio, EdD

Introductory Message For the facilitator: Welcome to the STATISTICS & PROBABILITY Grade 11 Alternative Delivery Mode (ADM) Module on Mean and Variance of a Discrete Random Variable and Normal Random Variable! This module was collaboratively designed, developed and reviewed by educators both from public and private institutions to assist you, the teacher or facilitator in helping the learners meet the standards set by the K to 12 Curriculum while overcoming their personal, social, and economic constraints in schooling. This learning resource hopes to engage the learners into guided and independent learning activities at their own pace and time. Furthermore, this also aims to help learners acquire the needed 21st century skills while taking into consideration their needs and circumstances. In addition to the material in the main text, you will also see this box in the body of the module: As a facilitator, you are expected to orient the learners on how to use this module. You also need to keep track of the learners' progress while allowing them to manage their own learning. Furthermore, you are expected to encourage and assist the learners as they do the tasks included in the module. Notes to the Teacher This contains helpful tips or strategies that will help you in guiding the learners.

Assessment This is a task which aims to evaluate your level of mastery in achieving the learning competency. Additional Activities In this portion, another activity will be given to you to enrich your knowledge or skill of the lesson learned. Answer Key This contains answers to all activities in the module. At the end of this module you will also find: The following are some reminders in using this module:

1. Use the module with care. Do not put unnecessary mark/s on any part of the module. Use a separate sheet of paper in answering the exercises.
2. Don’t forget to answer What I Know before moving on to the other activities included in the module.
3. Read the instruction carefully before doing each task.
4. Observe honesty and integrity in doing the tasks and checking your answers.
5. Finish the task at hand before proceeding to the next.
6. Return this module to your teacher/facilitator once you are through with it. If you encounter any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. Always bear in mind that you are not alone. We hope that through this material, you will experience meaningful learning and gain deep understanding of the relevant competencies. You can do it! References This is a list of all sources used in developing this module.

LEARNING COMPETENCIES: ▪ Interprets the mean and the variance of a discrete random variable. (M11/12SP-IIIb- 3 ) ▪ Solves problems involving mean and variance of probability distributions. (M11/12SP-IIIb- 4 ) ▪ Illustrates a normal random variable and its characteristics. (M11/12SP-IIIc- 1 )

PRETEST

Find what is asked. Write your answers on your activity notebook/activity sheet.

A. Consider the problem below.

A certain university medical research center finds out that treatment of skin cancer by the use of chemotherapy has a success rate of 70%. Suppose five patients are treated with chemotherapy. The probability distribution of x successful cures of the five patients is given in the table below: X 0 1 2 3 4 5 P(x) 0.002 0.029 0.132 0.309 0.360 0. Probability distribution of cancer cures of five patients.

1. Find 𝜇 2. Find 𝜎^2 3. Find 𝜎 4. Graph p(x) and explain how μ and σ can be used to describe p(x). I OBJECTIVES: K: Interpret the mean (and variance) of a discrete random variable and illustrate a normal random variable; S: Solve for the mean (and variance) of a discrete random variable; and A: Recognize the importance of solving the mean and variance of probability distributions, and the importance of the normal curve in statistical inference. I

In a specified discrete random variable X, the mean, denoted by μ, is the summation of the products formed from multiplying the possible values of X with their corresponding probabilities. It is also called the expected value of X, and given a symbol E(X). μ = E(X) = X 1 P 1 + X 2 P 2 +¨˙ + X k Pk = Σ Xi Pi where; Σ is Sigma notation, and means to sum up. Note, that the empirical probabilities lean towards theoretical probabilities and, in consequence, the mean is also a long-run average, or the expected average outcome over many observations. That is, as the number of trials of a statistical experiment increases, the empirical average also gets closer and closer to the value of the theoretical average. Mean is the value that we expect the long-run average to approach and it is not the value of the random variable X that we expect to observe. The variance of a discrete random variable X measures the spread, or variability, of the distribution. The variance, usually denoted by the symbol σ² , and is also denoted as Var (X) and formally defined as σ² = Var (X) = Σ ( Xi - μ)² Pi The variance gives a measure of how far the values of X are from the mean. Note, that in nontrivial cases (i.e. when there is more than one possible distinct value of X), the variance will be a positive value. The bigger the value of the variance, the farther the values of X get from the mean. The standard deviation is defined as the square root of the variance of X. That is, σ = √𝑽𝒂𝒓 (𝑿) ’s New

Since we already reviewed the equation for calculating the mean, variance and standard deviation, let us now solve some problems and interpret the results. Example 1. GROCERY ITEMS The probabilities that a customer will buy 1,2,3,4, and 5 items in a grocery store are 3 10

1 10

1 10

2 10

3 10 respectively. What is the average number of items that a customer will buy? Solution: μ = E(X) = X1 P1 + X2 P2 +¨˙ + Xk Pk = Σ Xi Pi μ = (1)( 3 10

1 10

1 10

2 10

3 10

μ = 3 10

2 10

3 10

8 10

15 10 μ = 3. So, the mean of the probability distribution is 3.1. This implies that the average number of items that the customer will buy is 3.1. Example 1. The probabilities that a surgeon operates on 3,4,5,6, or 7 patients in any day are 0.15, 0.10, 0.20, 0.25, and 0.30, respectively. Find the average number of patients that a surgeon operates on a day. μ = E(X) = X1 P1 + X2 P2 +¨˙ + Xk Pk = Σ Xi Pi μ = 3(0.15) + 4(0.10) + 5(0.20) + 6(0.25) + 7(0.30) μ = 0.45 + 0.40 + 1 + 1.50 + 2. μ = 5. So, the average number of patients that a surgeon will operate in a day is 5.45 or 6. What is It

LESSON 2 Plotting the graph, we have; Notice that μ = 2.2 is the center of probability distribution. We can conclude, that the number of cars sold per day in a local car dealership is 2.2 or approximately 3 cars. The standard deviation which is σ = 1. 25 in this case, measures the spread of the probability distribution p(x). The variance gives a measure of how far the values of X are from the mean. The bigger the value of the variance, the farther the values of X get from the mean. Normal Random Variable and its Characteristics Review of Continuous Random Variables Recall the definition of a continuous random variable. It is a random variable that can take any real value within a specified range whereas a discrete random variable takes some on a countable number of values). A continuous variable involves the measurement of something, such as height of a randomly selected student, the weight of a newborn baby, or the length of time that the battery of a cellphone lasts. ’s In 0

Number of Cars Sold, x

Number of Cars Sold, x

There are many events in real life that generate random variables that have the natural tendency to approximate the shape of a bell. For example, the heights of a large number of seedlings that we see in fields normally, consist of a few tall ones, a few short ones, and most of them having heights in between tall and short. If a well-prepared test is administered to a class of 100 students, there will be a few high scores, as well as a few low scores. Most of the scores will be found in between these two extremes scores. In reality, if a distribution consists of a very large number of cases and the three measures of averages (mean, median, mode) are equal, then the distribution is symmetrical and the skewness is zero. In Statistics, such distribution is called normal distribution or simply normal curve. The normal curve has a very important role in inferential statistics. It provides a graphical representation of statistical values that are needed in describing the characteristics of populations as well as in making decisions. It is defined by an equation that uses the population mean, μ and the standard deviation, σ. There is no single curve, but rather a whole family of normal curves that have the same basic characteristics but have different means and standard deviations. Consider the following data pertaining to hospital weights (in pounds) of all the 36 babies that were born in the maternity ward of a certain hospital. 4.94 4.69 5.16 7.29 7.19 9.47 6.61 5.84 6. 3.45 2.93 6.38 4.38 6.76 9.01 8.47 6.8 6. 8.6 3.99 7.68 2.24 5.32 6.24 6.19 5.63 5. 5.26 7.35 6.11 7.34 5.87 6.56 6.18 7.35 4. The data have an average of 6.11 pounds and a standard deviation of 1.61 pounds. Shown below is the histogram for this data set. Observe that the histogram is approximately bell- shaped. https://www.teacherph.com/statistics-&-probability-senior-high- school-teaching-guide/ ’s New

For symmetric distributions with a single peak, such as the normal curve, take note that; Mean = Median = Mode. The standard deviation determines the shape of the graphs (particularly, the height and width of the curve). When the standard deviation is large, the normal curve is short and wide, while a small value for the standard deviation yields a skinnier and taller graph. Figure 2: Height and Width of the Curve https://www.teacherph.com/statistics-&-probability-senior-high- school-teaching-guide/ The cure above on the left, is shorter and wider than the curve on the right, because the curve on the left has a bigger standard deviation. Note, that a normal curve is symmetric about its mean and is more concentrated in the middle rather than in the tails. Properties of the Normal Probability Distribution Use Figure 3 to understand the properties of a normal probability distribution.

• The total area under the normal curve is equal to 1.
• The probability that a normal random variable X equals any particular value a, P(X=a) is zero (0) (since it is a continuous random variable).
• The probability that X is less than a, equals the area under the normal curve bounded by a and minus infinity (as indicated by the shaded area in figure 3 below)
• The probability that X is greater than some value a, equals the area under the normal curve bounded by a and plus infinity (as indicated by the non-shaded area in figure 3)
• Since the normal curve is symmetric about the mean, the area under the curve to the right of μ equals the area under the curve to the left of μ which equals ½, i.e. the mean μis the median.
• The probability density function is maximized at μ, i.e. the mode is also the mean
• The normal curve has inflection points (i.e. point at which a change in the direction of curvature occurs) at μ - s and at μ + s
• As x increases without bound (gets larger and larger), the graph approaches but never reaches, the horizontal axis. As x decreases without bound (gets larger and larger in the negative direction), the graph approaches, but never reaches, the horizontal axis Figure 3 https://www.teacherph.com/statistics-&-probability-senior-high-school-teaching-guide/

• Take note that every normal curve (regardless of its mean or standard deviation) conforms to the following "empirical rule" (also called the 68- 95 - 99.7 rule): https://www.teacherph.com/statistics-&-probability-senior-high-school-teaching-guide/
• About 68% of the area under the curve falls within 1 standard deviation of the mean.
• About 95% of the area under the curve falls within 2 standard deviations of the mean.
• Nearly the entire distribution (About 99.7% of the area under the curve) falls within 3 standard deviations of the mean. https://www.teacherph.com/statistics-&-probability-senior-high-school-teaching-guide/

These are the weights of babies in bold face that are within/under 1 standard deviation from 4.5 lbs. to 7.72 lbs. 4.94 4.69 5.16 7.29 7.19 9.47 6.61 5.84 6. 3.45 2.93 6.38 4.38 6.76 9.01 8.47 6.8 6. 8.6 3.99 7.68 2.24 5.32 6.24 6.19 5.63 5. 5.26 7.35 6.11 7.34 5.87 6.56 6.18 7.35 4. Therefore, there are 26 babies out of 36 babies or about 72% of the babies are within 1 standard deviation from the mean. b) To find the frequencies and relative frequency within/under 2 standard deviation, we need to add 2 standard deviation from the mean and subtract 2 standard deviation from the mean to identify the range/area under 2 standard deviation. Range: 𝜇 − 2 𝜎 < 𝑥 < 𝜇 + 2 𝜎

6. 11 − 2 ( 1. 61 ) < 𝑥 < 6. 11 + 2 ( 1. 61 )
7. 11 − 3. 22 < 𝑥 < 6. 11 + 3. 22
8. 89 < 𝑥 < 9. 33 These are the weights of babies in bold face that are within/under 2 standard deviation from 2.89 lbs. to 9.33 lbs. 4.94 4.69 5.16 7.29 7.19 9.47 6.61 5.84 6. 3.45 2.93 6.38 4.38 6.76 9.01 8.47 6.8 6. 8.6 3.99 7.68 2.24 5.32 6.24 6.19 5.63 5. 5.26 7.35 6.11 7.34 5.87 6.56 6.18 7.35 4. Therefore, there are 34 babies out of 36 babies or about 94 % of the babies are within 2 standard deviation from the mean. c) To find the frequencies and relative frequency within/under 3 standard deviation, we need to add 3 standard deviation from the mean and subtract 3 standard deviation from the mean to identify the range/area under 3 standard deviation. Range: 𝜇 − 3 𝜎 < 𝑥 < 𝜇 + 3 𝜎
9. 11 − 3 ( 1. 61 ) < 𝑥 < 6. 11 + 3 ( 1. 61 )
10. 11 − 4. 83 < 𝑥 < 6. 11 + 4. 83
11. 28 < 𝑥 < 10. 94

These are the weights of babies in bold face that are within/under 3 standard deviation from 1. lbs. to 10.94 lbs. 4.94 4.69 5.16 7.29 7.19 9.47 6.61 5.84 6. 3.45 2.93 6.38 4.38 6.76 9.01 8.47 6.8 6. 8.6 3.99 7.68 2.24 5.32 6.24 6.19 5.63 5. 5.26 7.35 6.11 7.34 5.87 6.56 6.18 7.35 4. Therefore, there are 3 6 babies out of 36 babies or about 100 % of the babies are within 3 standard deviation from the mean.

TASK 2.1 Validating the Empirical Rule

Fifty students were asked to run a 100-meter dash. The data below represents the time it took to finish the dash, and the histogram. The mean time for the 50 students is 15.8 seconds, and the standard deviation s is approximately 3.29 seconds. 16 14 14 16 21 14 17 15 16 21 14 10 9 20 12 12 19 11 15 14 18 18 13 18 23 8 20 13 16 23 16 17 15 18 17 16 13 15 18 19 12 12 15 17 14 16 17 16 16 21 Draw the normal curve, find the probability/relative frequency and the number of students under/within: a) 1 standard deviation b) 2 standard deviation c) 3 standard deviation ’s More