🗊Презентация Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1)

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Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №1Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №2Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №3Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №4Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №5Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №6Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №7Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. 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Discrete Probability Distributions. Week 7 (1), слайд №31Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №32Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №33Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №34Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №35Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №36Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №37Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. 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BBA182 Applied Statistics Week 7 (1)Discrete random variables – expected variance and standard deviation Discrete Probability Distributions Dr Susanne Hansen Saral Email: susanne.saral@okan.edu.tr https://piazza.com/class/ixrj5mmox1u2t8?cid=4# www.khanacademy.org
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BBA182 Applied Statistics Week 7 (1)Discrete random variables – expected variance and standard deviation Discrete Probability Distributions Dr Susanne Hansen Saral Email: susanne.saral@okan.edu.tr https://piazza.com/class/ixrj5mmox1u2t8?cid=4# www.khanacademy.org

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Cumulative Probability Function, F() Practical application
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Cumulative Probability Function, F() Practical application

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Cumulative Probability Function, F(x0) Practical application: Car dealer
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Cumulative Probability Function, F(x0) Practical application: Car dealer

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Cumulative Probability Function, F(x0) Practical application
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Cumulative Probability Function, F(x0) Practical application

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Cumulative Probability Function, F(x0) Practical application
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Cumulative Probability Function, F(x0) Practical application

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Properties of discrete random variables: Expected value The expected value, E[X], also called the mean, of a discrete random variable is found by multiplying each possible value of the random variable by the probability that it occurs and then summing all the products: The expected value of tossing two coins simultaneously is :
Описание слайда:
Properties of discrete random variables: Expected value The expected value, E[X], also called the mean, of a discrete random variable is found by multiplying each possible value of the random variable by the probability that it occurs and then summing all the products: The expected value of tossing two coins simultaneously is :

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Expected value for a discrete random variable Exercise X is a discrete random variable. The graph below defines a probability distribution, P(X) for X. What is the expected value of X?  
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Expected value for a discrete random variable Exercise X is a discrete random variable. The graph below defines a probability distribution, P(X) for X. What is the expected value of X?  

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Expected value for a discrete random variable X is a discrete random variable. The graph below defines a probability distribution, P(X) for X. What is the expected value of X?  
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Expected value for a discrete random variable X is a discrete random variable. The graph below defines a probability distribution, P(X) for X. What is the expected value of X?  

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Expected variance of a Discrete Random Variables The measurements of central tendency and variation for discrete random variables: Expected value E[X] of a discrete random variable - expectations Expected Variance, of a discrete random variable Expected Standard deviation, of a discrete random variable
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Expected variance of a Discrete Random Variables The measurements of central tendency and variation for discrete random variables: Expected value E[X] of a discrete random variable - expectations Expected Variance, of a discrete random variable Expected Standard deviation, of a discrete random variable

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Variance of a discrete random variable The variance is the measure of the spread of a set of numerical observations to the expected value, E[X]. For a discrete random variable we define the variance as the weighted average of the squares of its possible deviations (x - ):
Описание слайда:
Variance of a discrete random variable The variance is the measure of the spread of a set of numerical observations to the expected value, E[X]. For a discrete random variable we define the variance as the weighted average of the squares of its possible deviations (x - ):

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Variance and Standard Deviation Let X be a discrete random variable. The expectation of the average of squared deviations about the mean, , is called the expected variance, denoted and given by: Expected Standard Deviation of a discrete random variable X
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Variance and Standard Deviation Let X be a discrete random variable. The expectation of the average of squared deviations about the mean, , is called the expected variance, denoted and given by: Expected Standard Deviation of a discrete random variable X

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Exercise: n n Expected value,E[X], and variance, of car sales At a car dealer the number of cars sold daily could vary between 0 and 5 cars, with the probabilities given in the table. Find the expected value and variance for this probability distribution
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Exercise: n n Expected value,E[X], and variance, of car sales At a car dealer the number of cars sold daily could vary between 0 and 5 cars, with the probabilities given in the table. Find the expected value and variance for this probability distribution

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Calculation of variance of discrete random variable. Car sales – example Calculating the expected value: E(x) = (0)(.15)+(1)(.3)+(2)(.2)+(3)(.2)+(4)(.1)+(5)(.05)= 1.95 rounded up to 2 (discrete random variable) Calculating the expected variance: =
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Calculation of variance of discrete random variable. Car sales – example Calculating the expected value: E(x) = (0)(.15)+(1)(.3)+(2)(.2)+(3)(.2)+(4)(.1)+(5)(.05)= 1.95 rounded up to 2 (discrete random variable) Calculating the expected variance: =

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Class exercise A car dealer calculates the proportion of new cars sold that have been returned a various number of times for the correction of defects during the guarantee period. The results are as follows: Graph the probability distribution function Calculate the cumulative probability distribution What is the probability that cars will be returned for corrections more than two times? P(x > 2) P(x < 2)? Find the expected value of the number of a car for corrections for defects during the guarantee period Find the expected variance
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Class exercise A car dealer calculates the proportion of new cars sold that have been returned a various number of times for the correction of defects during the guarantee period. The results are as follows: Graph the probability distribution function Calculate the cumulative probability distribution What is the probability that cars will be returned for corrections more than two times? P(x > 2) P(x < 2)? Find the expected value of the number of a car for corrections for defects during the guarantee period Find the expected variance

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Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the expected value of number of computer sold per day:
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Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the expected value of number of computer sold per day:

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Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the expected value of number of computer sold per day: E[x]= (0 x 0.05) + (1 x 0.1) + (2 x 0.2) + (3 x 0.2) + (4 x 0.2) + (5 x 0.15) + (6 x 0.1) = 3.25 rounded to 3
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Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the expected value of number of computer sold per day: E[x]= (0 x 0.05) + (1 x 0.1) + (2 x 0.2) + (3 x 0.2) + (4 x 0.2) + (5 x 0.15) + (6 x 0.1) = 3.25 rounded to 3

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Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the variance of number of computer sold per day:
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Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the variance of number of computer sold per day:

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Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the variance of number of computer sold per day: =(0.05) +(0.1)+ + (0.2)+ (0.2)+ (0.15) + (0.1) = 2.69 = 2.69
Описание слайда:
Dan’s computer Works – class exercise The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution: Calculate the variance of number of computer sold per day: =(0.05) +(0.1)+ + (0.2)+ (0.2)+ (0.15) + (0.1) = 2.69 = 2.69

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Quizz A small school employs 5 teachers who make between $40,000 and $70,000 per year. One of the 5 teachers, Valerie, decides to teach part-time which decreases her salary from $40,000 to $20,000 per year. The rest of the salaries stay the same. How will decreasing Valerie's salary affect the mean and median? Please choose from one of the following options: A) Both the mean and median will decrease. B) The mean will decrease, and the median will stay the same. C)The median will decrease, and the mean will stay the same. D) The mean will decrease, and the median will increase.
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Quizz A small school employs 5 teachers who make between $40,000 and $70,000 per year. One of the 5 teachers, Valerie, decides to teach part-time which decreases her salary from $40,000 to $20,000 per year. The rest of the salaries stay the same. How will decreasing Valerie's salary affect the mean and median? Please choose from one of the following options: A) Both the mean and median will decrease. B) The mean will decrease, and the median will stay the same. C)The median will decrease, and the mean will stay the same. D) The mean will decrease, and the median will increase.

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Khan Academy – Empirical Rule A company produces batteries with a mean life time of 1’300 hours and a standard deviation of 50 hours. Use the Empirical rule (68 – 95 – 99.7 %) to estimate the probability of a battery to have a lifetime longer than 1’150 hours. P (x > 1’150 hours) Which of the following is the right answer? 95 % 84% 73% 99.85%
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Khan Academy – Empirical Rule A company produces batteries with a mean life time of 1’300 hours and a standard deviation of 50 hours. Use the Empirical rule (68 – 95 – 99.7 %) to estimate the probability of a battery to have a lifetime longer than 1’150 hours. P (x > 1’150 hours) Which of the following is the right answer? 95 % 84% 73% 99.85%

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Stating that two events are statistically independent means that the probability of one event occurring is independent of the probability of the other event having occurred. TRUE FALSE
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Stating that two events are statistically independent means that the probability of one event occurring is independent of the probability of the other event having occurred. TRUE FALSE

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The time it takes a car to drive from Istanbul to Sinop is an example of a discrete random variable True False
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The time it takes a car to drive from Istanbul to Sinop is an example of a discrete random variable True False

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Probability is a numerical measure about the likelihood that an event will occur. TRUE FALSE
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Probability is a numerical measure about the likelihood that an event will occur. TRUE FALSE

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Suppose that you enter a lottery by obtaining one of 20 tickets that have been distributed. By using the relative frequency method, you can determine that the probability of your winning the lottery is 0.15. TRUE FALSE
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Suppose that you enter a lottery by obtaining one of 20 tickets that have been distributed. By using the relative frequency method, you can determine that the probability of your winning the lottery is 0.15. TRUE FALSE

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If we flip a coin three times, the probability of getting three heads is 0.125. TRUE FALSE
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If we flip a coin three times, the probability of getting three heads is 0.125. TRUE FALSE

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The number of products bought at a local store is an example of a discrete random variable. TRUE FALSE
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The number of products bought at a local store is an example of a discrete random variable. TRUE FALSE

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Empirical rule – Khan Academy a) Which shape does a distribution need to have to apply the Empirical Rule? b) The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives 20.5 years, the standard deviation is 3.9, years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living less than 32.2 years.
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Empirical rule – Khan Academy a) Which shape does a distribution need to have to apply the Empirical Rule? b) The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives 20.5 years, the standard deviation is 3.9, years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living less than 32.2 years.

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Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1), слайд №28
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Binomial Probability Distribution Bi-nominal (from Latin) means: Two-names
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Binomial Probability Distribution Bi-nominal (from Latin) means: Two-names

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Possible Binomial Distribution examples A manufacturing plant labels products as either defective or acceptable A firm bidding for contracts will either get a contract or not A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer or reject it A customer enters a store will either buy a product or will not buy a product
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Possible Binomial Distribution examples A manufacturing plant labels products as either defective or acceptable A firm bidding for contracts will either get a contract or not A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer or reject it A customer enters a store will either buy a product or will not buy a product

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The Binomial Distribution The binomial distribution is used to find the probability of a specific or cumulative number of successes in n trials
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The Binomial Distribution The binomial distribution is used to find the probability of a specific or cumulative number of successes in n trials

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The Binomial Distribution The binomial formula is:
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The Binomial Distribution The binomial formula is:

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Example: Calculating a Binomial Probability
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Example: Calculating a Binomial Probability

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Binomial probability - Calculating binomial probabilities Suppose that Ali, a real estate agent, has 5 people interested in buying a house in the area Ali’s real estate agent operates. Out of the 5 people interested how many people will actually buy a house if the probability of selling a house is 0.40. P(X = 4)?
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Binomial probability - Calculating binomial probabilities Suppose that Ali, a real estate agent, has 5 people interested in buying a house in the area Ali’s real estate agent operates. Out of the 5 people interested how many people will actually buy a house if the probability of selling a house is 0.40. P(X = 4)?

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Solving Problems with the Binomial Formula Find the probability of 4 people buying a house out of 5 people, when the probability of success is .40
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Solving Problems with the Binomial Formula Find the probability of 4 people buying a house out of 5 people, when the probability of success is .40

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Class exerise Find the probability of 3 people buying a house out of 5 people, when the probability of success is .40 P(X =3) ? n = 5, r = 3, p = 0.4, and q = 1 – 0.4 = 0.6
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Class exerise Find the probability of 3 people buying a house out of 5 people, when the probability of success is .40 P(X =3) ? n = 5, r = 3, p = 0.4, and q = 1 – 0.4 = 0.6

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P( X = 3) ? Find the probability of 3 people buying a house out of 5 people, when the probability of success is .40 n = 5, r = 3, p = 0.4, and q = 1 – 0.4 = 0.6
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P( X = 3) ? Find the probability of 3 people buying a house out of 5 people, when the probability of success is .40 n = 5, r = 3, p = 0.4, and q = 1 – 0.4 = 0.6

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Creating a probability distribution with the Binomial Formula – house sale example
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Creating a probability distribution with the Binomial Formula – house sale example

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Binomial Probability Distribution house sale example n = 5, P= .4
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Binomial Probability Distribution house sale example n = 5, P= .4

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The binomial distribution is used to find the probability of a specific or cumulative number of successes in n trials. Let’s look at the cumulative probability: P (x < 2 houses), P(x 3)
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The binomial distribution is used to find the probability of a specific or cumulative number of successes in n trials. Let’s look at the cumulative probability: P (x < 2 houses), P(x 3)

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The binomial distribution is used to find the probability of a specific or cumulative number of successes in n trials. Let’s look at the cumulative probability: P (x < 2 houses), P(x 3) P ( x < 2 houses) = P(0 house) + P(1 house) = 0.0778 + 0.2592 = .337 or 33.7% P(x 3 houses) = P(3 houses) + P(4 houses) + P(5 houses) = 0.2304 + 0.0768 + 0.0102 = 0.3174
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The binomial distribution is used to find the probability of a specific or cumulative number of successes in n trials. Let’s look at the cumulative probability: P (x < 2 houses), P(x 3) P ( x < 2 houses) = P(0 house) + P(1 house) = 0.0778 + 0.2592 = .337 or 33.7% P(x 3 houses) = P(3 houses) + P(4 houses) + P(5 houses) = 0.2304 + 0.0768 + 0.0102 = 0.3174

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Shape of Binomial Distribution The shape of the binomial distribution depends on the values of P and n
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Shape of Binomial Distribution The shape of the binomial distribution depends on the values of P and n

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Binomial Distribution shapes When P = .5 the shape of the distribution is perfectly symmetrical and resembles a bell-shaped (normal distribution) When P = .2 the distribution is skewed right. This skewness increases as P becomes smaller. When P = .8, the distribution is skewed left. As P comes closer to 1, the amount of skewness increases.
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Binomial Distribution shapes When P = .5 the shape of the distribution is perfectly symmetrical and resembles a bell-shaped (normal distribution) When P = .2 the distribution is skewed right. This skewness increases as P becomes smaller. When P = .8, the distribution is skewed left. As P comes closer to 1, the amount of skewness increases.

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Using Binomial Tables instead of to calculating Binomial probabilites
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Using Binomial Tables instead of to calculating Binomial probabilites

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Solving Problems with Binomial Tables MSA Electronics is experimenting with the manufacture of a new USB-stick and is looking into the Every hour a random sample of 5 USB-sticks is taken The probability of one USB-stick being defective is 0.15 What is the probability of finding 3, 4, or 5 defective USB-sticks ? P( x = 3), P(x = 4 ), P(x= 5)
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Solving Problems with Binomial Tables MSA Electronics is experimenting with the manufacture of a new USB-stick and is looking into the Every hour a random sample of 5 USB-sticks is taken The probability of one USB-stick being defective is 0.15 What is the probability of finding 3, 4, or 5 defective USB-sticks ? P( x = 3), P(x = 4 ), P(x= 5)

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Solving Problems with Binomial Tables
Описание слайда:
Solving Problems with Binomial Tables

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