🗊Презентация Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1)

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Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №1Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №2Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №3Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №4Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №5Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №6Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №7Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №8Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №9Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №10Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №11Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №12Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №13Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №14Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №15Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №16Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №17Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №18Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №19Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №20Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №21Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №22Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №23Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №24Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №25Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №26Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №27Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №28Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №29
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BBA182 Applied Statistics Week 9 (1) Calculating the probability of a continuous random variable – Normal Distribution 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 9 (1) Calculating the probability of a continuous random variable – Normal Distribution Dr Susanne Hansen Saral Email: susanne.saral@okan.edu.tr https://piazza.com/class/ixrj5mmox1u2t8?cid=4# www.khanacademy.org

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Mid-term exam statistics
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Mid-term exam statistics

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Mid-term exam statistics
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Mid-term exam statistics

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Continuous random variable A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. It is not possible to talk about the probability of the random variable assuming a particular value, because the probability will be close to 0. Instead, we talk about the probability of the random variable assuming a value within a given interval.
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Continuous random variable A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. It is not possible to talk about the probability of the random variable assuming a particular value, because the probability will be close to 0. Instead, we talk about the probability of the random variable assuming a value within a given interval.

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Calculating probabilities of continuous random variables
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Calculating probabilities of continuous random variables

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The Standard Normal Distribution – z-values Any normal distribution, F(x) (with any mean and standard deviation combination) can be transformed into the standardized normal distribution F(z), with mean 0 and standard deviation 1 We say that Z follows the standard normal distribution.
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The Standard Normal Distribution – z-values Any normal distribution, F(x) (with any mean and standard deviation combination) can be transformed into the standardized normal distribution F(z), with mean 0 and standard deviation 1 We say that Z follows the standard normal distribution.

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Procedure for calculating the probability of x using the Standard Normal Table For m = 100, s = 15, find the probability that X is less than 130 = P(x < 130) Transforming x - random variable into a z - standard random variable:
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Procedure for calculating the probability of x using the Standard Normal Table For m = 100, s = 15, find the probability that X is less than 130 = P(x < 130) Transforming x - random variable into a z - standard random variable:

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Procedure for calculating the probability of x using the Standard Normal Table (continued) Step 2 Look up the probability from the table of normal curve areas Column on the left is Z value Row at the top has second decimal places for Z values
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Procedure for calculating the probability of x using the Standard Normal Table (continued) Step 2 Look up the probability from the table of normal curve areas Column on the left is Z value Row at the top has second decimal places for Z values

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Using the Standard Normal Table
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Using the Standard Normal Table

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P(z < + 2) = P(z > -2) = .9772 In probability terms, a z-score of -2.0 and +2.0 has the same probability, because they are mirror images of each other. If we look for the z-score 2.0 in the table we find a value of 9772.
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P(z < + 2) = P(z > -2) = .9772 In probability terms, a z-score of -2.0 and +2.0 has the same probability, because they are mirror images of each other. If we look for the z-score 2.0 in the table we find a value of 9772.

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Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1), слайд №11
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Finding the probability of z-scores with two decimals and graph the probability P ( z < + 0.55) = 0.7088 or 70.88 % P (z > + .55) = 1.0 – 0.7088 = 0.2912 or 29.12% P ( z > - 0.55) = 0.7088 or 70.88 % P ( z < - 0.55) = 1.0 - .7088 = 0.2912 or 29.12 % P ( z < + 1.65) = 0.9505 or 95.05 % P (z > + 1.65) = 1.0 – 0.9505 = 0.0495 or 4.96 % P( z > - 2.36) = .9909 or 99.09 % P ( z < + 2.36) = .9909 or 99.09 %
Описание слайда:
Finding the probability of z-scores with two decimals and graph the probability P ( z < + 0.55) = 0.7088 or 70.88 % P (z > + .55) = 1.0 – 0.7088 = 0.2912 or 29.12% P ( z > - 0.55) = 0.7088 or 70.88 % P ( z < - 0.55) = 1.0 - .7088 = 0.2912 or 29.12 % P ( z < + 1.65) = 0.9505 or 95.05 % P (z > + 1.65) = 1.0 – 0.9505 = 0.0495 or 4.96 % P( z > - 2.36) = .9909 or 99.09 % P ( z < + 2.36) = .9909 or 99.09 %

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Determine for shampoo filling machine 1 the proportion of bottles that: = 500 ml = 10ml Contain less than 510 ml P(x < 510) Contain more than 515 ml P(x > 515) Contains more than 480 ml P(x > 480) Contain less than 490 ml P(x < 490) Contain more than 505 ml P(x > 505)
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Determine for shampoo filling machine 1 the proportion of bottles that: = 500 ml = 10ml Contain less than 510 ml P(x < 510) Contain more than 515 ml P(x > 515) Contains more than 480 ml P(x > 480) Contain less than 490 ml P(x < 490) Contain more than 505 ml P(x > 505)

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Solution: Contain more than 515 ml P(x 515ml) 1. Draw the graph to see which area we are looking for: 2. Z –score = = 1.5 = P(z > 1.5) 3. We can find P(z < 1.5) = .9332 directly from the table P(z 1.5) = 1 - .9332 = .0668 6.68% of the shampoo bottles contain more than 515 ml.
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Solution: Contain more than 515 ml P(x 515ml) 1. Draw the graph to see which area we are looking for: 2. Z –score = = 1.5 = P(z > 1.5) 3. We can find P(z < 1.5) = .9332 directly from the table P(z 1.5) = 1 - .9332 = .0668 6.68% of the shampoo bottles contain more than 515 ml.

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Solution: Contain more than 505 ml P(x > 505) ? 1. Draw the curve so you see which probability area we are looking for. 2. Z –score = = 0.5 = P(z < .5) = .6915 3. P(z 0.5) = 1 - . 6915= .3085 30.85 % of the shampoo bottles contain more than 505ml shampoo.
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Solution: Contain more than 505 ml P(x > 505) ? 1. Draw the curve so you see which probability area we are looking for. 2. Z –score = = 0.5 = P(z < .5) = .6915 3. P(z 0.5) = 1 - . 6915= .3085 30.85 % of the shampoo bottles contain more than 505ml shampoo.

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Exercise: Draw a graph of the below probabilities and find the probability of z in the standard normal table with = 0, =1 P ( z < + 1.05) = P (z > -1.05 ) = P (z < - 3.34) = P (z > - 3.34) = P (z > - 2.47) = P (z < + 1.87) = P (z > + 2.57) = P ( z < - 0.32) =
Описание слайда:
Exercise: Draw a graph of the below probabilities and find the probability of z in the standard normal table with = 0, =1 P ( z < + 1.05) = P (z > -1.05 ) = P (z < - 3.34) = P (z > - 3.34) = P (z > - 2.47) = P (z < + 1.87) = P (z > + 2.57) = P ( z < - 0.32) =

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P ( z < + 1.05) = 0.8531 or 85.31 % P (z > -1.05 ) = 0.8531 or 85.31 % P (z < - 3.34) = 1.0 – 0.9996 = 0.0004 or 0.04 % P (z > - 3.34) = 0.9996 or 99.96 % P (z > - 2.47) = 0.9932 or 99.32 % P (z < + 1.87) = 0.9693 or 96.93 % P (z > + 2.57) = 1.0 – 0.9949 = 0.0054 or 0.054 % P( z < - 0.32) = 1.0 – 0.6255 = 0.3745 or 37. 45 %
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P ( z < + 1.05) = 0.8531 or 85.31 % P (z > -1.05 ) = 0.8531 or 85.31 % P (z < - 3.34) = 1.0 – 0.9996 = 0.0004 or 0.04 % P (z > - 3.34) = 0.9996 or 99.96 % P (z > - 2.47) = 0.9932 or 99.32 % P (z < + 1.87) = 0.9693 or 96.93 % P (z > + 2.57) = 1.0 – 0.9949 = 0.0054 or 0.054 % P( z < - 0.32) = 1.0 – 0.6255 = 0.3745 or 37. 45 %

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Haynes Construction Company Example Builds three- and four-unit apartment buildings: Total construction time follows a normal distribution For triplexes, m = 100 days and = 20 days Contract calls for completion in 125 days Late completion will incur a severe penalty fee Calculate the probability of completing in less than 125 days P(x <125)
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Haynes Construction Company Example Builds three- and four-unit apartment buildings: Total construction time follows a normal distribution For triplexes, m = 100 days and = 20 days Contract calls for completion in 125 days Late completion will incur a severe penalty fee Calculate the probability of completing in less than 125 days P(x <125)

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Haynes Construction Company Compute Z:
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Haynes Construction Company Compute Z:

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Haynes Construction Company
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Haynes Construction Company

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Haynes Construction Company What is the probability that the company will not finish in 125 days and therefore will have to pay a penalty? m = 100 days and = 20 days P(z > 1.25) ?
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Haynes Construction Company What is the probability that the company will not finish in 125 days and therefore will have to pay a penalty? m = 100 days and = 20 days P(z > 1.25) ?

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Haynes Construction Company What is the probability that the company will not finish in 125 days and therefore will have to pay a penalty?
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Haynes Construction Company What is the probability that the company will not finish in 125 days and therefore will have to pay a penalty?

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Haynes Construction Company If finished in 75 days or less, Haynes will get a bonus of $5,000 What is the probability of a bonus? P ( x < 75)
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Haynes Construction Company If finished in 75 days or less, Haynes will get a bonus of $5,000 What is the probability of a bonus? P ( x < 75)

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Haynes Construction Company If finished in 75 days or less, bonus = $5,000 Probability of bonus? P ( x < 75)
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Haynes Construction Company If finished in 75 days or less, bonus = $5,000 Probability of bonus? P ( x < 75)

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Haynes Construction Company
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Haynes Construction Company

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Haynes Construction Company Probability of completing between 110 and 125 days?
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Haynes Construction Company Probability of completing between 110 and 125 days?

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Haynes Construction Company Probability of completing between 110 and 125 days?
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Haynes Construction Company Probability of completing between 110 and 125 days?

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Haynes Construction Company Probability of completing between 110 and 125 days?
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Haynes Construction Company Probability of completing between 110 and 125 days?

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Calculation procedure to find the probability of the area under the normal curve: 1. First draw the normal curve for the problem, to understand what area under the curve we are looking for. 2. Transform x-values to the standardized random variable, z 3. Use the standardized normal distribution table to find the probability of the calculated z-value
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Calculation procedure to find the probability of the area under the normal curve: 1. First draw the normal curve for the problem, to understand what area under the curve we are looking for. 2. Transform x-values to the standardized random variable, z 3. Use the standardized normal distribution table to find the probability of the calculated z-value

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