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COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-1 l Using Statistics l The Normal.

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Presentation on theme: "COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-1 l Using Statistics l The Normal."— Presentation transcript:

1 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-1 l Using Statistics l The Normal Probability Distribution l The Standard Normal Distribution l The Transformation of Normal Random Variables l The Inverse Transformation l More Complex Problems l The Normal Distribution as an Approximation to Other Probability Distributions l Using the Computer l Summary and Review of Terms The Normal Distribution 4

2 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-2 As n increases, the binomial distribution approaches a... n = 6n = 14n = 10 Normal Probability Density Function: 6543210 0.3 0.2 0.1 0.0 x P ( x ) Binomial Distribution: n=6, p=.5 109876543210 0.3 0.2 0.1 0.0 x P ( x ) Binomial Distribution: n=10, p=.5 50-5 0.4 0.3 0.2 0.1 0.0 x f ( x ) Normal Distribution:  = 0,  = 1 4-1 Introduction

3 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-3 The normal probability density function: 50-5 0.4 0.3 0.2 0.1 0.0 x f ( x ) Normal Distribution:  = 0,  = 1 4-2 The Normal Probability Distribution

4 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-4 The normal is a family of – Bell-shaped and symmetric distributions. because the distribution is symmetric, one-half (.50 or 50%) lies on either side of the mean. – Each is characterized by a different pair of mean, , and variance,  . That is: [X~N(   )]. – Each is asymptotic to the horizontal axis. – The area under any normal probability density function within k  of  is the same for any normal distribution, regardless of the mean and variance. The Normal Probability Distribution

5 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-5 All of these are normal probability density functions, though each has a different mean and variance. Z~N(0,1) 50-5 0.4 0.3 0.2 0.1 0.0 z f ( z ) Normal Distribution:  =0,  =1 W~N(40,1)X~N(30,25) 454035 0.4 0.3 0.2 0.1 0.0 w f ( w ) Normal Distribution:  =40,  =1 6050403020100 0.2 0.1 0.0 x f ( x ) Normal Distribution:  =30,  =5 Y~N(50,9) 65554535 0.2 0.1 0.0 y f ( y ) Normal Distribution:  =50,  =3 50 Consider: P(39  W  41) P(25  X  35) P(47  Y  53) P(-1  Z  1) The probability in each case is an area under a normal probability density function. Normal Probability Distributions

6 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-6 The standard normal random variable, Z, is the normal random variable with mean  = 0 and standard deviation  = 1: Z~N(0,1 2 ). 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution  =0  =1 { 4-3 The Standard Normal Distribution

7 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-7 z.00.01.02.03.04.05.06.07.08.09 0.00.00000.00400.00800.01200.01600.01990.02390.02790.03190.0359 0.10.03980.04380.04780.05170.05570.05960.06360.06750.07140.0753 0.20.07930.08320.08710.09100.09480.09870.10260.10640.11030.1141 0.30.11790.12170.12550.12930.13310.13680.14060.14430.14800.1517 0.40.15540.15910.16280.16640.17000.17360.17720.18080.18440.1879 0.50.19150.19500.19850.20190.20540.20880.21230.21570.21900.2224 0.60.22570.22910.23240.23570.23890.24220.24540.24860.25170.2549 0.70.25800.26110.26420.26730.27040.27340.27640.27940.28230.2852 0.80.28810.29100.29390.29670.29950.30230.30510.30780.31060.3133 0.90.31590.31860.32120.32380.32640.32890.33150.33400.33650.3389 1.00.34130.34380.34610.34850.35080.35310.35540.35770.35990.3621 1.10.36430.36650.36860.37080.37290.37490.37700.37900.38100.3830 1.20.38490.38690.38880.39070.39250.39440.39620.39800.39970.4015 1.30.40320.40490.40660.40820.40990.41150.41310.41470.41620.4177 1.40.41920.42070.42220.42360.42510.42650.42790.42920.43060.4319 1.50.43320.43450.43570.43700.43820.43940.44060.44180.44290.4441 1.60.44520.44630.44740.44840.44950.45050.45150.45250.45350.4545 1.70.45540.45640.45730.45820.45910.45990.46080.46160.46250.4633 1.80.46410.46490.46560.46640.46710.46780.46860.46930.46990.4706 1.90.47130.47190.47260.47320.47380.47440.47500.47560.47610.4767 2.00.47720.47780.47830.47880.47930.47980.48030.48080.48120.4817 2.10.48210.48260.48300.48340.48380.48420.48460.48500.48540.4857 2.20.48610.48640.48680.48710.48750.48780.48810.48840.48870.4890 2.30.48930.48960.48980.49010.49040.49060.49090.49110.49130.4916 2.40.49180.49200.49220.49250.49270.49290.49310.49320.49340.4936 2.50.49380.49400.49410.49430.49450.49460.49480.49490.49510.4952 2.60.49530.49550.49560.49570.49590.49600.49610.49620.49630.4964 2.70.49650.49660.49670.49680.49690.49700.49710.49720.49730.4974 2.80.49740.49750.49760.49770.49770.49780.49790.49790.49800.4981 2.90.49810.49820.49820.49830.49840.49840.49850.49850.49860.4986 3.00.49870.49870.49870.49880.49880.49890.49890.49890.49900.4990 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution 1.56 { Standard Normal Probabilities Look in row labeled 1.5 and column labeled.06 to find P(0  z  1.56) =.4406 Finding Probabilities of the Standard Normal Distribution: P(0 < Z < 1.56)

8 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-8 To find P(Z<-2.47): Find table area for 2.47 P(0 < Z < 2.47) =.4932 P(Z < -2.47) =.5 - P(0 < Z < 2.47) =.5 -.4932 = 0.0068 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution Table area for 2.47 P(0 < Z < 2.47) = 0.4932 Area to the left of -2.47 P(Z < -2.47) =.5 - 0.4932 = 0.0068 Finding Probabilities of the Standard Normal Distribution: P(Z < -2.47) z....06.07.08.. 2.3...0.49090.49110.4913 2.4...0.49310.49320.4934 2.5...0.49480.49490.4951.

9 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-9 z.00..... 0.90.3159... 1.00.3413... 1.10.3643..... 1.90.4713... 2.00.4772... 2.10.4821..... To find P(1  Z  2): 1. Find table area for 2.00 F(2) = P(Z  2.00) =.5 +.4772 =.9772 2. Find table area for 1.00 F(1) = P(Z  1.00) =.5 +.3413 =.8413 3. P(1  Z  2.00) = P(Z  2.00) - P(Z  1.00) =.9772 -.8413 =.1359 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution Area between 1 and 2 P(1  Z  2) =.4772 -.8413 = 0.1359 Finding Probabilities of the Standard Normal Distribution: P(1< Z < 2)

10 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-10 To find z such that P(0  Z  z) =.40: 1. Find a probability as close as possible to.40 in the table of standard normal probabilities. 2. Then determine the value of z from the corresponding row and column. P(0  Z  1.28) .40 Also, since P(Z  0) =.50 P(Z  1.28) .90 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution Area =.40 (.3997) Z = 1.28 Area to the left of 0 =.50 P(z  0) =.50 Finding Values of the Standard Normal Random Variable: P(0 < Z < z) = 0.40 z.00.01.02.03.04.05.06.07.08.09 0.00.00000.00400.00800.01200.01600.01990.02390.02790.03190.0359 0.10.03980.04380.04780.05170.05570.05960.06360.06750.07140.0753 0.20.07930.08320.08710.09100.09480.09870.10260.10640.11030.1141 0.30.11790.12170.12550.12930.13310.13680.14060.14430.14800.1517 0.40.15540.15910.16280.16640.17000.17360.17720.18080.18440.1879 0.50.19150.19500.19850.20190.20540.20880.21230.21570.21900.2224 0.60.22570.22910.23240.23570.23890.24220.24540.24860.25170.2549 0.70.25800.26110.26420.26730.27040.27340.27640.27940.28230.2852 0.80.28810.29100.29390.29670.29950.30230.30510.30780.31060.3133 0.90.31590.31860.32120.32380.32640.32890.33150.33400.33650.3389 1.00.34130.34380.34610.34850.35080.35310.35540.35770.35990.3621 1.10.36430.36650.36860.37080.37290.37490.37700.37900.38100.3830 1.20.38490.38690.38880.39070.39250.39440.39620.39800.39970.4015 1.30.40320.40490.40660.40820.40990.41150.41310.41470.41620.4177...........

11 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-11 z.04.05.06.07.08.09....... 2.4...0.49270.49290.49310.49320.49340.4936 2.5...0.49450.49460.49480.49490.49510.4952 2.6...0.49590.49600.49610.49620.49630.4964....... To have.99 in the center of the distribution, there should be (1/2)(1-.99) = (1/2)(.01) =.005 in each tail of the distribution, and (1/2)(.99) =.495 in each half of the.99 interval. That is: P(0  Z  z.005 ) =.495 Look to the table of standard normal probabilities to find that:  z.005  z.005  P(-.2575  Z  ) =.99 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) -z.005 z.005 Area in right tail =.005 Area in left tail =.005 Area in center right =.495 Area in center left =.495 2.575-2.575 Area in center =.99 99% Interval around the Mean: P(-z.005< Z < z.005) = 0.99

12 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-12 The area within k  of the mean is the same for all normal random variables. So an area under any normal distribution is equivalent to an area under the standard normal. In this example: P(40  X  P(-1  Z  since  and  1009080706050403020100 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 X f ( x ) Normal Distribution:  =50,  =10  = 10 { 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution 1.0 { Transformation (2) Division by  x ) The transformation of X to Z: The inverse transformation of Z to X: 4-4 The Transformation of Normal Random Variables (1) Subtraction: (X -  x )

13 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-13 Example 4-1 X~N(160,30 2 ) Example 4-2 X~N(127,22 2 ) Using the Normal Transformation

14 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-14 MTB > cdf 100; SUBC> normal 160,30. Cumulative Distribution Function Normal with mean = 160.000 and standard deviation = 30.0000 x P( X <= x) 100.0000 0.0228 MTB > cdf 180; SUBC> normal 160,30. Cumulative Distribution Function Normal with mean = 160.000 and standard deviation = 30.0000 x P( X <= x) 180.0000 0.7475 MTB > cdf 100; SUBC> normal 160,30. Cumulative Distribution Function Normal with mean = 160.000 and standard deviation = 30.0000 x P( X <= x) 100.0000 0.0228 MTB > cdf 180; SUBC> normal 160,30. Cumulative Distribution Function Normal with mean = 160.000 and standard deviation = 30.0000 x P( X <= x) 180.0000 0.7475 MTB > cdf 150; SUBC> normal 127,22. Cumulative Distribution Function Normal with  = 127.000 and  = 22.0000 x P( X <= x) 150.0000 0.8521 MTB > cdf 150; SUBC> normal 127,22. Cumulative Distribution Function Normal with  = 127.000 and  = 22.0000 x P( X <= x) 150.0000 0.8521 Using the Normal Transformation - Minitab Solutions for Examples 4-1 & 4-2

15 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-15 Example 4-3 X~N(383,12 2 ) 440390340 0.05 0.04 0.03 0.02 0.01 0.00 X f ( X ) Normal Distribution:  = 383,  = 12 MTB > cdf 394; SUBC> normal 383,12. Cumulative Distribution Function Normal with mean = 383.000 and standard deviation = 12.0000 x P( X <= x) 394.0000 0.8203 MTB > cdf 394; SUBC> normal 383,12. Cumulative Distribution Function Normal with mean = 383.000 and standard deviation = 12.0000 x P( X <= x) 394.0000 0.8203 MTB > cdf 399; SUBC> normal 383,12. Cumulative Distribution Function Normal with mean = 383.000 and standard deviation = 12.0000 x P( X <= x) 399.0000 0.9088 MTB > cdf 399; SUBC> normal 383,12. Cumulative Distribution Function Normal with mean = 383.000 and standard deviation = 12.0000 x P( X <= x) 399.0000 0.9088 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution Equivalent areas Using the Normal Transformation - Example 4-3

16 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-16 Do the same with X = 394 and subtract the two values to get 0.088447. Using the Normal Transformation - Excel Solution for Example 4-3

17 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-17 The transformation of X to Z: The inverse transformation of Z to X: The transformation of X to Z, where a and b are numbers:: The Transformation of Normal Random Variables

18 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-18 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution The probability that a normal random variable will be within 1 standard deviation from its mean (on either side) is 0.6826, or approximately 0.68. The probability that a normal random variable will be within 2 standard deviations from its mean is 0.9544, or approximately 0.95. The probability that a normal random variable will be within 3 standard deviation from its mean is 0.9974. Normal Probabilities

19 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-19 z.07.08.09..... 1.1... 0.37900.38100.3830 1.2... 0.39800.39970.4015 1.3... 0.41470.41620.4177..... The area within k  of the mean is the same for all normal random variables. To find a probability associated with any interval of values for any normal random variable, all that is needed is to express the interval in terms of numbers of standard deviations from the mean. That is the purpose of the standard normal transformation. If X~N(50,10 2 ), That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 =  + 2 . P(X > 70) is equivalent to P(Z > 2), an area under the standard normal distribution. The area within k  of the mean is the same for all normal random variables. To find a probability associated with any interval of values for any normal random variable, all that is needed is to express the interval in terms of numbers of standard deviations from the mean. That is the purpose of the standard normal transformation. If X~N(50,10 2 ), That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 =  + 2 . P(X > 70) is equivalent to P(Z > 2), an area under the standard normal distribution. Example 4-4 X~N(124,12 2 ) P(X > x) = 0.10 and P(Z > 1.28)  0.10 x =  + z  = 124 + (1.28)(12) = 139.36 18013080 0.04 0.03 0.02 0.01 0.00 X f ( x ) Normal Distribution:  = 124,  = 12 4-5 The Inverse Transformation

20 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-20 z.02.03.04..... 2.2... 0.48680.48710.4875 2.3... 0.48980.49010.4904 2.4... 0.49220.49250.4927..... z.05.06.07..... 1.8... 0.46780.46860.4693 1.9... 0.47440.47500.4756 2.0... 0.47980.48030.4808..... Example 4-5 X~N(5.7,0.5 2 ) P(X > x)=0.01 and P(Z > 2.33)  0.01 x =  + z  = 5.7 + (2.33)(0.5) = 6.865 Example 4-5 X~N(5.7,0.5 2 ) P(X > x)=0.01 and P(Z > 2.33)  0.01 x =  + z  = 5.7 + (2.33)(0.5) = 6.865 Example 4-6 X~N(2450,400 2 ) P(a<X<b)=0.95 and P(-1.96<Z<1.96)  0.95 x =  z  = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234) P(1666 < X < 3234) = 0.95 Example 4-6 X~N(2450,400 2 ) P(a<X<b)=0.95 and P(-1.96<Z<1.96)  0.95 x =  z  = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234) P(1666 < X < 3234) = 0.95 8.27.26.25.24.23.2 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 8.27.26.25.24.23.2 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 X f ( x ) Normal Distribution:  = 5.7  = 0.5 543210-1-2-3-4-5 z Z. 01 = 2.33 Area = 0.49 Area = 0.01 4000300020001000 0.0015 0.0010 0.0005 0.0000 X f ( x ) Normal Distribution:  = 2450  = 400 4000300020001000 0.0015 0.0010 0.0005 0.0000 543210-1-2-3-4-5 Z.4750.0250 -1.96 1.96 The Inverse Transformation (2) X. 01 =  +z  = 5.7 + (2.33)(0.5) = 6.865

21 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-21 4000300020001000 0.0012 0.0010 0.0008 0.0006 0.0004 0.0002 0.0000 X f ( x ) Normal Distribution:  = 2450,  = 400...... 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution 1.Draw pictures of the normal distribution in question and of the standard normal distribution. Finding Values of a Normal Random Variable, Given a Probability

22 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-22 1.Draw pictures of the normal distribution in question and of the standard normal distribution. 2.Shade the area corresponding to the desired probability. Finding Values of a Normal Random Variable, Given a Probability 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution.4750.9500

23 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-23 z.05.06.07..... 1.8... 0.46780.46860.4693 1.9... 0.47440.47500.4756 2.0... 0.47980.48030.4808..... 3.From the table of the standard normal distribution, find the z value or values. 1.Draw pictures of the normal distribution in question and of the standard normal distribution. 2.Shade the area corresponding to the desired probability. Finding Values of a Normal Random Variable, Given a Probability 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution.4750.9500 -1.96 1.96

24 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-24 4.Use the transformation from z to x to get value(s) of the original random variable. x =  z  = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234) x =  z  = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234) Finding Values of a Normal Random Variable, Given a Probability z.05.06.07..... 1.8... 0.46780.46860.4693 1.9... 0.47440.47500.4756 2.0... 0.47980.48030.4808..... 3.From the table of the standard normal distribution, find the z value or values. 1.Draw pictures of the normal distribution in question and of the standard normal distribution. 2.Shade the area corresponding to the desired probability. 543210-1-2-3-4-5 0.4 0.3 0.2 0.1 0.0 Z f ( z ) Standard Normal Distribution.4750.9500 -1.96 1.96

25 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-25 Using EXCEL to help solve Example 4-4. Finding Values of a Normal Random Variable, Given a Probability

26 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-26 1050 0.3 0.2 0.1 0.0 X f ( x ) Normal Distribution:  = 3.5,  = 1.323 76543210 0.3 0.2 0.1 0.0 X P ( x ) Binomial Distribution: n = 7, p = 0.50 The normal distribution with  = 3.5 and  = 1.323 is a close approximation to the binomial with n = 7 and p = 0.50. P(x<4.5) = 0.7749 MTB > cdf 4.5; SUBC> normal 3.5 1.323. Cumulative Distribution Function Normal with mean = 3.50000 and standard deviation = 1.32300 x P( X <= x) 4.5000 0.7751 MTB > cdf 4.5; SUBC> normal 3.5 1.323. Cumulative Distribution Function Normal with mean = 3.50000 and standard deviation = 1.32300 x P( X <= x) 4.5000 0.7751 MTB > cdf 4; SUBC> binomial 7,.5. Cumulative Distribution Function Binomial with n = 7 and p = 0.500000 x P( X <= x) 4.00 0.7734 MTB > cdf 4; SUBC> binomial 7,.5. Cumulative Distribution Function Binomial with n = 7 and p = 0.500000 x P( X <= x) 4.00 0.7734 P( x  4) = 0.7734 Finding Values of a Normal Random Variable, Given a Probability

27 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-27 1050 0.3 0.2 0.1 0.0 X f ( x ) Normal Distribution:  = 5.5,  = 1.6583 11109876543210 0.2 0.1 0.0 X P ( x ) Binomial Distribution: n = 11, p = 0.50 The normal distribution with  = 5.5 and  = 1.6583 is a closer approximation to the binomial with n = 11 and p = 0.50. P(x<4.5) = 0.2732 P(x  4) = 0.2744 MTB > cdf 4.5; SUBC> normal 5.5 1.6583. Cumulative Distribution Function Normal with mean = 5.50000 and standard deviation = 1.65830 x P( X <= x) 4.5000 0.2732 MTB > cdf 4.5; SUBC> normal 5.5 1.6583. Cumulative Distribution Function Normal with mean = 5.50000 and standard deviation = 1.65830 x P( X <= x) 4.5000 0.2732 MTB > cdf 4; SUBC> binomial 11,.5. Cumulative Distribution Function Binomial with n = 11 and p = 0.500000 x P( X <= x) 4.00 0.2744 MTB > cdf 4; SUBC> binomial 11,.5. Cumulative Distribution Function Binomial with n = 11 and p = 0.500000 x P( X <= x) 4.00 0.2744 The Normal Distribution as an Approximation to Other Probability Distributions (2)

28 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-28 Or: If p is either small (close to 0) or large (close to 1), use the Poisson approximation. Approximating a Binomial Probability Using the Normal Distribution

29 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-29 In EXCEL, the command NORMSDIST(number) will return cumulative probability for a standard normal random variable. The command NORMDIST(number, mean, standard deviation) will return the cumulative probability for a general random variable. 4-7 Using the Computer

30 COMPLETE f o u r t h e d i t i o n BUSINESS STATISTICS Aczel Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 1999 4-30 For example, NORMSDIST(1.0) = 0.8413. NORMDIST(10.0, 5, 2) = 0.9938. The inverse commands are NORMSINV(number) and NORMINV(number, mean, standard deviation). NORMSINV(0.975) = 1.96. NORMINV(0.975, 20, 10) = 39.6. 4-7 Using the Computer

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