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The normal distribution

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Name                                       Shakeel Nouman
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The normal distribution

  1. 1. The Normal Distribution Slide 1 Shakeel Nouman M.Phil Statistics The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  2. 2. 4        Slide 2 The Normal Distribution Using Statistics The Normal Probability Distribution The Standard Normal Distribution The Transformation of Normal Random Variables The Inverse Transformation The Normal Distribution as an Approximation to Other Probability Distributions Summary and Review of Terms The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  3. 3. 4-1 Introduction Slide 3 As n increases, the binomial distribution approaches a ... n=6 n = 10 Bino mial Dis trib utio n: n=6, p =.5 n = 14 Bino mial Distrib utio n: n=1 0 , p =.5 Bino mial Dis trib utio n: n=1 4 , p =.5 0.3 0.2 0.2 0.2 0.1 P(x) 0.3 P(x) P(x) 0.3 0.1 0.0 0.1 0.0 0 1 2 3 4 5 6 0.0 0 1 x 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x x Normal Probability Density Function: f ( x)  1      x 2 2p where e  2.7182818... and p  314159265... . 0.4 0.3 f(x) x  2  e 2 2 for      Normal Distribution:  = 0,= 1 0.2 0.1 0.0 -5 0 x The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 5
  4. 4. The Normal Probability Distribution Slide 4 The normal probability density function: 1 e 0.4 x  2     2 2 0.3 for  x 2p 2 where e  2.7182818... and p  314159265... . f(x) f ( x)        Normal Dis tribution:  = 0,= 1 0.2 0.1 0.0 -5 0 x The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 5
  5. 5. Properties of the Normal Probability Distribution • Slide 5 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 Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  6. 6. Properties of the Normal Probability Distribution (continued) • • • Slide 6 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). The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  7. 7. Properties of the Normal Probability Distribution (continued) • • • • Slide 7 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. The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  8. 8. Properties of the Normal Probability Distribution – Example 4-1 Slide 8 Example 4.1: Let X1, X2, and X3 be independent random variables that are normally distributed with means and variances as shown. Mean Variance X1 10 1 X2 20 2 X3 30 3 Let S = X1 + X2 + X3. Then E(S) = 10 + 20 + 30 = 60 and V(S) = 1 + 2 + 3 = 6. The standard deviation of S 6 is = 2.45. The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  9. 9. Properties of the Normal Probability Distribution (continued) • • • • Slide 9 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. The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  10. 10. Properties of the Normal Probability Distribution – Example 4-3 Slide 10 Example 4.3: Let X1 , X2 , X3 and X4 be 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 Mean Variance X1 12 4 X2 -5 2 X3 8 5 X4 10 1 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) = 73  8.544 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  11. 11. Computing the Mean, VarianceSlide 11 and Standard Deviation for the Sum of Independent Random Variables Using the Template The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  12. 12. Normal Probability Distributions Slide 12 All of these are normal probability density functions, though each has a different mean and variance. Normal Distribution: =40,  =1 Normal Distribution: =30,  =5 0.4 Normal Distribution: =50,  =3 0.2 0.2 0.2 f(y) f(x) f(w) 0.3 0.1 0.1 0.1 0.0 0.0 35 40 45 0.0 0 w 10 20 30 40 50 x W~N(40,1) X~N(30,25) 60 35 45 50 55 y Y~N(50,9) Normal Distribution:   =0, =1 Consider: 0.4 f(z) 0.3 0.2 0.1 0.0 -5 0 5 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 Z~N(0,1) The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 65
  13. 13. Computing Normal Probabilities Using the Template Slide 13 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  14. 14. 4-3 The Standard Normal Distribution Slide 14 The standard normal random variable, Z, is the normal random variable with mean = 0 and standard deviation = 1: Z~N(0,12). Standard Normal Distribution 0 .4  =1 { f( z) 0 .3 0 .2 0 .1 0 .0 -5 -4 -3 -2 -1 0 1 2 3 4 5  =0 Z The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  15. 15. Finding Probabilities of the Standard Normal Distribution: P(0 < Z < 1.56) Slide 15 Standard Normal Probabilities Standard Normal Distribution 0.4 f(z) 0.3 0.2 0.1 { 1.56 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Z Look in row labeled 1.5 and column labeled .06 to find P(0 z 1.56) = .4406 z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 .00 0.0000 0.0398 0.0793 0.1179 0.1554 0.1915 0.2257 0.2580 0.2881 0.3159 0.3413 0.3643 0.3849 0.4032 0.4192 0.4332 0.4452 0.4554 0.4641 0.4713 0.4772 0.4821 0.4861 0.4893 0.4918 0.4938 0.4953 0.4965 0.4974 0.4981 0.4987 .01 0.0040 0.0438 0.0832 0.1217 0.1591 0.1950 0.2291 0.2611 0.2910 0.3186 0.3438 0.3665 0.3869 0.4049 0.4207 0.4345 0.4463 0.4564 0.4649 0.4719 0.4778 0.4826 0.4864 0.4896 0.4920 0.4940 0.4955 0.4966 0.4975 0.4982 0.4987 .02 0.0080 0.0478 0.0871 0.1255 0.1628 0.1985 0.2324 0.2642 0.2939 0.3212 0.3461 0.3686 0.3888 0.4066 0.4222 0.4357 0.4474 0.4573 0.4656 0.4726 0.4783 0.4830 0.4868 0.4898 0.4922 0.4941 0.4956 0.4967 0.4976 0.4982 0.4987 .03 0.0120 0.0517 0.0910 0.1293 0.1664 0.2019 0.2357 0.2673 0.2967 0.3238 0.3485 0.3708 0.3907 0.4082 0.4236 0.4370 0.4484 0.4582 0.4664 0.4732 0.4788 0.4834 0.4871 0.4901 0.4925 0.4943 0.4957 0.4968 0.4977 0.4983 0.4988 .04 0.0160 0.0557 0.0948 0.1331 0.1700 0.2054 0.2389 0.2704 0.2995 0.3264 0.3508 0.3729 0.3925 0.4099 0.4251 0.4382 0.4495 0.4591 0.4671 0.4738 0.4793 0.4838 0.4875 0.4904 0.4927 0.4945 0.4959 0.4969 0.4977 0.4984 0.4988 .05 0.0199 0.0596 0.0987 0.1368 0.1736 0.2088 0.2422 0.2734 0.3023 0.3289 0.3531 0.3749 0.3944 0.4115 0.4265 0.4394 0.4505 0.4599 0.4678 0.4744 0.4798 0.4842 0.4878 0.4906 0.4929 0.4946 0.4960 0.4970 0.4978 0.4984 0.4989 .06 0.0239 0.0636 0.1026 0.1406 0.1772 0.2123 0.2454 0.2764 0.3051 0.3315 0.3554 0.3770 0.3962 0.4131 0.4279 0.4406 0.4515 0.4608 0.4686 0.4750 0.4803 0.4846 0.4881 0.4909 0.4931 0.4948 0.4961 0.4971 0.4979 0.4985 0.4989 .07 0.0279 0.0675 0.1064 0.1443 0.1808 0.2157 0.2486 0.2794 0.3078 0.3340 0.3577 0.3790 0.3980 0.4147 0.4292 0.4418 0.4525 0.4616 0.4693 0.4756 0.4808 0.4850 0.4884 0.4911 0.4932 0.4949 0.4962 0.4972 0.4979 0.4985 0.4989 .08 0.0319 0.0714 0.1103 0.1480 0.1844 0.2190 0.2517 0.2823 0.3106 0.3365 0.3599 0.3810 0.3997 0.4162 0.4306 0.4429 0.4535 0.4625 0.4699 0.4761 0.4812 0.4854 0.4887 0.4913 0.4934 0.4951 0.4963 0.4973 0.4980 0.4986 0.4990 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer .09 0.0359 0.0753 0.1141 0.1517 0.1879 0.2224 0.2549 0.2852 0.3133 0.3389 0.3621 0.3830 0.4015 0.4177 0.4319 0.4441 0.4545 0.4633 0.4706 0.4767 0.4817 0.4857 0.4890 0.4916 0.4936 0.4952 0.4964 0.4974 0.4981 0.4986 0.4990
  16. 16. Finding Probabilities of the Standard Normal Distribution: P(Z < -2.47) 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) z ... . . . 2.3 ... 2.4 ... 2.5 ... Slide 16 .06 .07 .08 . . . . . . . . . 0.4909 0.4911 0.4913 0.4931 0.4932 0.4934 0.4948 0.4949 0.4951 . . . = .5 - .4932 = 0.0068 Standard Normal Distribution 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 f(z) 0.3 0.2 0.1 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Z The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  17. 17. Finding Probabilities of the Standard Normal Distribution: P(1< Z < 2) 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) z .00 . . . 0.9 1.0 1.1 . . . 1.9 2.0 2.1  .9772  .8413  .1359 . . . . . . 0.3159 0.3413 0.3643 . . . 0.4713 0.4772 0.4821 . . . Slide 17 ... ... ... ... ... ... ... Standard Normal Distribution 0.4 Area between 1 and 2 P(1  Z  2)  .9772  .8413  0.1359 f(z) 0.3 0.2 0.1 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Z The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  18. 18. Finding Values of the Standard Normal Random Variable: P(0 < Z < z) = 0.40 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. z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 . . . .00 0.0000 0.0398 0.0793 0.1179 0.1554 0.1915 0.2257 0.2580 0.2881 0.3159 0.3413 0.3643 0.3849 0.4032 . . . .01 0.0040 0.0438 0.0832 0.1217 0.1591 0.1950 0.2291 0.2611 0.2910 0.3186 0.3438 0.3665 0.3869 0.4049 . . . .02 0.0080 0.0478 0.0871 0.1255 0.1628 0.1985 0.2324 0.2642 0.2939 0.3212 0.3461 0.3686 0.3888 0.4066 . . . 2. Then determine the value of z from the corresponding row and column. .03 0.0120 0.0517 0.0910 0.1293 0.1664 0.2019 0.2357 0.2673 0.2967 0.3238 0.3485 0.3708 0.3907 0.4082 . . . .04 0.0160 0.0557 0.0948 0.1331 0.1700 0.2054 0.2389 0.2704 0.2995 0.3264 0.3508 0.3729 0.3925 0.4099 . . . .05 0.0199 0.0596 0.0987 0.1368 0.1736 0.2088 0.2422 0.2734 0.3023 0.3289 0.3531 0.3749 0.3944 0.4115 . . . .06 0.0239 0.0636 0.1026 0.1406 0.1772 0.2123 0.2454 0.2764 0.3051 0.3315 0.3554 0.3770 0.3962 0.4131 . . . Slide 18 .07 0.0279 0.0675 0.1064 0.1443 0.1808 0.2157 0.2486 0.2794 0.3078 0.3340 0.3577 0.3790 0.3980 0.4147 . . . .08 0.0319 0.0714 0.1103 0.1480 0.1844 0.2190 0.2517 0.2823 0.3106 0.3365 0.3599 0.3810 0.3997 0.4162 . . . Standard Normal Distribution 0.4 Area to the left of 0 = .50 P(0 Z 1.28)  .40P(z 0) = .50 f(z) Also, since P(Z 0) = .50 Area = .40 (.3997) 0.3 0.2 0.1 P(Z 1.28)  .90 0.0 -5 -4 -3 -2 -1 0 Z 1 2 3 4 5 Z = 1.28 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer .09 0.0359 0.0753 0.1141 0.1517 0.1879 0.2224 0.2549 0.2852 0.3133 0.3389 0.3621 0.3830 0.4015 0.4177 . . .
  19. 19. 99% Interval around the Mean 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 z .04 . . . . 2.4 ... 2.5 ... 2.6 ... . . . .05 .08 . . . 0.4931 0.4948 0.4961 . . . .09 . . . 0.4932 0.4949 0.4962 . . . . . . 0.4934 0.4951 0.4963 . . . . . 0.4936 0.4952 0.4964 . . . Total area in center = .99 Area in center left = .495 0.4 Area in center right = .495 0.3 f(z) P(-.2575   .99 Z )= .07 . . . 0.4929 0.4946 0.4960 . . . Look to the table of standard normal probabilities to find that:    z.005 z.005  .06 . . . 0.4927 0.4945 0.4959 . . . Slide 19 0.2 Area in right tail = .005 Area in left tail = .005 0.1 0.0 -5 -4 -3 -2 -z.005 -2.575 -1 0 Z 1 2 3 4 5 z.005 2.575 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  20. 20. 4-4 The Transformation of Normal Random Variables Slide 20 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: X x Z x Normal Distribution:=50, =10 0.07 0.06 Transformation f(x) (1) Subtraction: (X -  ) x 0.05 0.04 0.03  10 = { 0.02 Standard Normal Distribution 0.01 0.00 0.4 0 20 30 40 50 60 70 80 90 100 X 0.3 0.2 (2) Division by  ) x { f(z) 10 1.0 0.1 0.0 -5 -4 -3 -2 -1 0 Z 1 2 3 4 5 The inverse transformation of Z to X: X  x + Z x The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  21. 21. Using the Normal Transformation Example 4-9 X~N(160,302) Slide 21 Example 4-10 X~N(127,222) P (100  X  180)  100    X    180     P  P ( X  150)  X    150     P          100  160  Z  180  160 P   30 30  ( )  P 2  Z  .6667  0.4772 + 0.2475  0.7247      150  127  P Z    22  ( )  P Z  1.045  0.5 + 0.3520  0.8520 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  22. 22. Using the Normal Transformation - Example 4-11 Normal Dis tribution:  = 383, = 12 Example 4-11 X~N(383,122) 0.05 0.04 (   399  383 ) 12  P 0.9166  Z  1.333  0.4088  0.3203  0.0885   0.03 0.02 0.01 Standard Normal Distribution 0.00 340 0.4 390 X 0.3 f(z)  f( ) X P ( 394  X  399)  394   X   399     P        394  383 P Z   12 Slide 22 0.2 0.1 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Z Template solution The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 440
  23. 23. The Transformation of Normal Random Variables The transformation of X to Z: Z  X  x x Slide 23 The inverse transformation of Z to X: X   + Z x x The transformation of X to Z, where a and b are numbers:: a    P( X  a)  P Z      b    P( X  b)  P Z      b   a  P(a  X  b)  P Z      The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  24. 24. Normal Rule) Probabilities (Empirical S t a n d a rd N o rm a l D is trib u tio n • The probability that a normal • • 0 .4 0 .3 f(z) 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. Slide 24 0 .2 0 .1 0 .0 -5 -4 -3 -2 -1 0 1 2 3 Z The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 4 5
  25. 25. 4-5 The Inverse Transformation Slide 25 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), 70  50   x   70     P( X  70)  P    P Z    P( Z  2)      10  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. Normal Distribution:  = 124, = 12 Example 4-12 X~N(124,122) P(X > x) = 0.10 and P(Z > 1.28)  0.10 x = + z= 124 + (1.28)(12) = 139.36 . . . 1.1 1.2 1.3 . . . .07 . . . 0.3790 0.3980 0.4147 . . . ... ... ... . . . . . . .08 . . . 0.3810 0.3997 0.4162 . . . .09 . . . 0.3830 0.4015 0.4177 . . . 0.03 f(x) z 0.04 0.02 0.01 0.01 0.00 80 130 X 139.36 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 180
  26. 26. Template Solution for Example 4-12 Slide 26 Example 4-12 X~N(124,122) P(X > x) = 0.10 and P(Z > 1.28)  0.10 x = + z= 124 + (1.28)(12) = 139.36 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  27. 27. The Inverse Transformation (Continued) Example 4-13 X~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 z . . . 2.2 2.3 2.4 . . . .02 . . . 0.4868 0.4898 0.4922 . . . ... ... ... . . . .03 . . . 0.4871 0.4901 0.4925 . . . . . . Example 4-14 X~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 .04 . . . 0.4875 0.4904 0.4927 z . . . 1.8 1.9 2.0 . . . . Normal Distribution: = 5.7  0.5 = ... ... ... . . .06 . . . 0.4686 0.4750 0.4803 . . . 0.0015 Area = 0.49 .4750 .4750 0.0010 f(x) 0.5 0.4 X.01 =  +z= 5.7 + (2.33)(0.5) = 6.865 0.3 0.0005 0.2 .0250 .0250 Area = 0.01 0.1 0.0 0.0000 3.2 4.2 5.2 6.2 7.2 8.2 1000 2000 X -5 -4 -3 -2 -1 0 z 3000 4000 X 1 2 3 4 5 Z.01 = 2.33 -5 -4 -3 -2 -1.96 -1 0 Z 1 2 3 .07 . . . 0.4693 0.4756 0.4808 . . Normal Distribution:  = 2450 = 400 0.6 f(x) .05 . . . 0.4678 0.4744 0.4798 . . . . . 0.8 0.7 Slide 27 4 5 1.96 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  28. 28. Finding Values of a Normal Random Variable, Given a Probability Norm al Distribution:  = 24 50, = 40 0 0.0012 . 0.0010 . 0.0008 . f(x) 1. Draw pictures of the normal distribution in question and of the standard normal distribution. Slide 28 0.0006 . 0.0004 . 0.0002 . 0.0000 1000 2000 3000 4000 X S ta nd a rd N o rm al D is trib utio n 0.4 f(z) 0.3 0.2 0.1 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Z The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  29. 29. Finding Values of a Normal Random Variable, Given a Probability Norm al Distribution:  = 24 50, = 400 0.0012 . .4750 0.0010 . .4750 0.0008 . f(x) 1. Draw pictures of the normal distribution in question and of the standard normal distribution. Slide 29 0.0006 . 0.0004 . 0.0002 . .9500 0.0000 1000 2000 3000 4000 X 2. Shade the area corresponding to the desired probability. S ta nd a rd No rm al D is trib utio n 0.4 .4750 .4750 f(z) 0.3 0.2 0.1 .9500 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Z The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  30. 30. Finding Values of a Normal Random Variable, Given a Probability Norm al Distribution:  = 2450, = 400 3. From the table of the standard normal distribution, find the z value or values. 0.0012 . .4750 0.0010 . .4750 0.0008 . f(x) 1. Draw pictures of the normal distribution in question and of the standard normal distribution. Slide 30 0.0006 . 0.0004 . 0.0002 . .9500 0.0000 1000 2000 3000 4000 X 2. Shade the area corresponding to the desired probability. S ta nd a rd No rm al D is trib utio n 0.4 .4750 f(z) z . . . 1.8 1.9 2.0 . . .05 . . . 0.4678 0.4744 0.4798 . . . . ... ... ... . . . .06 . . . 0.4686 0.4750 0.4803 . . .4750 0.3 .07 . . . 0.4693 0.4756 0.4808 . 0.2 0.1 .9500 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Z . -1.96 1.96 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  31. 31. Finding Values of a Normal Random Variable, Given a Probability Norm al Distribution:  = 24 50, = 400 3. From the table of the standard normal distribution, find the z value or values. 0.0012 . .4750 0.0010 . .4750 0.0008 . f(x) 1. Draw pictures of the normal distribution in question and of the standard normal distribution. Slide 31 0.0006 . 0.0004 . 0.0002 . .9500 0.0000 1000 2000 3000 4000 X 2. Shade the area corresponding to the desired probability. 0.4 .4750 . . . ... ... ... . . .05 . . . 0.4678 0.4744 0.4798 . . .06 . . . 0.4686 0.4750 0.4803 . . .4750 0.3 f(z) z . . . 1.8 1.9 2.0 . . 4. Use the transformation from z to x to get value(s) of the original random variable. S ta nd a rd No rm al D is trib utio n .07 . . . 0.4693 0.4756 0.4808 . . 0.2 0.1 .9500 0.0 -5 -4 -3 -2 -1 0 1 2 Z -1.96 3 4 5 x = z= 2450 (1.96)(400) = 2450 784=(1666,3234) 1.96 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  32. 32. Finding Values of a Normal Random Variable, Given a Probability Slide 32 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 Normal Distribution:  = 3.5, = 1.323 Binomial Distribution: n = 7, p = 0.50 0.3 0.3 P( x  = 0.7734 4) 0.2 f(x) P(x) 0.2 0.1 0.1 0.0 0.0 0 5 10 X 0 1 2 3 4 5 6 7 X MTB > cdf 4.5; SUBC> normal 3.5 1.323. Cumulative Distribution Function MTB > cdf 4; SUBC> binomial 7,.5. Cumulative Distribution Function Normal with mean = 3.50000 and standard deviation = 1.32300 Binomial with n = 7 and p = 0.500000 x P( X <= x) 4.5000 0.7751 x P( X <= x) 4.00 0.7734 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  33. 33. 4-6 The Normal Approximation of Binomial Distribution Slide 33 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 Normal Distribution: = 5.5, = 1.6583 Binomial Distribution: n = 11, p = 0.50 P(x 4) = 0.2744 0.3 0.2 f(x) P(x) 0.2 0.1 0.1 0.0 0.0 0 5 10 X 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 0 1 2 3 4 5 6 7 8 9 10 11 X 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 By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  34. 34. Approximating a Binomial Probability Using the Normal Distribution Slide 34 b  np   a  np P ( a  X  b)  P  Z    np(1  p) np(1  p)  for n large (n  50) and p not too close to 0 or 1.00 or: b + 0.5  np   a  0.5  np P ( a  X  b)  P  Z   np(1  p)   np(1  p) for n moderately large (20  n < 50). If p is either small (close to 0) or large (close to 1), use the Poisson approximation. The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  35. 35. Using the Template for Normal Approximation of the Binomial Distribution Slide 35 The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  36. 36. Slide 36 Name Religion Domicile Contact # E.Mail M.Phil (Statistics) Shakeel Nouman Christian Punjab (Lahore) 0332-4462527. 0321-9898767 sn_gcu@yahoo.com sn_gcu@hotmail.com GC University, . (Degree awarded by GC University) M.Sc (Statistics) Statitical Officer (BS-17) (Economics & Marketing Division) GC University, . (Degree awarded by GC University) Livestock Production Research Institute Bahadurnagar (Okara), Livestock & Dairy Development Department, Govt. of Punjab The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

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