Презентация Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1) онлайн
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Презентации » Математика » Discrete random variables – expected variance and standard deviation. Discrete Probability Distributions. Week 7 (1)
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- Тип файла:ppt / pptx (powerpoint)
- Всего слайдов:46 слайдов
- Для класса:1,2,3,4,5,6,7,8,9,10,11
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Слайды и текст к этой презентации:
№6 слайд
![Properties of discrete random](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img5.jpg)
Содержание 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 :
№9 слайд
![Expected variance of a](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img8.jpg)
Содержание слайда: 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
№10 слайд
![Variance of a discrete random](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img9.jpg)
Содержание 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 - ):
№14 слайд
![Class exercise A car dealer](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img13.jpg)
Содержание слайда: 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
№16 слайд
![Dan s computer Works class](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img15.jpg)
Содержание 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 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
№18 слайд
![Dan s computer Works class](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img17.jpg)
Содержание 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
№19 слайд
![Quizz A small school employs](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img18.jpg)
Содержание слайда: 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.
№20 слайд
![Khan Academy Empirical Rule A](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img19.jpg)
Содержание слайда: 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%
№27 слайд
![Empirical rule Khan Academy a](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img26.jpg)
Содержание слайда: 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.
№30 слайд
![Possible Binomial](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img29.jpg)
Содержание слайда: 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
№34 слайд
![Binomial probability -](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img33.jpg)
Содержание слайда: 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)?
№41 слайд
![The binomial distribution is](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img40.jpg)
Содержание слайда: 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
№43 слайд
![Binomial Distribution shapes](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img42.jpg)
Содержание слайда: 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.
№45 слайд
![Solving Problems with](/documents_6/316090a3b741a80aebc2910b84e4bbfc/img44.jpg)
Содержание слайда: 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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