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Unformatted text preview: Topic #2 Basic Statistics – Part 2 Measures of Relative Location • Zscores Continuous Probability Distributions
o o Measures of Relative Location o o zScores The Empirical Rule • NOTE: gives you an idea/sense of how values and probabilities are spread over the normal distribution distribution • NOT a substitute for the normal distribution
table Slide 1 Slide 2 zScores
o o x _ zScores EXAMPLES
o o The zscore is often called the standardized value. It denotes the number of standard deviations a data value xi is from the mean. Suppose x = .250 and s = .035 (batting averages) How many standard deviations above the mean is .285? xi − x .285 −.250 x −x zi = i s zi = o
x
_ A data value less than the sample mean will have a less zscore less than zero. o A data value greater than the sample mean will have a zscore greater than zero. zo A data value equal to the sample mean will have a zscore of zero.
o How many standard deviations below the mean is .215? s = .035 =1 zi =
o xi − x .215−.250 = =−1 s .035
xi − x .320 −.250 = =2 .035 s How many standard deviations above the mean is .320? zi = Slide 3 Slide 4 Example: Apartment Rents
o o Example: Apartment Rents
o o Actual data values for apartment rents. Compare with zscores on next slide. z Recall x ≈ 491 & s ≈ 55 Notice how zscore values change as approach zsample mean 425 4 25 440 450 465 480 510 575 430 440 450 470 485 515 575 430 440 450 470 490 525 580 435 445 450 472 490 525 590 435 445 450 475 490 525 600 435 445 460 475 500 535 600 435 445 460 475 500 549 600 435 445 460 480 500 550 600 440 450 465 480 500 570 615 440 450 465 480 510 570 615
Slide 5 1.20 0.93 0.75 0.47 0.20 0.35 1.54 1.11 1.11 1 .02 1.02 1.02 1.02 1.02 0.93 Standardized Values 0.84 0.84 0.84 0.75 0.93 0.93 0.84 0.84 for Apartment Rents 0.75 0.75 0.75 0.75 0.56 0.56 0.56 0.47 0.38 0.38 0.34 0.29 0.29 0.29 0.20 0.20 0.11 0.01 0.01 0.01 0.17 0.17 0.17 0.17 0.44 0.62 0.62 0.62 0.81 1.06 1.08 1.45 1.54 1.63 1.81 1.99 1.99 1.99 1.99 2.27 0.93 0.75 0.47 0.20 0.35 1.45 2.27
Slide 6 1 Example: Apartment Rents
o The Empirical Rule  NOT a substitute for the normal distribution table
For data having a normal distribution: zScore of Smallest Value (425)
z= xi − x 425 − 490.80 = = −1. 20 s 54. 74 Standardized Values for Apartment Rents 1.20 0.93 0.75 0.47 0.20 0.35 1.54 1.11 0.93 0.75 0.38 0.11 0.44 1.54 1.11 0.93 0.75 0.38 0.01 0.62 1.63 1 .02 0.84 0.75 0.34 0.01 0.62 1.81 1.02 0.84 0.75 0.29 0.01 0.62 1.99 1.02 0.84 0.56 0.29 0.17 0.81 1.99 1.02 0.84 0.56 0.29 0.17 1.06 1.99 1.02 0.84 0.56 0.20 0.17 1.08 1.99 0.93 0.75 0.47 0.20 0.17 1.45 2.27 0.93 0.75 0.47 0.20 0.35 1.45 2.27
Slide 7 • Approximately 68% of the data values will be within one standard deviation of the mean. one • Approximately 95% of the data values will be within two standard deviations of the mean. two • Almost all of the items (99.7%) will be of within three standard deviations of the mean. three
Slide 8 Example: Apartment Rents
o Example: Apartment Rents
The Empirical Rule ( x ≈ 491 , s ≈ 55) % in Interval Interval Within +/ 1s +/436 436 to 546 48/70 = 69% Within +/ 2s +/381 381 to 601 68/70 = 97% Within +/ 3s +/326 326 to 656 70/70 = 100%
o The Empirical Rule ( x ≈ 491 , s ≈ 55, so I’ll use these rounded values) Interval 491+1*55=546 4911*55=436 491+1*55=546 & 4911*55=436 491+2*55=601 & 4912*55=381 491+2 491+3*55=656 4913*55=326 491+3*55=656 & 4913*55=326 436 to 546 381 to 601 326 to 656 Within +/ 1s +/Within +/ 2s Within +/ 3s +/ Actual 440535 440425600 425425615 425440 450 465 480 500 570 615 440 450 465 480 510 570 615
Slide 10 425 440 450 465 480 510 575 430 440 450 470 485 515 575 430 440 450 470 490 525 580 435 445 450 472 490 525 590 435 445 450 475 490 525 600 435 445 460 475 500 535 600 435 445 460 475 500 549 600 435 445 460 480 500 550 600 Slide 9 Detecting Outliers
An outlier is an unusually small or unusually large value in a data set. o A data value with a zscore less than 3 or greater zthan +3 might be considered an outlier. o Outlier might be . . . • incorrectly recorded data value • data value that was incorrectly included in the data set. • correctly recorded data value that belongs in the data data set ! • investigate to see if this reflects a problem!
o o Example: Apartment Rents
Detecting Outliers The most extreme zscores are 1.20 and 2.27. zUsing z > 3 as the criterion for an outlier, there are no outliers in this data set. Standardized Standardized Values for Apartment Rents
1.20 0.93 0.75 0.47 0.20 0.35 1.54 1.11 0.93 0.75 0.38 0.11 0.44 1.54 1.11 0.93 0.75 0.38 0.01 0.62 1.63 1.02 0.84 0.75 0.34 0.01 0.62 1.81 1.02 0.84 0.75 0.29 0.01 0.62 1.99 1.02 0.84 0.56 0.29 0.17 0.81 1.99 1.02 0.84 0.56 0.29 0.17 1.06 1.99 1.02 0.84 0.56 0.20 0.17 1.08 1.99 0.93 0.75 0.47 0.20 0.17 1.45 2.27 0.93 0.75 0.47 0.20 0.35 1.45 2.27
Slide 12 Slide 11 2 Continuous Probability Distributions
o Continuous Probability Distributions
o A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. • Person’s height • Worker’s hourly wage • Speed of NASCAR vehicle • Revenues • GPA • Average points scored per game It is not possible to talk about the probability of the random random variable assuming a particular value. • Suppose soccer team scores 15 goals in seven games. • Average goals per game = 2.142857143. . . Instead, we talk about the probability of the random variable variable assuming a value within a given interval. o Slide 13 13 Slide 14 14 Continuous Probability Distributions
The probability of the random variable assuming a value within some given interval from x1 to x2 is o area under the graph of the probability density function function between x1 and x2.
o o Continuous Probability Distributions
The probability of the random variable assuming a value within some given interval from x1 to x2 is o area under the graph of the probability density function function between x1 and x2. Area = .2967 f(x) P(12 < x < 15) = 0.3 1/10 5 10 12 15 Area = 0.3 x P(0 < x < 0.83) = 29.67% 0
Slide 15 .83 z
Slide 16 The Normal Probability Distribution
o o o o o Graph of the Normal Probability Density Function Many business variables have a bellshaped belldistribution of values. Another name for this commonlyseen shape is the commonlyNORMAL NORMAL PROBABILITY DISTRIBUTION. OR more simply NORMAL DISTRIBUTION. See following figure Slide 17 Slide 18 3 Who Uses This
The exchange rate between the Euro and the dollar is approximately a normal random variable. normal o IQ scores have a normal distribution with a mean of 100 points and a standard deviation of 15 points, not the widely accepted 10 points. (Source: Newsweek, Newsweek Vol. 124, No. 17, 1994, pp.5355.) pp.53o The SAT Verbal test has a mean of 500 and a standard deviation of 90 points. (Source: The College Board, www.collegeboard.com, 2002)
o o Who Uses This
In project management, the industry standard since 1940s is to use a normal distribution to model activity normal times and project completions times. o The daily high temperature in Chicago in February has very close to a normal distribution with a mean of normal and maximum 35º F and a maximum daily high temperature of 70º F and a minimum daily high temperature of 14º F. minimum F. (Source: Chicago Tribune, February 25, 1999, Section 2, p. 12.) Slide 19 Slide 20 20 For example, a student’s weight is the result of many factors:. . . . maybe even last night’s party. The The Morning After Last Night’s Party That explains why the Normal is everywhere: stock markets, student weights, yearly temperatures, SAT scores, etc. Slide 21 Slide 22 Graph of the Distribution of Talent in College Football Graph of the Normal Probability Density Function
f(x) = height of curve less average more x = variable
mean of x = μ Slide 23 Slide 24 4 Normal Distribution Characteristics
o Normal Distribution Characteristics (cont.)
• The highest point on the normal curve is at the mean, which is also the median and mode. • The mean can be any numerical value: negative, zero, or positive. f(x) The normal distribution can be fully described by two two values: • its mean µ (mu) • its standard deviation σ (sigma) or s • or, equivalently, its variance σ2 or s2 f(x) f(x) x
µ µσ µ+σ x MEAN
Slide 25 Slide 26 µ Normal Distribution Characteristics (cont.)
• The standard deviation determines the width of the curve: • LARGER values result in wider, flatter curves. f(x)
LARGER S Normal Distribution Characteristics (cont.)
• The standard deviation determines the width of the curve: • SMALLER values result in narrower, more peaked curves. f(x) SMALLER S µσ µ µ+σ x x
µσ Slide 27 µ µ+σ Slide 28 Normal Distribution Characteristics (cont.)
Each and every variable with a normal distribution has certain characteristics. o These characteristics are worth covering since they are the basis for answering many questions about normal variables.
o o The Normal Distribution (cont.)
Characteristics • symmetric around vertical line at mean • probabilities for the normal random variable are given by areas under the curve. • total area under the curve is = 1 or 100% • area to right of mean is = 0.5, or 50% of total right 0.5, area area under curve • area to left of mean is = 0.5, or 50% of total area under curve • see next slides Slide 29 Slide 30 30 5 The Normal Distribution (cont.) The Normal Distribution (cont.)
50% of total area 50% of total area f(x) f(x) x
µ x
µ symmetric around vertical line at x = µ
Slide 31 area to left of mean is 50% of total area; area to right of mean is 50% of total area
Slide 32 The Normal Distribution (cont.)
o 68% 68% of the values, within one standard deviation of the mean f(x) Areas Under Normal Curve Characteristics (cont.) • approximately 68% of area under normal curve is within one standard deviation of mean • approximately 95% of area under normal curve is within two standard deviations of mean • approximately 99.7% of area under normal curve is within three standard deviations of mean ithi th • NOTE: standard deviation is denoted by “σ” on following following slides, mean of 600, s.d. of 65. s.d. x
µσ µ 68% µ+σ 535 665 Slide 33 33 Slide 34 95% of the values, within 2 standard deviations of the mean f(x) Areas Under Normal Curve 99.7% of the values, within 3 standard deviations of the mean f(x) Areas Under Normal Curve x
µ  2σ µσ µ 68% 95% µ+σ µ + 2σ µ  3σ µ  2σ µσ µ 68% 95% µ+σ µ + 2σ µ + 3σ x 470 730 405
Slide 35 99.7% 795
Slide 36 6 Example
o o Let X be a random variable whose values are the test scores obtained on a nationwide test given to college seniors. Suppose that X is normally distributed with a mean of 600 and a standard deviation (sigma ) of 65. Then the probability that X lies within 2 sigma = 2(65) = 130 points of 600 is approx. 95%. In other words, • approx. 95% of all test scores lie between 470 and 730. Similarly, • approx. 99.7% of the scores are within 3 sigma = 3(65) = 195 points of 600 • between 405 and 795. f(x) Areas Under Normal Curve 405 470 730 795 x o 600 3(65) 600 2(65) 600 600 + 2(65) 600 + 3(65) 95% 99.7% Slide 37 37 Slide 38 What is area below 600? f(x) Area Under Normal Curve = Probability What is probability of getting a score below 600? f(x) What is area below 730? Area Under Normal Curve = Probability What is probability of getting a score below 730? 405 470 730 795 x 405 470 730 795 x 600 3(65) 600 2(65) 600 600 + 2(65) 600 + 3(65) 600 3(65) 600 2(65) 600 600 + 2(65) 600 + 3(65) 95% 99.7% 95% 99.7% Slide 39 Slide 40 How are percentiles (on test results) calculated? f(x) Area Under Normal Curve = Probability f(x) If you score 730, what percentile are you in? Area Under Normal Curve = Probability 405 470 730 795 x 405 470 730 795 x 600 3(65) 600 2(65) 600 600 + 2(65) 600 + 3(65) 600 3(65) 600 2(65) 600 600 + 2(65) 600 + 3(65) 95% 99.7% 95% 99.7% Slide 41 Slide 42 7 What is area between 600 & 730? f(x) Area Under Normal Curve = Probability What is probability of getting a score 600  730? f(x) What is area between 470 & 600? Area Under Normal Curve = Probability What is probability of getting a score 470  600? 405 470 730 795 x 405 470 730 795 x 600 3(65) 600 2(65) 600 600 + 2(65) 600 + 3(65) 600 3(65) 600 2(65) 600 600 + 2(65) 600 + 3(65) 95% 99.7% 95% 99.7% Slide 43 Slide 44 The Normal Distribution (cont.)
o The Normal Distribution (cont.) Normal Probability Density Function f ( x) =
where: 2 2 1 e −( x −μ ) / 2σ 2 πσ o μ = mean σ = standard deviation π = 3.14159
e = 2.71828 B. Standard Normal Variable • 1. standard normal random variable: is a normal variable with • mean = 0 and • standard deviation (σ) = 1.
• See figure on next slide. • This variable is usually designated Z. • Any "regular" normal variable can be designated as X. o This standard normal variable Z can be useful in answering business questions. Slide 45 Slide 46 What is area between 2 & +2? f(z)
NOTICE “Z” RATHER THAN “X”. f(z)
NOTICE “Z” RATHER THAN “X”. z
3 2 1 0 +1 +2 +3 3 2 1 0 +1 +2 +3 z Standard Normal Curve: mean = 0 and standard deviation σ =1 Standard Normal Curve: mean = 0 and standard deviation σ=1 Slide 47 Slide 48 8 The Standard Normal Distribution (cont.)
• 2. The values of Z are in units of standard deviations deviations away from Z's mean. • See the table in a few slides. • area under standard normal curve between z = 0 & z = z0 where z0 => 0. How many standard deviations is this away from Z’s mean? f(z) z
3 2 1 0 +1 +2 +3 The values of Z are in units of standard deviations deviations away from Z's mean.
Slide 49 49 Slide 50 How many standard deviations is this away from Z’s mean? f(z) The Standard Normal Distribution (cont.)
• What’s Probability of 0 ≤ Z ≤ 2.54? • See the next slide. • At the intersection of the row for 2.5 and the column column under 0.04, • meaning Z = 2.54, • you find a value of 0.9945. • This means that the area under the standard normal normal curve between to the left of z = 2.54 z
3 2.54 2 1 0 +1 +2 +3 The values of Z are in units of standard deviations deviations away from Z's mean.
Slide 51 • is 0.9945. • The area to the left of z=0 is 0.5. • So the area between z= 0 and z=2.54 is 0.99450.99450.5=0.4945. • probability of 0 ≤ Z ≤ 2.54 = 49.45%
Slide 52 52 Table Table z
2.0 2.1 2.2 2.3 2.4 .00 .9772 .9821 .9861 .9893 .9918 .9938 .9953 .9965 .9974 .9981 .01 .9778 .9826 .9864 .9896 .9920 .9940 .9955 .9966 .9975 .9982 .02 .9783 .9830 .9868 .9898 .9922 .9941 .9956 .9967 .9976 .9982 .03 .9788 .9834 .9871 .9901 .9925 .9943 .9957 .9968 .9977 .9983 .04
.9793 .9838 .9875 .9904 .9927 .9945 .9959 .9969 .9977 .9984 z
2.0 2.1 2.2 2.3 2.4 . .00 .01 .02 .03 .04 2.5
2.6 2.7 2.8 2.9 2.5
2.6 2.7 2.8 2.9 .4945 . . 99.45% of area under standard normal curve is below (left of) z=2.54 for row value of 2.5 & column under 0.04, meaning Z = 2.54, 2.54, value = 0.9945
Slide 53 for row value of 2.5 & column under 0.04, meaning Z = 2.54, 2.54, value = 0.9945
Slide 54 9 f(z) 49.45% of area under standard normal curve is between z=0 and z=2.54 The Standard Normal Distribution (cont.)
• What’s Probability of 0 ≤ Z ≤ 2.54? • You have to be careful with how a particular Z table is set up. • Some tables give P(Z<a particular value), i.e. the area under the curve to the left of a particular zzvalue: previous example. • Other tables give P(0<Z<a particular value), i.e. Oth P(0<Z< the the area between z=0 and z=the value. • See the next slide. z
3 2 1 0 +1 +2 +3 +2.54 Slide 55 Slide 56 56 The The Standard Normal Distribution (cont.)
• What’s Probability of 0 ≤ Z ≤ 2.54? • See the next slide for the other table setup. set• At the intersection of the row for 2.5 and the column column under 0.04, • meaning Z = 2.54, • you find a value of 0.4945. • This means that the area under the standard normal normal curve between z = 0 & z = 2.54 Table z
2.0 2.1 2.2 2.3 2.4 .00 .4772 .4821 .4861 .4893 .4918 .4938 .4953 .4965 .4974 .4981 .01 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .02 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .03 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .04
.4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 • is 0.4945 • 49.45% of total area under curve • probability of 0 ≤ Z ≤ 2.54 = 49.45% 2.5
2.6 2.7 2.8 2.9 for row value of 2.5 & column under 0.04, meaning Z = 2.54, 2.54, value = 0.4945
Slide 57 57 Slide 58 The Standard Normal Distribution (cont.)
• area under standard normal curve between • z = 0 & z = 2.54 (note minus 2.54) Table • also 0.4945 (49.45% of total area under curve)
• area under curve between z = 0 and some z0 is same to right and left of z = 0 f(z) z
2.0 2.1 2.2 2.3 2.4 .00 .4772 .4821 .4861 .4893 .4918 .4938 .4953 .4965 .4974 .4981 .01 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .02 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .03 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .04
.4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 2.5
2.6 2.7 2.8 2.9 z
3 2 1 0 +1 +2 +3 Slide 59 2.54 for row value of 2.5 & column under 0.04, meaning Z = 2.54, 2.54, value = 0.4945
Slide 60 10 Transforming X into a Standard Normal Variable Z
o Two points 1. Can transform normally distributed X with µ ≠ 0 & σ ≠ 1 into standard normal Z (with µ = 0 and σ = 1) 2. Whatever probability is true for Z is true for corresponding X It's merely some algebra: ⎛ x − μ 170 − 150 ⎞ > Prob( X > 170) = Prob ⎜ ⎟= 20 ⎝σ ⎠ 20 ⎞ ⎛ = Prob ( Z > 1) = 0.1587. Prob ⎜ Z > ⎟ 20 ⎠ ⎝ Slide 61 Slide 62 Transforming X into a Standard Normal Variable Z (cont.) (cont.)
o Transforming X into a Standard Normal Variable Z (cont.) (cont.)
o 1. can transform normally distributed X with µ ≠ 0 & σ ≠ 1 into standard normal Z (with µ = 0 and σ = 1) using formula 1. can transform normally distributed X with µ ≠ 0 & σ ≠ 1 into standard normal Z (with µ = 0 and σ = 1) using formula • Z = (X  µ) / σ
• Example: batting averages: µ = .265, σ = .025 • So, BA of .290 equivalent to Z = 1. • (.290  .265) / .025 = 1 • Z = (X  µ) / σ
• Example: batting averages: µ = .265, σ = .025 • So, BA of .290 equivalent to Z = 1. • (.290  .265) / .025 = 1 How many standard deviations away from µ is Z = 1?
Slide 63 How many standard deviations away from µ is X = .290?
Slide 64 Example
Benny Bonds is a professional baseball player. His contract states that he will get a $400,000 bonus if his season batting average is in the top 3% of all hitters. o Suppose μ = .250 and σ = .040 (batting averages) o Benny hit .330. o Does he get his bonus?
o o o Example Example
Your assignment is to find how likely it is that a variable variable X can have a value between 0 and 6. o Suppose you know that X is normally distributed with • mean = 4 & standard deviation = 2. (from (from your data) data) o This can be stated as "find the probability that X "find can can be between 0 and 6." o This can also be stated as"find P(0 < X < 6)." Slide 65 Slide 66 11 Example (cont.)
•Here are the steps to find the answer: •Step (1) convert the X value of 0 to a Zvalue: Z• z1 = 0  4 / 2 =  2 • so, x=0 equivalent to z=2 z=•Step (2) convert the X value of 6 to a Zvalue: Z• z2 = 6  4 / 2 = 1 • so, x=6 equivalent to z=1 •Step (3) it is true in this case that •P(0 < X < 6) = P(2 < Z < 1) P( Example Example
• HINT: draw picture step (4) P(2 < Z < 1) is area between z = 2 and P(z=1 • this area is sum of two areas: • area between z = 2 & z =0 (area A1) & (area • area between z = 0 & z = 1 (area A2) (area • step (5) P(2 < Z < 1) = area A1 + area A2 P(area = 0.4772 + 0.3413 = 0.8185 •This answer means there is 81.85% probability that X will be between 0 & 6 • (or 81.85% of all X values fall between 0 & 6). Slide 67 Slide 68 The Standard Normal Distribution (cont.)
Area A1 = 0.4772 Area A2 = 0.3413 The Standard Normal Distribution (cont.) NOTE: NOTE: P(0 < X < 6) = P(2 < Z < 1) P( (Z’s sigma = 1) (X’s sigma = 2) 2 0 0 4 1 6 z values x values 2 0 0 4 1 6 z values x values Z = (X  µ) / σ
Slide 69 Z = (X  µ) / σ
Slide 70 The Standard Normal Distribution (cont.) NOTE: NOTE: 81.85% of all values The Standard Normal Distribution (cont.) •Here’s how to use the table of Z values to find the two areas I showed you earlier. earlier.
P(Z<P(Z<2) = 0.0228. P(Z<0) = 0.5. So, P(2<Z<0)=0.5P(2<Z<0)=0.50.228=0.4772. This is area A1. Thi A1 P(Z<1) P(Z<1) =0.8413. P(Z>0)=0.5. So, P(0<Z<1)=0.84130.5=0.3413. P(0<Z<1)=0.8413This is area A2. 2 0 0 4 1 6 z values x values
Finally, P(2<Z<1) = P(2<Z<0) + P(0<Z<1) = 0.4772+0.3413 = P(P(0.8185. This is area A1 + area A2. Z = (X  µ) / σ
Slide 71 Slide 72 72 12 The Normal Distribution (cont.)
Area A1 Area A2 The Normal Distribution (cont.)
Area A1 = 0.4772 Area A2 = 0.3413 (Z’s sigma = 1) (X’s sigma = 2) 2 0 0 4 1 6 z values x values (Z’s sigma = 1) (X’s sigma = 2) 2 0 0 4 1 6 z values x values Z = (X  µ) / σ
Slide 73 Z = (X  µ) / σ
Slide 74 The Normal Distribution (cont.) Example
You are negotiating with the union on your company’s next labor contract. o The union representative has claimed that too many employees (650) of your company receive low weekly earnings (under $300) and because of that (under your company must give its lowerpaid employees a lowerra raise to eliminate that inequity. li th it o You sample the salaries of your company’s 5000 hourly workers and find that their earnings are normally distributed with a mean of $350 and a standard deviation of $40.
o • P(2 < Z < 1) = area A1 + area A2 = 0.4772 + P(0.3413 = 0.8185; • means there is 81.85% probability that X will be between 0 & 6 (or 81.85% of all X values fall between 0 & 6) 6) Slide 75 Slide 76 76 Example Example HINT: draw picture HINT: draw picture Example
The weekly salaries of 5000 employees of a large corporation are normally distributed with a mean of $350 and a standard deviation of $40. How many employees earn less than $300 per week? The weekly salaries of 5000 hourly employees are normally distributed with a mean of $350 and a standard deviation of $40. How many employees earn less than $300 per week? STEPS: o convert x values into equivalent z values o calculate probability using the z values o whatever probability you find for the z values is also true for the x values in the problem Converting X values to Z values by the formula Z = (X  µ) / σ where X=300, µ=350, & σ =40 gives Z = (300  350) / 40 =  1.25 (300 350)
o Calculating the probability P(X < 300) = P(Z <  1.25) gives (next slide) Slide 77 Slide 78 13 Example (cont.) Example (cont.) 10.56% z
1.25 $300 0 z z
1.25 $300 0 z • The probability P(Z < 1.25) is the area shown. • • The probability P(Z < 1.25) is the area shown. •So, P(X < 300) = P(Z < 1.25) = 0.1056. So, 0.1056. • This means that 10.56% of the employees have salaries less than $300. This corresponds to 0.1056(5000) = 528 employees. employees. Slide 79 Slide 80 Example (cont.) The union representative claimed that too many employees (650) of your company receive low weekly (650 earnings (under $300) and because of that your (under company must give its employees a raise to eliminate that inequity. o You found that 528 employees earn less than $300 per per week.
o o o Exercise
o Exercises about Normal distributions • SHORT exercise – in class • LONG exercise  homework Is the union representative correct? What do you tell her during the negotiations? Slide 81 Slide 82 #1. What’s the Probability When Know X Value?
Pep Zone Pep Zone sells auto parts and supplies including a popular multigrade motor oil. When the stock of this multioil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for an order. being It It has been determined that leadtime demand (purchases (purchases after replenishment order placed) is normally normally distributed with a mean of 15 gallons and a standard standard deviation of 6 gallons. The manager would like to know the probability of of a stockout, P(x > 20).
Slide 83 83 #1. What’s the Probability When Know X Value?
Example: Pep Zone The manager would like to know the probability of of a stockout, P(x > 20). That is, what ‘s the probability that after a replenishment replenishment order is placed, total customer demand will EXCEED 20 gallons of oil. will oil. (Pep Zone will lose sales if that happens before the replenishment order arrives.)
o o Example: Slide 84 84 14 #1. What’s the Probability When Know X Value?
• This graph shows the distribution of oil purchases AFTER AFTER a replenishment order has been placed. z = (x  μ)/σ = (20  15)/6 = .83 #1. What’s the Probability When Know X Value?
o Standard Normal Probability Distribution • The Standard Normal table shows an area of .7967 for the region to the left of z = .83 line below. .83 • The shaded tail area is 1  .7967 = .2033. • The probability of a stockout is .2033. z = (x  μ)/σ = (20  15)/6 = .83 Area = .2967 Area = 10.7967 =.5 =.5  .2967 = .2033 .2033 0 .83 z
Slide 86 0 .83 z 1. How many gallons does z = 0 correspond to? 2. How many gallons does z = .83 correspond to?
Slide 85 Area = .5 #1. What’s the Probability When Know X Value?
o #1. What’s the Probability When Know X Value?
o Using the Standard Normal Probability Table
.00 .01 z .0 .5000 .5040 .1 .5398 .5438 .2 .5793 .5832 .3 .6179 .6217 .4 .6554 .6591 .5 .6 .7 .8 .9 .6915 .6950 .7257 .7291 .7580 .7881 .8159 .02 .03 .5120 .5517 .5910 .6293 .6664 .7019 .7357 .7642 .7673 .7704 .7939 .7967 .7995 .8023 .8212 .8238 .8264
Slide 87 Standard Normal Probability Distribution • The probability of a stockout is .2033. • That is, there is about a 20% chance that Pep Zone will lose sales before the replenishment oil arrives. .04 .05 .06 .07 .08 .09 If you’re the manager, would that result concern you? Slide 88 #2. What’s the X Value When Know Probability?
o #2. #2. What’s the X Value When Know Probability?
o Standard Normal Probability Distribution If the manager of Pep Zone wants the probability of a stockout stockout to be no more than .05, what should the reorder reorder point be? How many gallons? Area Area = .05 Area = .5 Area = .45 Using the Standard Normal Probability Table We now lookup the .9500 area in the lookStandard Normal Probability table to find the corresponding z.05 value. Use z = 1.645 How many σ above the mean is the X value?
z . 1.5 1.6 1.7 1.8 1.9 . .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .9382 .9394 .9495 .9505 .9591 .9599 .9738 .9744
Slide 90 z.05 0 Let z.05 represent the z value cutting the .05 tail area.
Slide 89 z.05 = 1.645 is a reasonable estimate. .9671 .9678 15 #2. #2. What’s the X Value When Know Probability?
o Standard Normal Probability Distribution The corresponding value of x is given by x = μ + z.05σ = 15 + 1.645(6) 15 = 24.87 A reorder point of 24.87 gallons will place the probability probability of a stockout during leadtime at .05. Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability under .05. If the store reorders oil when current stock = 25 gals., what’s the prob. of running out of oil before the new oil arrives?
Slide 91 16 ...
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This note was uploaded on 11/15/2010 for the course ECO 6416 taught by Professor Staff during the Spring '08 term at University of Central Florida.
 Spring '08
 Staff

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