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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 1
Elementary Statistics
M. Ghamsary, Ph.D.
Chapter 05
Normal Distribution
1
Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 2
Normal Distribution
One of the most common continuous distributions in the world of statistics is normal distribution. It is
also called Gaussian distribution. It has the following characteristics:
1. Symmetric about the mean and it has Bell shape
2. The mean, median, and mode are all the same and located at the center of the distribution.
3.The total area under the curve is equal to 100% or 1.
4. The area under the curve of normal distribution represent the probability.
So P[ a< X < b]= The shaded area in the following:
 1   x   2 
exp   
4. It has the following density: f ( x;  , ) 
 ,
2
 2  
2 


1
where,
2

is the mean

is the standard deviation
Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 3
Standard Normal Distribution:
If a normal distribution has
  0 and   1, then it is called standard normal distribution. For
standard normal distribution we have a table to calculate the probability.
Example1: Find P[ Z <1.20].
P( Z < 1.20) = .8849
Which means that 88.49% of all z-scores
are less than 1.20.
And, 11.51% of all z-scores are greater than 1.20.
If you want area to the RIGHT of a z-score you must either:
Look up the given Z-score and subtract the shown area from 1.00
OR
Multiply the given Z-score by -1, then look up that (new) Z-score. Since the curve is symmetric
the area to the right of Z* will be the same as the area to the left of -Z*.
Example2: P(Z > 1.20 )
P(Z > 1.20 ) = 1 - .8849 = .1151
So, 11.51% of all z-scores are
greater than 1.20
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Elementary Statistics
Chapter 5
Normal Distribution
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P(Z > 1.20) = P (Z < -1.20) = .1151
and 11.51% of all z-scores are
less than -1.20. This means that 11.51%
of all Z-scores are also greater than +1.20
Example3A: Find P[ -1.20< Z <0].
P[ -1.20< Z <0]= 0.50- 0.1151=0.3849
Example3B: Find P[-1.20< Z <1.20].
P[-1.20< Z <1.20]= P[ Z <1.20]- P[Z <-1.20]=0.8849-0.1151=0.7698
If you need to find the area between two Z-scores, you should look up the larger Z-score and record
the area to the left of this number. Then look up the smaller Z-score and again record the area to the
left. Finally subtract the smaller area from the larger area
Example4: Find P[ -0.81<Z <1.42].
P(-0.81 < Z < 1.42)
=
=
4
P (Z < 1.42)
-
-
P (Z < -0.81)
Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 5
Example5: Determine the area under the standard normal curve that lies between
a. –2.18 and 1.44
b. –2 and –1.5
c. 0.59 and 1.51
d. 1.1 and 4.2
e. –1.28 and 1.28.
Solution:
a. P (-2.18 < Z < 1.44) = P (Z < 1.44) – P (Z < -2.18) = 0.9251 – 0.0146 = 0.9105.
b. P (-2 < Z < -1.5) = P (Z < -1.5) – P (Z < -2) = 0.0668 – 0.0228 = 0.044.
c. P (0.59 < Z < 1.51) = P (Z < 1.51) – P (Z < 0.59) = 0.9345 – 0.7224 = 0.2121.
d. P (1.1 < Z < 4.2) = P (Z < 4.2) – P (Z < 1.1) = 1.0000 – 0.8643 = 0.1357.
e. P (-1.28 < Z < 1.28) = P (Z < 1.28) – P (Z < -1.28) = 0.8997 – 0.1003 = 0.7994.
Example6:
Use Table to find
a. Z0.025 and
b. Z0.05.
Solution:
(a) P (Z > z) = 0.025  P (Z < z) = 0.975  P (Z < 1.96) = 0.975.
Hence, Z0.025 = 1.96.
(b) P (Z > z) = 0.05  P (Z < z) = 0.95  P (Z < 1.645) = 0.95.
Hence, Z0.05= 1.645.
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Elementary Statistics
Chapter 5
Normal Distribution
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Example7: Find the valve of z with probability between 0 and z equal 0.4525.
Example8: Find the valve of z with probability above z equal 0.0041.
Example9: Find the valve of z with probability below z equal 0.0749.
Example10: Find the valve of z with probability below z equal 0.9222.
Example11: Find P95.
Example12: Find P5.
Example13: Find P50.
Example14: Find P35.
Example15: Find Q1.
Example16: Find D2.
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Elementary Statistics
Chapter 5
Normal Distribution
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Nonstandard Normal Distribution:
In order to find the probability in this case, we can convert any nonstandard normal random variable
to a standard normal one by using the formula:
z  x ,

where:
x : is the nonstandard score
 : is the mean
 : is the standard deviation.
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 8
Example17: Suppose the weight of new babies delivered at LLU is normally distributed with
mean of 3250 grams and standard deviation of 550 grams. What proportion of babies would the
normal distribution predict to weigh more than 4536 grams at birth?
Solution:
We have a distribution that is Normal with  =3250 grams and   550 grams.
We want P( X > 4536).
First convert 4536 to its Z-score: z 
x


Then, we have P  X  4536  P  Z  2.34
4536  3250
 2.34
550
Now use the table to draw the following picture:
Thus, P ( X > 4536) = P(Z > 2.34) = 1 -.9904 = .0096 = 0.96%. In other words, less than 1% of all
babies should weigh more than 4536 grams at birth.
Example18A- In Example 17, Determine the probability that a randomly selected baby weighs
between 3000 and 4000 grams at birth.
Solution:
We want P( 3000 < X < 4000 ). To find this, we convert both numbers to their Z-scores.
Then we find the area to the left of each Z-score and subtract the smaller area from the larger.
x   3000  3250
x   4000  3250

 -0.45 and X=4000: z 

 1.36
X=3000: z 

550

550
P  3000  X  4000  P  0.45  Z  1.36 =.9131 - .3264 = .5867.
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 9
Then, using the table,
P( Z< 1.36) = .9131 and Pr( Z<-.45) = .3264.
So, the probability we want is .9131 - .3264 = .5867.
So there is a 58.67% chance that a randomly
selected newborn will weigh between 3000 and 4000 grams.
Example18B- In Example 17 How little would a baby have to weigh to be among the lightest
2.5% of all newborns?
Solution:
To be among the lightest 2.5% = 0.025 of all newborns, a baby would have to have a z-score of -1.96.
To see this, realize that we want a Standard Normal curve with area 0.0250 to the left of our Z-score.
So read the table "backwards" that is look in the center (area portion) of the table for an area of
0.0250. Once you find it, read "backwards" to see the Z-score which is -1.96.
Then use the formula z 
x

 1.96 
x  3250
to determine that the babies weight (X)
550
would need to be 2172 grams: X = -1.96(550) + 3250=2172
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 10
Example18C- In Example 17 How much would a baby have to weigh to be among the heaviest
10% of all newborns?
Solution:
To be among the heaviest 10% = 0.10 of all newborns, a baby would have to have a z-score of +1.28.
To see this, realize that we want a Standard Normal curve
with area 0.9000 to the LEFT of our Z-score
(and an area of 0.1000 to the RIGHT).
So read the table "backwards" -- that is look in the center
(area portion) of the table for an area of 0.9000.
Once you find it, read "backwards" to see the Z-score
which is 1.28.
x
x  3250
to find that a baby would need to weigh

550
X = 1.28(550) + 3250=3954 grams in order to be among the heaviest 10% of all newborns.
Then use the formula z 
 1.28 
Example19A: Suppose X is normally distributed with the mean of   70 and standard deviation of
  8 , then find the following probabilities:
a. P [70 < X < 78]
b. P [60 < X < 80]
c. P [55 < X < 95]
d. P [76 < X< 88].
Solution:
a. to find P[70 < X < 78], first we sketch the normal curve and then we find the z score for each X as
follows:
X=70: z 
70  70
70  78
 0 , X=78: z 
1
8
8
So now use the table and we get:
P[70 < X < 78]= P[0 < Z < 1] =0.8414-0.50=0.3414.
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 11
b. to find P[60< X < 80], first we sketch the normal curve and then we find the z score for each X as
follows:
X=60: z 
60  70 10
80  70 10

 1.25 , X=80: z 

 1.25
8
8
8
8
So now use the table and we get:
P[60 < X < 80]= P[-1.25 < Z < 1.25] =0.8944-0.1056 = 0.7888
c. to find P[55< X < 95], first we sketch the normal curve and then we find the z score for each X as
follows:
X=55: z 
55  70 15
95  70 25

 1.875  1.88 , X=95: z 

 3.125  3.13
8
8
8
8
So now use the table and we get:
P[55 < X < 95]= P[-1.88 < Z < 3.13]= 0.8999-0.0301
= 0.9698.
d. to find P[76 < X< 88], first we sketch the normal curve and then we find the z score for each X as
follows:
X=76: z 
76  70 6
88  70 18
  0.75 , X=88: z 

 2.25
8
8
8
8
So now use the table and we get:
P[76 < X < 88]= P[0.75 < Z < 2.25] =0.9878- 0.7266 = 0.2612.
Example19B: Suppose X is normally distributed with the mean of   100 and standard deviation of
  15 , then find the following probabilities:
e. P [70 < X < 130]
f. P [120 < X < 132]
g. P [X > 95]
h. P [X< 88].
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Elementary Statistics
Chapter 5
Normal Distribution
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Example20A: Suppose the salary of professors at certain school is normally distributed with an
average of $70,000 and standard deviation of $10,000.
a. What percent of them are making above $90,000?
b. What percent of them are making below $50,000?
c. What percent of them are making between $60,000and 80,000?
Solution: As we did in example19, first we sketch the normal curve and then we find the z score for
each nonstandard score. We also have   $70,000 , and   $10,000 .
a. X= 90,000: z 
90,000  70,000
2
10,000
P[ X > 90,000]= P[ Z > 2] = 1-0.9772= 0.0228, which is about 2.3%.
b. X= 50,000: z 
50,000  70,000
 2
10,000
P[ X < 90,000]= P[ Z <- 2] =0.0228, which is again 2.3%.
c. Or we can use the table as follows:
X= 60,000: z 
60,000  70,000
 1
10,000
X= 80,000: z 
80,000  70,000
1
10,000
P[60,000< X < 80,000]= P[ -1 < Z <1] = 0.8413-1587 =0.6826, which is about 68%.
Example20B: Suppose the age of students at certain school is normally distributed with an average of
25 years and standard deviation of 5 years.
a. What percent of them are above 30?
b. What percent of them are making below 20?
c. What percent of them are making between 15 and 35?
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 13
Example21A: The total cholesterol value for population of certain city is approximately normal with a
mean of 200mg/100ml and a standard deviation of 20mg/100ml. If we randomly select a subject from
this population, find the probability that the individual will have a cholesterol level
a. between 180 and 220
b. between 160 and 240
c. above 230
d. below 175
Solution: We also have   200 and   20 . We can use the empirical rule to get the answer part a
and b. For a it is about 68% and for b is about 95%.
a.
X=180: z 
180  200
 1
20
X =220: z 
220  200
1
20
P[180 < X < 220] = P[-1 < Z < 1] =0.8413-0.1587 =0.6826, which is about 68%.
b.
X=160: z 
160  200
 2
20
X =240: z 
240  200
2
20
P[160 < X < 240] = P[-2 < Z < 2] = 0.8772-0.0228 = 0.9544, which is about 95%.
c.
X =230: z 
230  200
 1.5
20
P[ X > 230] = P[ Z > 1.5] = 1-0.9332 = 0.0668, which is about 7%.
d.
X =175: z 
175  200
 1.25
20
P[ X < 175] = P[ Z <-1.25] = 0.1056, which is about 11%.
Example21B: The LDL value for population of certain city is approximately normal with a mean of
52mg/100ml and a standard deviation of 8mg/100ml. If we randomly select a subject from this
population, find the probability that the individual will have a cholesterol level
a. between 44 and 60
b. below 40
c. above 62
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 14
Example22A: Let Z be a standard normal variable (=0, =1). Verify each of the following
probabilities.
a. P(1.25 < Z < 2.23)=0.0927
b. P(Z > 1.68)=0.0465
c. P(|Z| < 1.43)= 0.8472
Example22B: Let Z be a standard normal variable (=0, =1). Verify each of the following
probabilities.
a. P (Z < 1.28) = 0.8994
b. P (-1.58 < Z < 2.06) = 0.9232
c. P (Z > 1.87) = 0.0307
d. P (|Z| < 1.25) = 0.7888
c. P(|Z| > 1.645) = 0.10
Example22C: Let Z be a standard normal variable (=0, =1). Find the value of each of the following
probabilities.
a. P (-2.23 < Z <-1.25) =
b. P (Z <-1.68) =
c. P (|Z| >1.43) =
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 15
Example23A: As part of a physical fitness exam, individuals walk on a treadmill until tired. Suppose
the length of time for individuals to complete a treadmill test is normally distributed with the mean of
22.5 minutes and standard deviation of 3.0 minutes.
a. What is the probability an individual will complete the treadmill test in a time between 18
minutes and 30 minutes?
b. What time is exceeded by only 10% of the population of fitness exam takers?
Example23B: Assume the numerical grades from an hour examination for a large class are
normally distributed with  = 65 and  = 10. If the top 12% percent of the examination scores are
correspond to an A, what is the lowest score that will receive an A?
Example23C: Assume the numerical grades from an hour examination for a large class are
normally distributed with  = 70 and  = 8 If the bottom 10% percent of the examination scores
are correspond to an F, what is the highest score that will receive an F?
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 16
Example24A: The density of the lean body tissue of a human is approximately normally distributed
with mean µ = 1.10 g/cm3 and standard deviation
 = 0.02 g/cm3.
(a) What percentage of the people have lean-body-tissue densities greater than 1.13 g/cm3?
(b) Find the 25th percentile point of the lean-body-tissue density.
Example24B: Octane rating of 87 for a gasoline means that the burning characteristic of the gasoline
is like that of a fuel that contains 87% octane (burns slowly) and 13% heptanes (burns quickly).
However, gasoline’s are made up of many different chemicals and octane ratings are just
approximations. Suppose that the true octane ratings of all gasoline’s labeled 87 are normally
distributed with mean µ = 87.0 and standard deviation

= 0.4.
(a) What is the probability that the gasoline labeled 87 sold at a gas station near your residence will
have a true octane rating higher than 87.6?
(b) Thirty-three percent (33%) of all gasoline’s labeled 87 have true octane ratings
below x. Find the value of x.
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 17
Example25A: The heating cost for a company office in the winter season varies according to a normal
distribution with mean µ = 450 dollars and standard deviation

= 15.
(a) What proportion of the time will the heating cost be higher than 425 dollars?
(b) How much should the company budget for heating so that the actual cost will be within the budget
90% of the time?
Example25B: The measurement error for a household scale has a normal distribution with µ = 50.00
and
 = 0.02.
(a) What percent of the time will the scale reading be higher than 50.03 lb.?
(b) Twenty percent of the time the scale reading will be less than x pounds. Find
the value of x.
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Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 18
The Central Limit Theorem (CLT)
The central limit theorem states that given any distribution with a mean μ and finite variance σ², the
sampling distribution of the mean approaches a normal distribution as n, the sample size, increases.
x  
x 

n
The real advantage of the central limit theorem is that sample data drawn from populations not
normally distributed or from populations of unknown shape also can be analyzed by using the normal
distribution, because the sample means are normally distributed for sample sizes of n>=30.
Column 1 of the following figure shows four different population distributions. Each ensuing column
displays the shape of the distribution of the sample means for a particular sample size. Note that the
distribution of the sample means begins to approximate the normal curve as the sample size, n, gets
larger.
18
n=2
n=5
n=30
Elementary Statistics
Chapter 5
Normal Distribution
Dr.Ghamsary Page 19
Example26A: A random sample of n = 36 is selected from a population with  = 100 and  = 15.
a. What is the sampling distribution of sample mean
b. What are the mean and standard deviation of the sampling distribution of the sample mean?
c. If we randomly select 36subjects, approximately what is the probability that their sample
mean is over 106?
Solution:
a. The sampling distribution of sample mean is normal, by CLT.
b. Mean:  x    100 and Standard deviation:  x 
c.
P  X  106 : z 
x  x
x


n

15
3
25
106  100
2
3
P  X  106  P  z  2  1  0.0228  0.9772
Example26B: A random sample of n = 100 is selected from a population with  = 1000 and  = 100.
a. What is the sampling distribution of sample mean
b. What are the mean and standard deviation of the sampling distribution of the sample mean?
c. If we randomly select 100 subjects, approximately what is the probability that their sample
mean is between 980 and 1020?
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Elementary Statistics
Chapter 5
Normal Distribution
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Example27A: A random sample of n = 64 is selected from a population with  = 70 and  = 8.
a. What is the sampling distribution of sample mean
b. What are the mean and standard deviation of the sampling distribution of the sample mean?
c. If we randomly select 64 subjects, approximately what is the probability that their sample
mean is over 68?
Example27B: A random sample of n = 49 is selected from a population with  = 70 and  = 14
a. What is the sampling distribution of sample mean
b. What are the mean and standard deviation of the sampling distribution of the sample mean?
c. If we randomly select 49 subjects, approximately what is the probability that their sample
mean is over between 67 and 71?
20
Elementary Statistics
z
Chapter 5
Normal Distribution
Dr.Ghamsary Page 21
0
Cumulative probabilities for NEGATIVE z-values are shown in the following table:
21
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-3.0
.0013
.0013
.0013
.0012
.0012
.0011
.0011
.0011
.0010
.0010
-2.9
.0019
.0018
.0018
.0017
.0016
.0016
.0015
.0015
.0014
.0014
-2.8
.0026
.0025
.0024
.0023
.0023
.0022
.0021
.0021
.0020
.0019
-2.7
.0035
.0034
.0033
.0032
.0031
.0030
.0029
.0028
.0027
.0026
-2.6
.0047
.0045
.0044
.0043
.0041
.0040
.0039
.0038
.0037
.0036
-2.5
.0062
.0060
.0059
.0057
.0055
.0054
.0052
.0051
.0049
.0048
-2.4
.0082
.0080
.0078
.0075
.0073
.0071
.0069
.0068
.0066
.0064
-2.3
.0107
.0104
.0102
.0099
.0096
.0094
.0091
.0089
.0087
.0084
-2.2
.0139
.0136
.0132
.0129
.0125
.0122
.0119
.0116
.0113
.0110
-2.1
.0179
.0174
.0170
.0166
.0162
.0158
.0154
.0150
.0146
.0143
-2.0
.0228
.0222
.0217
.0212
.0207
.0202
.0197
.0192
.0188
.0183
-1.9
.0287
.0281
.0274
.0268
.0262
.0256
.0250
.0244
.0239
.0233
-1.8
.0359
.0351
.0344
.0336
.0329
.0322
.0314
.0307
.0301
.0294
-1.7
.0446
.0436
.0427
.0418
.0409
.0401
.0392
.0384
.0375
.0367
-1.6
.0548
.0537
.0526
.0516
.0505
.0495
.0485
.0475
.0465
.0455
-1.5
.0668
.0655
.0643
.0630
.0618
.0606
.0594
.0582
.0571
.0559
-1.4
.0808
.0793
.0778
.0764
.0749
.0735
.0721
.0708
.0694
.0681
-1.3
.0968
.0951
.0934
.0918
.0901
.0885
.0869
.0853
.0838
.0823
-1.2
.1151
.1131
.1112
.1093
.1075
.1056
.1038
.1020
.1003
.0985
-1.1
.1357
.1335
.1314
.1292
.1271
.1251
.1230
.1210
.1190
.1170
-1.0
.1587
.1562
.1539
.1515
.1492
.1469
.1446
.1423
.1401
.1379
-0.9
.1841
.1814
.1788
.1762
.1736
.1711
.1685
.1660
.1635
.1611
-0.8
.2119
.2090
.2061
.2033
.2005
.1977
.1949
.1922
.1894
.1867
-0.7
.2420
.2389
.2358
.2327
.2296
.2266
.2236
.2206
.2177
.2148
-0.6
.2743
.2709
.2676
.2643
.2611
.2578
.2546
.2514
.2483
.2451
-0.5
.3085
.3050
.3015
.2981
.2946
.2912
.2877
.2843
.2810
.2776
-0.4
.3446
.3409
.3372
.3336
.3300
.3264
.3228
.3192
.3156
.3121
-0.3
.3821
.3783
.3745
.3707
.3669
.3632
.3594
.3557
.3520
.3483
-0.2
.4207
.4168
.4129
.4090
.4052
.4013
.3974
.3936
.3897
.3859
-0.1
.4602
.4562
.4522
.4483
.4443
.4404
.4364
.4325
.4286
.4247
0.0
.5000
.4960
.4920
.4880
.4840
.4801
.4761
.4721
.4681
.4641
Elementary Statistics
0
Chapter 5
Normal Distribution
Dr.Ghamsary Page 22
z
Cumulative probabilities for POSITIVE z-values are in the following table:
22
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
.5000
.5040
.5080
.5120
.5160
.5199
.5239
.5279
.5319
.5359
0.1
.5398
.5438
.5478
.5517
.5557
.5596
.5636
.5675
.5714
.5753
0.2
.5793
.5832
.5871
.5910
.5948
.5987
.6026
.6064
.6103
.6141
0.3
.6179
.6217
.6255
.6293
.6331
.6368
.6406
.6443
.6480
.6517
0.4
.6554
.6591
.6628
.6664
.6700
.6736
.6772
.6808
.6844
.6879
0.5
.6915
.6950
.6985
.7019
.7054
.7088
.7123
.7157
.7190
.7224
0.6
.7257
.7291
.7324
.7357
.7389
.7422
.7454
.7486
.7517
.7549
0.7
.7580
.7611
.7642
.7673
.7704
.7734
.7764
.7794
.7823
.7852
0.8
.7881
.7910
.7939
.7967
.7995
.8023
.8051
.8078
.8106
.8133
0.9
.8159
.8186
.8212
.8238
.8264
.8289
.8315
.8340
.8365
.8389
1.0
.8413
.8438
.8461
.8485
.8508
.8531
.8554
.8577
.8599
.8621
1.1
.8643
.8665
.8686
.8708
.8729
.8749
.8770
.8790
.8810
.8830
1.2
.8849
.8869
.8888
.8907
.8925
.8944
.8962
.8980
.8997
.9015
1.3
.9032
.9049
.9066
.9082
.9099
.9115
.9131
.9147
.9162
.9177
1.4
.9192
.9207
.9222
.9236
.9251
.9265
.9279
.9292
.9306
.9319
1.5
.9332
.9345
.9357
.9370
.9382
.9394
.9406
.9418
.9429
.9441
1.6
.9452
.9463
.9474
.9484
.9495
.9505
.9515
.9525
.9535
.9545
1.7
.9554
.9564
.9573
.9582
.9591
.9599
.9608
.9616
.9625
.9633
1.8
.9641
.9649
.9656
.9664
.9671
.9678
.9686
.9693
.9699
.9706
1.9
.9713
.9719
.9726
.9732
.9738
.9744
.9750
.9756
.9761
.9767
2.0
.9772
.9778
.9783
.9788
.9793
.9798
.9803
.9808
.9812
.9817
2.1
.9821
.9826
.9830
.9834
.9838
.9842
.9846
.9850
.9854
.9857
2.2
.9861
.9864
.9868
.9871
.9875
.9878
.9881
.9884
.9887
.9890
2.3
.9893
.9896
.9898
.9901
.9904
.9906
.9909
.9911
.9913
.9916
2.4
.9918
.9920
.9922
.9925
.9927
.9929
.9931
.9932
.9934
.9936
2.5
.9938
.9940
.9941
.9943
.9945
.9946
.9948
.9949
.9951
.9952
2.6
.9953
.9955
.9956
.9957
.9959
.9960
.9961
.9962
.9963
.9964
2.7
.9965
.9966
.9967
.9968
.9969
.9970
.9971
.9972
.9973
.9974
2.8
.9974
.9975
.9976
.9977
.9977
.9978
.9979
.9979
.9980
.9981
2.9
.9981
.9982
.9982
.9983
.9984
.9984
.9985
.9985
.9986
.9986
3.0
.9987
.9987
.9987
.9988
.9988
.9989
.9989
.9989
.9990
.9990