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NORMAL
       DISTRIBUTION

(A Continuous Probability Distribution)
Normal Distribution
    In probability theory, the normal (or Gaussian) distribution is a
        continuous probability distribution. The graph of the
associated probability density function is "bell"-shaped, and is known as
the Gaussian function or bell curve
Definition: A continuous random variable X is said to follow Normal
Distribution if its probability density function is given as




                                                       -∞ < x < ∞
where parameter μ is the mean (location of the peak) and
σ  2 is the variance (the measure of the width of the distribution).

                      X ~ N (μ, σ  2 )
Area property of Normal Distribution
The Standard Normal Distribution:
If X is a normal variable with mean μ and variance σ 2 then
  the variable z = (x- μ )/σ is called standard normal
  variable A standard normal distribution is a normal
  distribution with zero mean (          ) and unit variance (
          ), given by the probability density function

               P(Z=z) =

The distribution of            with μ = 0 and σ 2 = 1 is called the
  standard normal.

•      If             X ~ N (μ, σ  2 )



                          ~ N ( 0, 1 )
For normal variable infinite values of mean and variance
  are possible that’s why it is to be standardize…
Properties of Normal Curve:
The normal curve with mean µ and standard deviation σ has following
     properties
1.   The equation of the curve is




2. The curve is symmetric about the line x = µ and x ranges from -∞ to
     +∞.
3.Mean, mode and median coincide at x = µ as distribution is symmetrical.

4.The points of inflection of the curve are x = µ + σ and x = µ - σ.

5. In normal distribution Q.D.: M.D. : S.D. = 10 : 12 : 15

6.The M.D. about mean = 4/5 * S.D.

7.The normal curve has single peak i.e. unimodal.
8.Area property of Normal Distribution
Area property of standard normal distribution
• Standard Normal Distribution Table




•   Z     0.00      0.01      0.02    0.03     0.04      0.05    0.06    0.07     0.08      0.09
•   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
•   Z     0.00      0.01      0.02    0.03      0.04       0.05     0.06   0.07     0.08      0.09
•   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
                                              *************************
Normal Probability:
1] As normal distribution is continuous distribution,
   probability at a point is zero i.e. P ( z = a ) = 0

2] As the standard normal distribution is symmetric about
  mean (=0).     P (z < 0 ) = P ( z > 0 ) = ½

3] P ( 0 < z < a) then table value for a gives the required
  probability.

4] P (-a < z < 0) = P ( 0 < z < a) as the curve is symmetric
  about 0.

5] P( z < -a) = P ( z > a) = P ( z > 0 ) - P ( 0 < z < a)
• 1]Question: What is the relative frequency of observations
  below 1.18?
• That is, find the relative frequency of the event Z < 1.18. (Here
  small z is 1.18.) (Or P(Z < 1.18 )




• P( z < 1.18) = P( z < 0 ) + P ( 0 < z < 1.18)
•              = 0.5 + 0.3810
•              = 0.8810
2] P(z<-0.63) = P(z>0.63) = P(z>0) – P(0<z<063) = 0.2643




3] P(z>-1.48) = P(z>0) + P(0<z<1.48) = 0.9306
4] P( -1.65 < z < 1.65 ) = 2 P ( 0< z < 1.65 )




5] P(z<0.84) = 0.8000
• Normal tables can be of this type P ( z < a )
• z.      00 .01....                            08.      09


•   0.0   .5000 .5040....                      .5319    .5359
•   0.1   .5398 .5438....                      .5714    .5753
•   |
•   |
•   1.0   .8413 .8438...                        .8599    .8621
•   1.1   .8643 .8665...                       .8810    .8830
•   1.2   .8849 .8869...                       .8997    .9015
• Example 1
  An average light bulb manufactured by the Acme
  Corporation lasts 300 days with a standard deviation of
  50 days. Assuming that bulb life is normally distributed,
  what is the probability that an Acme light bulb will last
  at most 365 days?

• Example 2
  Suppose scores on an IQ test are normally distributed.
  If the test has a mean of 100 and a standard deviation
  of 10, what is the probability that a person who takes
  the test will score between 90 and 110?
 [P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )
  P( 90 < X < 110 ) = 0.84 - 0.16
  P( 90 < X < 110 ) = 0.68
• Thus, about 68% of the test scores will fall between 90
  and 110.]
• Problem 3
• Molly earned a score of 940 on a national achievement
  test. The mean test score was 850 with a standard
  deviation of 100. What proportion of students had a
  higher score than Molly? (Assume that test scores are
  normally distributed.)
• [z = (X - μ) / σ = (940 - 850) / 100 = 0.90


• we find P(Z < 0.90) = 0.8159.
  Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159
  = 0.1841. Thus, we estimate that 18.41 percent of the
  students tested had a higher score than Molly. ]
4.The Acme Light Bulb Company has found that an average light bulb
   lasts 1000 hours with a standard deviation of 100 hours. Assume
   that bulb life is normally distributed. What is the probability that
   a randomly selected light bulb will burn out in 1200 hours or less?

5. Bill claims that he can do more push-ups than 90% of the boys in
   his school. Last year, the average boy did 50 push-ups, with a
   standard deviation of 10 pushups. Assume push-up performance is
   normally distributed. How many pushups would Bill have to do to
   beat 90% of the other boys?


                           ***************
• Example
• You and your friends have just measured the heights of your dogs
  (in millimeters):




• The heights (at the shoulders) are: 600mm, 470mm, 170mm,
  430mm and 300mm.
• Find out the Mean, the Variance, and the Standard Deviation.
• Your first step is to find the Mean:
• Answer:
• Mean =  [600 + 470 + 170 + 430 + 300]/5  =  1970/5  = 394
•




• so the mean (average) height is 394 mm. Let's plot this on the
  chart:

• Now, we calculate each dogs difference from the Mean:
•   To calculate the Variance, take each difference, square it, and then average the
    result:
•   Variance: σ2 =  [2062 + 762 + (-224)2 + 362 + (-94)2 ]/5 =  108,520/5  = 21,704
•     So, the Variance is 21,704.
•   And the Standard Deviation is just the square root of Variance, so:
•   Standard Deviation: σ = √21,704 = 147.32... = 147 (to the nearest mm)
•    
•   And the good thing about the Standard Deviation is that it is useful. Now we can
    show which heights are within one Standard Deviation (147mm) of the Mean:




•   So, using the Standard Deviation we have a "standard" way of knowing what is
    normal, and what is extra large or extra small.
• Area under the Normal Curve from 0 to X
X 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.00000 0.00399 0.00798 0.01197 0.01595 0.01994 0.02392 0.02790 0.03188 0.03586
0.1 0.03983 0.04380 0.04776 0.05172 0.05567 0.05962 0.06356 0.06749 0.07142 0.07535
0.2 0.07926 0.08317 0.08706 0.09095 0.09483 0.09871 0.10257 0.10642 0.11026 0.114090

0.3 0.11791 0.12172 0.12552 0.12930 0.13307 0.13683 0.14058 0.14431 0.14803 0.15173
0.4 0.15542 0.15910 0.16276 0.16640 0.17003 0.17364 0.17724 0.18082 0.18439 0.18793
0.5 0.19146 0.19497 0.19847 0.20194 0.20540 0.20884 0.21226 0.21566 0.21904 0.22240
0.6 0.22575 0.22907 0.23237 0.23565 0.23891 0.24215 0.24537 0.24857 0.25175 0.25490
0.7 0.25804 0.26115 0.26424 0.26730 0.27035 0.27337 0.27637 0.27935 0.28230 0.28524
0.8 0.28814 0.29103 0.29389 0.29673 0.29955 0.30234 0.30511 0.30785 0.31057 0.31327
0.9 0.31594 0.31859 0.32121 0.32381 0.32639 0.32894 0.33147 0.33398 0.33646 0.33891
1.0 0.34134 0.34375 0.34614 0.34849 0.35083 0.35314 0.35543 0.35769 0.35993 0.36214
1.1 0.36433 0.36650 0.36864 0.37076 0.37286 0.37493 0.37698 0.37900 0.38100 0.38298
1.2 0.38493 0.38686 0.38877 0.39065 0.39251 0.39435 0.39617 0.39796 0.39973 0.40147
1.3 0.40320 0.40490 0.40658 0.40824 0.40988 0.41149 0.41308 0.41466 0.41621 0.41774
Normal distri
Normal distri
Normal distri
Normal distri

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Normal distri

  • 1. NORMAL DISTRIBUTION (A Continuous Probability Distribution)
  • 2. Normal Distribution In probability theory, the normal (or Gaussian) distribution is a continuous probability distribution. The graph of the associated probability density function is "bell"-shaped, and is known as the Gaussian function or bell curve Definition: A continuous random variable X is said to follow Normal Distribution if its probability density function is given as -∞ < x < ∞ where parameter μ is the mean (location of the peak) and σ  2 is the variance (the measure of the width of the distribution). X ~ N (μ, σ  2 )
  • 3. Area property of Normal Distribution
  • 4. The Standard Normal Distribution: If X is a normal variable with mean μ and variance σ 2 then the variable z = (x- μ )/σ is called standard normal variable A standard normal distribution is a normal distribution with zero mean ( ) and unit variance ( ), given by the probability density function P(Z=z) = The distribution of with μ = 0 and σ 2 = 1 is called the standard normal. • If X ~ N (μ, σ  2 ) ~ N ( 0, 1 )
  • 5. For normal variable infinite values of mean and variance are possible that’s why it is to be standardize…
  • 6. Properties of Normal Curve: The normal curve with mean µ and standard deviation σ has following properties 1. The equation of the curve is 2. The curve is symmetric about the line x = µ and x ranges from -∞ to +∞. 3.Mean, mode and median coincide at x = µ as distribution is symmetrical. 4.The points of inflection of the curve are x = µ + σ and x = µ - σ. 5. In normal distribution Q.D.: M.D. : S.D. = 10 : 12 : 15 6.The M.D. about mean = 4/5 * S.D. 7.The normal curve has single peak i.e. unimodal.
  • 7. 8.Area property of Normal Distribution
  • 8. Area property of standard normal distribution
  • 9. • Standard Normal Distribution Table • Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 • 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
  • 10. Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 • 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 *************************
  • 11. Normal Probability: 1] As normal distribution is continuous distribution, probability at a point is zero i.e. P ( z = a ) = 0 2] As the standard normal distribution is symmetric about mean (=0). P (z < 0 ) = P ( z > 0 ) = ½ 3] P ( 0 < z < a) then table value for a gives the required probability. 4] P (-a < z < 0) = P ( 0 < z < a) as the curve is symmetric about 0. 5] P( z < -a) = P ( z > a) = P ( z > 0 ) - P ( 0 < z < a)
  • 12. • 1]Question: What is the relative frequency of observations below 1.18? • That is, find the relative frequency of the event Z < 1.18. (Here small z is 1.18.) (Or P(Z < 1.18 ) • P( z < 1.18) = P( z < 0 ) + P ( 0 < z < 1.18) • = 0.5 + 0.3810 • = 0.8810
  • 13. 2] P(z<-0.63) = P(z>0.63) = P(z>0) – P(0<z<063) = 0.2643 3] P(z>-1.48) = P(z>0) + P(0<z<1.48) = 0.9306
  • 14. 4] P( -1.65 < z < 1.65 ) = 2 P ( 0< z < 1.65 ) 5] P(z<0.84) = 0.8000
  • 15. • Normal tables can be of this type P ( z < a ) • z. 00 .01.... 08. 09 • 0.0 .5000 .5040.... .5319 .5359 • 0.1 .5398 .5438.... .5714 .5753 • | • | • 1.0 .8413 .8438... .8599 .8621 • 1.1 .8643 .8665... .8810 .8830 • 1.2 .8849 .8869... .8997 .9015
  • 16. • Example 1 An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days? • Example 2 Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110? [P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 ) P( 90 < X < 110 ) = 0.84 - 0.16 P( 90 < X < 110 ) = 0.68 • Thus, about 68% of the test scores will fall between 90 and 110.]
  • 17. • Problem 3 • Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.) • [z = (X - μ) / σ = (940 - 850) / 100 = 0.90 • we find P(Z < 0.90) = 0.8159. Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159 = 0.1841. Thus, we estimate that 18.41 percent of the students tested had a higher score than Molly. ]
  • 18. 4.The Acme Light Bulb Company has found that an average light bulb lasts 1000 hours with a standard deviation of 100 hours. Assume that bulb life is normally distributed. What is the probability that a randomly selected light bulb will burn out in 1200 hours or less? 5. Bill claims that he can do more push-ups than 90% of the boys in his school. Last year, the average boy did 50 push-ups, with a standard deviation of 10 pushups. Assume push-up performance is normally distributed. How many pushups would Bill have to do to beat 90% of the other boys? ***************
  • 19. • Example • You and your friends have just measured the heights of your dogs (in millimeters): • The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm. • Find out the Mean, the Variance, and the Standard Deviation. • Your first step is to find the Mean:
  • 20. • Answer: • Mean =  [600 + 470 + 170 + 430 + 300]/5  =  1970/5  = 394 • • so the mean (average) height is 394 mm. Let's plot this on the chart: • Now, we calculate each dogs difference from the Mean:
  • 21. To calculate the Variance, take each difference, square it, and then average the result: • Variance: σ2 =  [2062 + 762 + (-224)2 + 362 + (-94)2 ]/5 =  108,520/5  = 21,704 • So, the Variance is 21,704. • And the Standard Deviation is just the square root of Variance, so: • Standard Deviation: σ = √21,704 = 147.32... = 147 (to the nearest mm) •   • And the good thing about the Standard Deviation is that it is useful. Now we can show which heights are within one Standard Deviation (147mm) of the Mean: • So, using the Standard Deviation we have a "standard" way of knowing what is normal, and what is extra large or extra small.
  • 22. • Area under the Normal Curve from 0 to X X 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.00000 0.00399 0.00798 0.01197 0.01595 0.01994 0.02392 0.02790 0.03188 0.03586 0.1 0.03983 0.04380 0.04776 0.05172 0.05567 0.05962 0.06356 0.06749 0.07142 0.07535 0.2 0.07926 0.08317 0.08706 0.09095 0.09483 0.09871 0.10257 0.10642 0.11026 0.114090 0.3 0.11791 0.12172 0.12552 0.12930 0.13307 0.13683 0.14058 0.14431 0.14803 0.15173 0.4 0.15542 0.15910 0.16276 0.16640 0.17003 0.17364 0.17724 0.18082 0.18439 0.18793 0.5 0.19146 0.19497 0.19847 0.20194 0.20540 0.20884 0.21226 0.21566 0.21904 0.22240 0.6 0.22575 0.22907 0.23237 0.23565 0.23891 0.24215 0.24537 0.24857 0.25175 0.25490 0.7 0.25804 0.26115 0.26424 0.26730 0.27035 0.27337 0.27637 0.27935 0.28230 0.28524 0.8 0.28814 0.29103 0.29389 0.29673 0.29955 0.30234 0.30511 0.30785 0.31057 0.31327 0.9 0.31594 0.31859 0.32121 0.32381 0.32639 0.32894 0.33147 0.33398 0.33646 0.33891 1.0 0.34134 0.34375 0.34614 0.34849 0.35083 0.35314 0.35543 0.35769 0.35993 0.36214 1.1 0.36433 0.36650 0.36864 0.37076 0.37286 0.37493 0.37698 0.37900 0.38100 0.38298 1.2 0.38493 0.38686 0.38877 0.39065 0.39251 0.39435 0.39617 0.39796 0.39973 0.40147 1.3 0.40320 0.40490 0.40658 0.40824 0.40988 0.41149 0.41308 0.41466 0.41621 0.41774