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COMPLETE BUSINESS STATISTICS

COMPLETE BUSINESS STATISTICS. by AMIR D. ACZEL & JAYAVEL SOUNDERPANDIAN 6 th edition (SIE). Chapter 4. The Normal Distribution. Using Statistics Properties of the Normal Distribution The Template The Standard Normal Distribution The Transformation of Normal Random Variables

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COMPLETE BUSINESS STATISTICS

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  1. COMPLETE BUSINESS STATISTICS by AMIR D. ACZEL & JAYAVEL SOUNDERPANDIAN 6th edition (SIE)

  2. Chapter 4 The Normal Distribution

  3. Using Statistics Properties of the Normal Distribution The Template The Standard Normal Distribution The Transformation of Normal Random Variables The Inverse Transformation The Normal Approximation of Binomial Distributions 4 The Normal Distribution

  4. Identify when a random variable will be normally distributed Use the properties of normal distributions Explain the significance of the standard normal distribution Use normal distribution tables to compute probabilities Transform a normal distribution into a standard normal distribution Convert a binomial distribution into an approximate normal distribution Use spreadsheet templates to solve normal distribution problems 4 LEARNING OBJECTIVES After studying this chapter, you should be able to:

  5. B i n o m i a l D i s t r i b u t i o n : n = 6 , p = . 5 B i n o m i a l D i s t r i b u t i o n : n = 1 0 , p = . 5 B i n o m i a l D i s t r i b u t i o n : n = 1 4 , p = . 5 0 . 3 0 . 3 0 . 3 0 . 2 0 . 2 0 . 2 ) ) ) x x x ( ( ( P P P 0 . 1 0 . 1 0 . 1 0 . 0 0 . 0 0 . 0 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 1 0 0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 x x x N o r m a l D i s t r i b u t i o n :  = 0 ,  = 1 0 . 4 2 - m x 0 . 3 æ ö - ç ÷ ç ÷ ) è ø x 0 . 2 1 2 ( s f 2 = - ¥ < < ¥ f ( x ) e for x 0 . 1 2 ps 2 0 . 0 - 5 0 5 = p = where e 2 . 7182818 ... and 3 . 14159265 ... x 4-1 Introduction As n increases, the binomial distribution approaches a ... n = 6 n = 10 n = 14 Normal Probability Density Function:

  6. N o r m a l D i s t r i b u t i o n :  = 0 ,  = 1 0 . 4 2 - m x æ ö - ç ÷ 0 . 3 ç ÷ è ø 1 2 s 2 = - ¥ < < ¥ ) f ( x ) e for x 0 . 2 x ( f 2 ps 2 0 . 1 0 . 0 = p = where e 2 . 7182818 . . . and 3 . 14159265 . . . - 5 0 5 x The Normal Probability Distribution The normal probability density function:

  7. The normal is a familyof Bell-shaped and symmetricdistributions. 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 isasymptotic 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. 4-2 Properties of the Normal Distribution

  8. If several independent random variables are normally distributed then their sum will also be normally distributed. The mean of the sum will be the sum of all the individual means. The variance of the sum will be the sum of all the individual variances (by virtue of the independence). 4-2 Properties of the Normal Distribution (continued)

  9. If X1, X2, …, Xn are independent normal random variable, then their sum S will also be normally distributed with E(S) = E(X1) + E(X2) + … + E(Xn) V(S) = V(X1) + V(X2) + … + V(Xn) Note: It is the variances that can be added above and not the standard deviations. 4-2 Properties of the Normal Distribution (continued)

  10. Example 4.1: Let X1, X2, and X3be independent random variables that are normally distributed with means and variances as shown. 4-2 Properties of the Normal Distribution – Example 4-1 Let S = X1 + X2 + X3. Then E(S) = 10 + 20 + 30 = 60 and V(S) = 1 + 2 + 3 = 6. The standard deviation of S is = 2.45.

  11. If X1, X2, …, Xn are independent normal random variable, then the random variable Q defined as Q = a1X1 + a2X2 + … + anXn + b will also be normally distributed with E(Q) = a1E(X1) + a2E(X2) + … + anE(Xn) + b V(Q) = a12 V(X1) + a22 V(X2) + … + an2 V(Xn) Note: It is the variances that can be added above and not the standard deviations. 4-2 Properties of the Normal Distribution (continued)

  12. Example 4.3: Let X1 , X2 , X3and X4be independent random variables that are normally distributed with means and variances as shown. Find the mean and variance of Q = X1 - 2X2 + 3X2 - 4X4 + 5 4-2 Properties of the Normal Distribution – Example 4-3 E(Q) = 12 – 2(-5) + 3(8) – 4(10) + 5 = 11 V(Q) = 4 + (-2)2(2) + 32(5) + (-4)2(1) = 73 SD(Q) =

  13. Computing the Mean, Variance and Standard Deviation for the Sum of Independent Random Variables Using the Template

  14. N o r m a l D i s t r i b u t i o n :  = 4 0 ,  = 1 N o r m a l D i s t r i b u t i o n :  = 3 0 ,  = 5 N o r m a l D i s t r i b u t i o n :  = 5 0 ,  = 3 0 . 4 0 . 2 0 . 2 0 . 3 ) ) ) w x y 0 . 2 0 . 1 0 . 1 ( ( ( f f f 0 . 1 0 . 0 0 . 0 0 . 0 3 5 4 0 4 5 0 1 0 2 0 3 0 4 0 5 0 6 0 3 5 4 5 5 5 6 5 5 0 w x y N o r m a l D i s t r i b u t i o n :  = 0 ,  = 1 0 . 4 0 . 3 ) z 0 . 2 ( f 0 . 1 0 . 0 - 5 0 5 z Normal Probability Distributions All of these are normal probability density functions, though each has a different mean and variance. W~N(40,1) X~N(30,25) Y~N(50,9) 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. Z~N(0,1)

  15. 4-3 Computing Normal Probabilities Using the Template

  16. Standard Normal Distribution 0 . 4 0 . 3 { ) =1 z ( 0 . 2 f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5  = 0 Z 4-4 The Standard Normal Distribution The standard normal random variable, Z, is the normal random variable with mean  = 0 and standard deviation  = 1: Z~N(0,12).

  17. S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 0 . 3 ) z 0 . 2 ( f 0 . 1 1.56 { 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Finding Probabilities of the Standard Normal Distribution: P(0 < Z < 1.56) Standard Normal Probabilities z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 Look in row labeled 1.5 and column labeled .06to find P(0  z  1.56) = 0.4406

  18. 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 S t a n d a r d N o r m a l D i s t r i b u t i o n Area to the left of -2.47 P(Z < -2.47) = .5 - 0.4932 = 0.0068 0 . 4 Table area for 2.47 P(0 < Z < 2.47) = 0.4932 0 . 3 ) z 0 . 2 ( f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Finding Probabilities of the Standard Normal Distribution: P(Z < -2.47) z ... .06 .07 .08 . . . . . . . . . . . . 2.3 ... 0.4909 0.4911 0.4913 2.4 ... 0.4931 0.4932 0.4934 2.5 ... 0.4948 0.4949 0.4951 . . .

  19. 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 = 0.1359 z .00 ... . . . . . . 0.9 0.3159 ... 1.0 0.3413 ... 1.1 0.3643 ... . . . . . . 1.9 0.4713 ... 2.0 0.4772 ... 2.1 0.4821 ... . . . . . . S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 Area between 1 and 2 P(1  Z  2) = .9772 - .8413 = 0.1359 0 . 3 ) z ( 0 . 2 f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Finding Probabilities of the Standard Normal Distribution: P(1< Z < 2)

  20. z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 Area to the left of 0 = .50 P(z  0) = .50 Area = .40 (.3997) 0 . 3 ) z 0 . 2 ( f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z = 1.28 Z Finding Values of the Standard Normal Random Variable: P(0 < Z < z) = 0.40

  21. 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 z .04 .05 .06 .07 .08 .09 . . . . . . . . . . . . . . . . . . . . . 2.4 ... 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 ... 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 ... 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 . . . . . . . . . . . . . . . . . . . . . Total area in center = .99 Area in center left = .495 0 . 4 0 . 3 Area in center right = .495 ) z 0 . 2 ( f 0 . 1 Area in right tail = .005 Area in left tail = .005 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z -z.005 z.005 -2.575 2.575 99% Interval around the Mean

  22. m - X x = Z s x = + m X Z s x x 4-5 The Transformation of Normal Random Variables 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 m = 50 and s = 10. The transformation of X to Z: N o r m a l D i s t r i b u t i o n :  = 5 0 ,  = 1 0 0 . 0 7 0 . 0 6 Transformation 0 . 0 5 ) 0 . 0 4 x ( f (1) Subtraction: (X - x) 0 . 0 3 { =10 0 . 0 2 S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 0 1 0 . 0 0 0 . 4 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 1 0 0 X 0 . 3 ) z 0 . 2 ( f (2) Division by x) { The inverse transformation of Z to X: 1.0 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z

  23. £ £ < P ( 100 X 180 ) P ( X 150 ) æ ö æ ö - m - m - m - m - m 100 X 180 X 150 ç ÷ ç ÷ = £ £ = < P P è ø è ø s s s s s æ ö æ ö - - - 150 127 100 160 180 160 ç ÷ ç ÷ = < = £ £ P Z P Z è ø è ø 22 30 30 ( ) ( ) = < = - £ £ P Z 1 . 045 P 2 Z . 6667 = + = = + = 0 . 5 0 . 3520 0 . 8520 0 . 4772 0 . 2475 0 . 7247 Using the Normal Transformation Example 4-9 X~N(160,302) Example 4-10 X~N(127,222)

  24. £ £ P ( 394 X 399 ) æ ö - m - m - m 394 X 399 ç ÷ = £ £ P è ø S t a n d a r d N o r m a l D i s t r i b u t i o n s s s 0 . 4 æ ö - - 394 383 399 383 ç ÷ 0 . 3 = £ £ P Z è ø ) z 0 . 2 12 12 ( f ( ) 0 . 1 = £ £ P 0 . 9166 Z 1 . 333 0 . 0 = - = - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 0 . 4088 0 . 3203 0 . 0885 Z Using the Normal Transformation - Example 4-11 Example 4-11 X~N(383,122) N o r m a l D i s t r i b u t i o n :  = 3 8 3 ,  = 1 2 0 . 0 5 0 . 0 4 0 . 0 3 ) X ( f 0 . 0 2 0 . 0 1 0 . 0 0 340 390 440 X Equivalent areas Template solution

  25. - m X = m + s x X Z = Z x x s x - m æ a ö < = < ç ÷ P ( X a ) P Z è ø s - m æ b ö > = > ç ÷ P ( X b ) P Z è ø s - m - m æ a b ö < < = < < ç ÷ P ( a X b ) P Z è ø s s The Transformation of Normal Random Variables 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::

  26. S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 0 . 3 ) z 0 . 2 ( f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Normal Probabilities (Empirical Rule) • The probability that a normal random variable will be within 1 standard deviationfrom its mean (on either side) is 0.6826, or approximately 0.68. • The probability that a normal random variable will be within 2 standard deviationsfrom its mean is 0.9544, or approximately 0.95. • The probability that a normal random variable will be within 3 standard deviationfrom its mean is 0.9974.

  27. - m - m - æ ö æ ö x 70 70 50 > = > = > = > ç ÷ ç ÷ P ( X 70 ) P P Z P ( Z 2 ) è ø è ø s s 10 N o r m a l D i s t r i b u t i o n :  = 1 2 4 ,  = 1 2 0 . 0 4 0 . 0 3 ) x 0 . 0 2 ( f 0 . 0 1 0 . 0 0 8 0 1 3 0 1 8 0 X 4-6 The Inverse Transformation 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(50,102), 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 4-12X~N(124,122) P(X > x) = 0.10 and P(Z > 1.28) 0.10 x =  + z = 124 + (1.28)(12) = 139.36 z .07 .08 .09 . . . . . . . . . . . . . . . 1.1 . . . 0.3790 0.3810 0.3830 1.2 . . . 0.3980 0.3997 0.4015 1.3 . . . 0.4147 0.4162 0.4177 . . . . . . . . . . . . . . . 0.01 139.36

  28. Template Solution for Example 4-12 Example 4-12X~N(124,122) P(X > x) = 0.10 and P(Z > 1.28) 0.10 x =  + z = 124 + (1.28)(12) = 139.36

  29. z .02 .03 .04 . . . . . . . . . . . . . . . 2.2 . . . 0.4868 0.4871 0.4875 2.3 . . . 0.4898 0.4901 0.4904 2.4 . . . 0.4922 0.4925 0.4927 . . . . . . . . . . . . . . . z .05 .06 .07 . . . . . . . . . . . . . . . 1.8 . . . 0.4678 0.4686 0.4693 1.9 . . . 0.4744 0.4750 0.4756 2.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . . N o r m a l D i s t r i b u t i o n :  = 5 . 7  = 0 . 5 N o r m a l D i s t r i b u t i o n :  = 2 4 5 0  = 4 0 0 0 0 . . 8 8 0 0 . . 0 0 0 0 1 1 5 5 Area = 0.49 0 0 . . 7 7 0 0 . . 6 6 .4750 .4750 0 0 . . 0 0 0 0 1 1 0 0 0 0 . . 5 5 ) ) x x 0 0 . . 4 4 ( f ( f 0 0 . . 3 3 X.01 = +z = 5.7 + (2.33)(0.5) = 6.865 0 0 . . 0 0 0 0 0 0 5 5 0 0 . . 2 2 .0250 .0250 Area = 0.01 0 0 . . 1 1 0 0 . . 0 0 0 0 . . 0 0 0 0 0 0 0 0 3 3 . . 2 2 4 4 . . 2 2 5 5 . . 2 2 6 6 . . 2 2 7 7 . . 2 2 8 8 . . 2 2 1 1 0 0 0 0 0 0 2 2 0 0 0 0 0 0 3 3 0 0 0 0 0 0 4 4 0 0 0 0 0 0 X X - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 z Z -1.96 1.96 Z.01 = 2.33 The Inverse Transformation (Continued) Example 4-14X~N(2450,4002) 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 4-13X~N(5.7,0.52) P(X > x)=0.01 and P(Z > 2.33) 0.01 x =  + z = 5.7 + (2.33)(0.5) = 6.865

  30. 1. Draw pictures of the normal distribution in question and of the standard normal distribution. N o r m a l D i s t r i b u t i o n :  = 2 4 5 0 ,  = 4 0 0 0 . 0 0 1 2 . 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 X S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 0 . 3 ) z ( 0 . 2 f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Finding Values of a Normal Random Variable, Given a Probability

  31. N o r m a l D i s t r i b u t i o n :  = 2 4 5 0 ,  = 4 0 0 0 . 0 0 1 2 . .4750 .4750 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . .9500 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 X S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 .4750 .4750 0 . 3 ) z ( 0 . 2 f 0 . 1 .9500 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Finding Values of a Normal Random Variable, Given a Probability 1. Draw pictures of the normal distribution in question and of the standard normal distribution. 2. Shade the area corresponding to the desired probability.

  32. N o r m a l D i s t r i b u t i o n :  = 2 4 5 0 ,  = 4 0 0 0 . 0 0 1 2 . .4750 .4750 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . .9500 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 X S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 .4750 .4750 0 . 3 ) z ( 0 . 2 f 0 . 1 .9500 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Finding Values of a Normal Random Variable, Given a Probability 1. Draw pictures of the normal distribution in question and of the standard normal distribution. 3. From the table of the standard normal distribution, find the z value or values. 2. Shade the area corresponding to the desired probability. z .05 .06 .07 . . . . . . . . . . . . . . . 1.8 . . . 0.4678 0.4686 0.4693 1.9 . . . 0.4744 0.4750 0.4756 2.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . . -1.96 1.96

  33. N o r m a l D i s t r i b u t i o n :  = 2 4 5 0 ,  = 4 0 0 0 . 0 0 1 2 . .4750 .4750 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . .9500 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 X S t a n d a r d N o r m a l D i s t r i b u t i o n 0 . 4 .4750 .4750 0 . 3 ) z ( 0 . 2 f 0 . 1 .9500 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Finding Values of a Normal Random Variable, Given a Probability 1. Draw pictures of the normal distribution in question and of the standard normal distribution. 3. From the table of the standard normal distribution, find the z value or values. 2. Shade the area corresponding to the desired probability. 4. Use the transformation from z to x to get value(s) of the original random variable. z .05 .06 .07 . . . . . . . . . . . . . . . 1.8 . . . 0.4678 0.4686 0.4693 1.9 . . . 0.4744 0.4750 0.4756 2.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . . x =   z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234) -1.96 1.96

  34. 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 N o r m a l D i s t r i b u t i o n :  = 3 . 5 ,  = 1 . 3 2 3 B i n o m i a l D i s t r i b u t i o n : n = 7 , p = 0 . 5 0 0 . 3 0 . 3 P( x 4) = 0.7734 0 . 2 0 . 2 ) ) x x ( ( P f 0 . 1 0 . 1 0 . 0 0 . 0 0 5 1 0 0 1 2 3 4 5 6 7 X X 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 4.5; 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 Finding Values of a Normal Random Variable, Given a Probability

  35. 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 B i n o m i a l D i s t r i b u t i o n : n = 1 1 , p = 0 . 5 0 P(x  4) = 0.2744 N o r m a l D i s t r i b u t i o n :  = 5 . 5 ,  = 1 . 6 5 8 3 0 . 3 0 . 2 0 . 2 ) x ( ) P x ( f 0 . 1 0 . 1 0 . 0 0 . 0 0 1 2 3 4 5 6 7 8 9 1 0 1 1 0 5 1 0 X X 4-7 The Normal Approximation of Binomial Distribution

  36. - - æ ö a np b np £ £ = £ £ P ( a X b ) P Z ç ÷ & è ø - - np ( 1 p ) np ( 1 p ) ³ for n large (n 50) and p not too c lose to 0 or 1.00 - - + - æ ö a 0 . 5 np b 0 . 5 np £ £ = £ £ P ( a X b ) P Z ç ÷ & - - è ø np ( 1 p ) np ( 1 p ) £ for n moderatel y large (2 0 n < 50). Approximating a Binomial Probability Using the Normal Distribution or: NOTE: If p is either small (close to 0) or large (close to 1), use the Poisson approximation.

  37. Using the Template for Normal Approximation of the Binomial Distribution

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