IS 310 Business Statistics CSU Long Beach

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IS 310 – Business StatisticsIS 310 – Business Statistics

IS 310

Business Statistic

sCSU

Long Beach

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Continuous Probability DistributionsContinuous Probability Distributions

A A continuous random variablecontinuous random variable can assume any can assume any value in an interval on the real line or in a value in an interval on the real line or in a collection of intervals.collection of intervals.

It is not possible to talk about the probability It is not possible to talk about the probability of the random variable assuming a particular of the random variable assuming a particular value.value. Instead, we talk about the probability of the Instead, we talk about the probability of the random variable assuming a value within a random variable assuming a value within a given interval.given interval.

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Continuous Probability DistributionsContinuous Probability Distributions

The probability of the random variable The probability of the random variable assuming a value within some given interval assuming a value within some given interval from from xx11 to to xx22 is defined to be the is defined to be the area under area under the graphthe graph of the of the probability density functionprobability density function betweenbetween x x11 andand x x22..

f (x)f (x)

x x

UniformUniform

xx11 xx11 xx22 xx22

xx

f f ((xx)) NormalNormal

xx11 xx11 xx22 xx22

xx11 xx11 xx22 xx22

ExponentialExponential

xx

f (x)f (x)

xx11

xx11

xx22 xx22

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Normal Probability DistributionNormal Probability Distribution

The The normal probability distributionnormal probability distribution is the most is the most important distribution for describing a important distribution for describing a continuous random variable.continuous random variable.

It is widely used in statistical inference.It is widely used in statistical inference.

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HeightsHeightsof peopleof peopleHeightsHeights

of peopleof people


It has been used in a wide variety of It has been used in a wide variety of applications:applications:

ScientificScientific measurementsmeasurements

ScientificScientific measurementsmeasurements

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AmountsAmounts

of rainfallof rainfall

AmountsAmounts

of rainfallof rainfall


It has been used in a wide variety of It has been used in a wide variety of applications:applications:

TestTest scoresscoresTestTest

scoresscores

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Normal Probability Density FunctionNormal Probability Density Function

2 2( ) / 21( )

2xf x e

2 2( ) / 21( )

2xf x e

= mean= mean

= standard deviation= standard deviation

= 3.14159= 3.14159

ee = 2.71828 = 2.71828

where:where:

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The distribution is The distribution is symmetricsymmetric; its skewness; its skewness measure is zero.measure is zero. The distribution is The distribution is symmetricsymmetric; its skewness; its skewness measure is zero.measure is zero.


CharacteristicsCharacteristics

xx

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The entire family of normal probabilityThe entire family of normal probability distributions is defined by itsdistributions is defined by its meanmean and its and its standard deviationstandard deviation . .

The entire family of normal probabilityThe entire family of normal probability distributions is defined by itsdistributions is defined by its meanmean and its and its standard deviationstandard deviation . .



Standard Deviation Standard Deviation

Mean Mean xx

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The The highest pointhighest point on the normal curve is at the on the normal curve is at the meanmean, which is also the , which is also the medianmedian and and modemode.. The The highest pointhighest point on the normal curve is at the on the normal curve is at the meanmean, which is also the , which is also the medianmedian and and modemode..



xx

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-10-10 00 2020

The mean can be any numerical value: negative,The mean can be any numerical value: negative, zero, or positive.zero, or positive. The mean can be any numerical value: negative,The mean can be any numerical value: negative, zero, or positive.zero, or positive.

xx

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= 15= 15

= 25= 25

The standard deviation determines the width of theThe standard deviation determines the width of thecurve: larger values result in wider, flatter curves.curve: larger values result in wider, flatter curves.The standard deviation determines the width of theThe standard deviation determines the width of thecurve: larger values result in wider, flatter curves.curve: larger values result in wider, flatter curves.

xx

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Probabilities for the normal random variable areProbabilities for the normal random variable are given by given by areas under the curveareas under the curve. The total area. The total area under the curve is 1 (.5 to the left of the mean andunder the curve is 1 (.5 to the left of the mean and .5 to the right)..5 to the right).

Probabilities for the normal random variable areProbabilities for the normal random variable are given by given by areas under the curveareas under the curve. The total area. The total area under the curve is 1 (.5 to the left of the mean andunder the curve is 1 (.5 to the left of the mean and .5 to the right)..5 to the right).



.5.5 .5.5

xx

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Since the area under the curve represents Since the area under the curve represents probability, the probability of a normal random probability, the probability of a normal random variable at one specific value is zero . With a variable at one specific value is zero . With a single value, one can’t find the area since the single value, one can’t find the area since the area must be bound by two values. Thus, area must be bound by two values. Thus,

P(x = 10) = 0 P(x = 3) = 0 P(x = 7.5) P(x = 10) = 0 P(x = 3) = 0 P(x = 7.5) = 0= 0

However, one can find the following probabilities:However, one can find the following probabilities:

P( 1 < x < 3) P(2.2 < x < 3.7) P(x > 3)P( 1 < x < 3) P(2.2 < x < 3.7) P(x > 3)

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of values of a normal random variableof values of a normal random variable are within of its mean.are within of its mean. of values of a normal random variableof values of a normal random variable are within of its mean.are within of its mean.68.26%68.26%68.26%68.26%

+/- 1 standard deviation+/- 1 standard deviation+/- 1 standard deviation+/- 1 standard deviation


+/- 2 standard deviations+/- 2 standard deviations+/- 2 standard deviations+/- 2 standard deviations


+/- 3 standard deviations+/- 3 standard deviations+/- 3 standard deviations+/- 3 standard deviations

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xx – – 33 – – 11

– – 22 + 1+ 1

+ 2+ 2 + 3+ 3

68.26%68.26%95.44%95.44%99.72%99.72%

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There may be thousands of normal distribution There may be thousands of normal distribution curves, each with a different mean and a curves, each with a different mean and a different standard deviation. Since the shapes different standard deviation. Since the shapes are different, the areas under the curves are different, the areas under the curves between any two points are also different. between any two points are also different.

To make life easier, all normal distributions can To make life easier, all normal distributions can be converted to a standard normal be converted to a standard normal distribution. A standard normal distribution distribution. A standard normal distribution has a mean of 0 and a standard deviation of 1.has a mean of 0 and a standard deviation of 1.

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00zz

The letter The letter z z is used to designate the standardis used to designate the standard normal random variable.normal random variable. The letter The letter z z is used to designate the standardis used to designate the standard normal random variable.normal random variable.

Standard Normal Probability DistributionStandard Normal Probability Distribution

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Converting to the Standard Normal Converting to the Standard Normal Distribution requires the use Distribution requires the use

of this formulaof this formula


zx

zx

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Applications of Standard Normal Applications of Standard Normal DistributionDistribution

Example ProblemExample Problem

Test scores of a special examination administered to all potential Test scores of a special examination administered to all potential employees of a firm are normally distributed with a mean of 500 employees of a firm are normally distributed with a mean of 500 points and a standard deviation of 100 points. What is the points and a standard deviation of 100 points. What is the probability that a score selected at random will be higher than 700?probability that a score selected at random will be higher than 700?

P(x > 700) = ?P(x > 700) = ?

If we convert this normal variable, x, to a standard normal variable, z, If we convert this normal variable, x, to a standard normal variable, z,

z = (x - µ) / z = (x - µ) / σσ = (700 – 500) / 100 = 2 = (700 – 500) / 100 = 2

--------------500----------700 --------------500----------700 x-scale x-scale

P(x > 700) = P(z > 2) ----------------0-----------2 P(x > 700) = P(z > 2) ----------------0-----------2 z-scalez-scale

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The area under the normal curve between any The area under the normal curve between any two points represents the probability.two points represents the probability.

Refer to pages 235 through 238 (10Refer to pages 235 through 238 (10thth edition) edition) or pages 242-245 (11or pages 242-245 (11thth edition) of your text edition) of your text book and get a clear understanding of how to book and get a clear understanding of how to calculate probabilities .calculate probabilities .

Be proficient in using Table 1 of Appendix B Be proficient in using Table 1 of Appendix B (10-Pages 918 and 919; 11-Pages 978-979) (10-Pages 918 and 919; 11-Pages 978-979)

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Sample ProblemsSample Problems

Problem # 10 (10-Page 241; 11-Page 248)Problem # 10 (10-Page 241; 11-Page 248)

Using Table 1 (10-Page 919; 11-Page 979)Using Table 1 (10-Page 919; 11-Page 979)

a.a. P(z ≤ 1.5) = 0.9332P(z ≤ 1.5) = 0.9332

b.b. P(z ≤ 1.0) = 0.8413P(z ≤ 1.0) = 0.8413

c.c. P(1 ≤ z ≤ 1.5) = 0.9332 – 0.8413 = 0.09P(1 ≤ z ≤ 1.5) = 0.9332 – 0.8413 = 0.09

d.d. P(0 < z < 2.5) = 0.9938 – 0.5000 = 0.4938P(0 < z < 2.5) = 0.9938 – 0.5000 = 0.4938

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More Sample ProblemsMore Sample Problems


a.a. P(z ≤ - 1.0) = 0.1587P(z ≤ - 1.0) = 0.1587

b.b. P(z ≥ - 1) = 1 – P(z ≤ - 1) = 1 – 0.1587 = P(z ≥ - 1) = 1 – P(z ≤ - 1) = 1 – 0.1587 = 0.84130.8413

c.c. P(z ≥ - 1.5) = 1 – P(z ≤ - 1.5) = 1 – 0.0668 = P(z ≥ - 1.5) = 1 – P(z ≤ - 1.5) = 1 – 0.0668 = 0.93320.9332

d.d. Do it yourselfDo it yourself

e.e. P(- 3 < z ≤ 0) = 0.5 – 0.0013 = 0. 4987P(- 3 < z ≤ 0) = 0.5 – 0.0013 = 0. 4987

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More Sample ProblemMore Sample Problem

Problem # 15 (10-2; 11-Page 249)Problem # 15 (10-Page 242; 11-Page 249)

Let’s do this problem in class!Let’s do this problem in class!

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More Sample ProblemsMore Sample Problems


Given: µ = 77 Given: µ = 77 σσ = 20 = 20

a.a. P(x < 50) = ? P(x < 50) = ?

Convert to z: z = (x - µ) / Convert to z: z = (x - µ) / σσ = (50 – 77) / 20 = - = (50 – 77) / 20 = - 1.351.35

P(x < 50) = P(z < - 1.35) = 0.0885P(x < 50) = P(z < - 1.35) = 0.0885

b.b. P(x > 100) = ? z = (100 – 77) / 20 = 1.15P(x > 100) = ? z = (100 – 77) / 20 = 1.15

P(x > 100) = P(z > 1.15) = 1 – P(z ≤ 1.15) = 1 – P(x > 100) = P(z > 1.15) = 1 – P(z ≤ 1.15) = 1 – 0.8749 = 0.1251 or 12.51 %0.8749 = 0.1251 or 12.51 %

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Continuation of Sample ProblemContinuation of Sample Problem

c.c. x = ? to be considered a heavy userx = ? to be considered a heavy user

Upper 20% of the area is in the right tail of the Upper 20% of the area is in the right tail of the normal curve. 80% of the area is to the left. normal curve. 80% of the area is to the left. Go to Table 1 and locate 0.8 (or 80%) as the Go to Table 1 and locate 0.8 (or 80%) as the table entry. The closest entry is 0.7995. That table entry. The closest entry is 0.7995. That point represents a z-value of 0.84. Use this point represents a z-value of 0.84. Use this value of z in the following equation:value of z in the following equation:

z = (x - µ) / z = (x - µ) / σσ

0.84 = (x – 77)/ 20 x = 93.8 hours 0.84 = (x – 77)/ 20 x = 93.8 hours

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Sample ProblemsSample Problems

The service life of a certain brand of automobile The service life of a certain brand of automobile battery is normally distributed with a mean of battery is normally distributed with a mean of 1000 days and a standard deviation of 100 days. 1000 days and a standard deviation of 100 days. The manufacturer of the battery wants to offer a The manufacturer of the battery wants to offer a guarantee, but does not know the length of the guarantee, but does not know the length of the warranty. It does not want to replace more than warranty. It does not want to replace more than 10 percent of the batteries sold. 10 percent of the batteries sold.

What should be the length of the warranty? What should be the length of the warranty?

z = (x - µ) / z = (x - µ) / σσ

- 1.2817 = (x – 1000) / 100- 1.2817 = (x – 1000) / 100

x = 871.83 or 872 days x = 871.83 or 872 days

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More Sample Problems More Sample Problems

A statistics instructor grades on a curve. He A statistics instructor grades on a curve. He does not want to give more than 15 percent A does not want to give more than 15 percent A in his class. If test scores of students in in his class. If test scores of students in statistics are normally distributed with a mean statistics are normally distributed with a mean of 75 and a standard deviation of 10, what of 75 and a standard deviation of 10, what should be the cut-off point for an A?should be the cut-off point for an A?

z = (x - µ) / z = (x - µ) / σσ

1.04 = (x – 75) / 101.04 = (x – 75) / 10

x = 85.4 or 85x = 85.4 or 85

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End of Chapter 6End of Chapter 6

IS 310 Business Statistics CSU Long Beach

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