1. NADEEM UDDIN
ASSOCIATE PROFESSOR
OF STATISTICS
NORMAL PROBABILITY
DISTRIBUTION

2. A continuous random variable is a random variable where
the data can take infinite values.
For example
measuring the time taken for something to be done is
continuous random variable. A continuous random variable
can be describe by a graph and by a function.
A normal distribution, sometimes called the bell curve, The
bell curve is symmetrical. Half of the data will fall to the left
of the mean and half will fall to the right.

3. Normal Distribution:
The normal distribution refers to a family of continuous
probability distributions described by the normal equation.
The normal distribution is defined by the following equation.
 2
2
2
1
f(x)= ,
2
x
e x


 
 


  

4. Standard Normal Distribution:
The standard normal distribution is a special case of the normal
distribution. It is the distribution that occurs when a normal
random variable has a mean of zero and a standard deviation of
one. The normal random variable of a standard normal
distribution is called a standard score or a z-score. Every normal
random variable X can be transformed into a z score via the
following equation:
. .
x
S N V Z



 
Where X is a normal random variable, μ is the mean of X, and σ is the standard
deviation of X.

6. Using Standard Normal Variable
Example-1
Given a normal distribution with µ = 40 and σ = 6 , find
(a). the area below 32,
(b). the area above 27,
(c). the area between 42 and 51.

7. Solution :
Since µ = 40 and σ = 6, X ̴ N(40,62)
You need to find p(x < 32).To be able to use the standard normal
table, standardize the x variable by subtracting the mean 40 then
dividing by the standard deviation, 6.Appling this to both sides of the
inequality x<32.
(a). 𝑃 𝑥 < 32 = 𝑃
𝑥−𝜇
σ <
32−𝜇
σ
= 𝑃 𝑍 <
32 − 40
6
= 𝑃 𝑍 < −1.33 = 0.0918
32 40

8. (b). 𝑃(𝑥 > 27) = 𝑃
𝑥−𝜇
σ >
27−𝜇
σ
= 𝑃 𝑍 >
27−40
6
= 𝑃 𝑍 > −2.17
=1 - 𝑃 𝑍 < −2.17
= 1 − 0.0150
=0.9850
27 40

9. (c).
𝑃(42 < 𝑥 < 51) = 𝑃
42 − 𝜇
σ
<
𝑥 − 𝜇
σ
<
51 − 𝜇
σ
= 𝑃
42−40
6
< 𝑍 <
51−40
6
= 𝑃 0.33 < 𝑍 < 1.83
= 𝑃 0 < 𝑍 < 1.83 − 𝑃 0 < 𝑍 < 0.33
= 0.4664 − 0.1293
= 0.3371
40 42 51

10. Example-2
The life of a certain make of saver bulb is known to be
normally distributed with a mean life of 2000 hours and a
standard deviation of 120 hours. In a batch of 5000 saver
bulbs, estimate the number of saver bulbs that the life of
such a bulb will be
(a)Less than 1950 hours’
(b)Greater than 2150 hours’
(c)From 1850 hours to 2090 hours.

11. 1950 2000
Solution :
(a). 𝑃(𝑥 < 1950) = 𝑃
𝑥−𝜇
σ <
1950−𝜇
σ
= 𝑃 𝑍 <
1950−2000
120
= 𝑃 𝑍 < −0.42
= 0.3372
Number of saver bulb life less than 1950 hours is ;
0.3372×5000 = 1686 bulbs.

12. 2000 2150
(b). 𝑃(𝑥 > 2150) = 𝑃
𝑥−𝜇
σ >
2150−𝜇
σ
= 𝑃 𝑍 >
2150−2000
120
= 𝑃 𝑍 > 1.25
=1 - 𝑃 𝑍 < 1.25
= 1 − 0.8944
= 0.1056
Number of saver bulb life greater than 2150 hours is ;
0.1056×5000 = 528 bulbs.

13. 1850 2000 2090
(c).
𝑃(1850 < 𝑥 < 2090) = 𝑃
1850 − 𝜇
σ
<
𝑥 − 𝜇
σ
<
2090 − 𝜇
σ
= 𝑃
1850−2000
120
< 𝑍 <
2090−2000
120
= 𝑃 −1.25 < 𝑍 < 0.75
= 𝑃 0 < 𝑍 < 0.75 − 𝑃 −1.25 < 𝑍 < 0
= 0.7734 − 0.1056
= 0.6678
Number of saver bulb life between 1850 hours to 2090 hours is
0.6678×5000 = 3339 bulbs.

14. Example-3
If X is normally distributed with parameters µ = 6 and 𝜎2 = 16 find
(a) 𝑃 𝑥 − 4 ˂ 8
(b) 𝑃( 𝑥 − 4 > 8)
(c) 𝑃( 𝑥 − 4 > 8) by using complementary law.
Solution: (a)
𝑃 𝑥 − 4 < 8 = 𝑃(−4 ˂ 𝑥˂12)
𝑃 𝑥 − 4 < 8 = 𝑃(
−4−𝜇
𝜎
˂
𝑥−𝜇
𝜎
˂
12−𝜇
𝜎
)
𝑃 𝑥 − 4 < 8 = 𝑃(
−4−6
4
˂ 𝑍 ˂
12−6
4
)
𝑃 𝑥 − 4 < 8 = 𝑃(−2.50˂ 𝑍 ˂1.50)

15. 𝑃 𝑥 − 4 < 8 = 𝑃 𝑍 ˂ 1.50 − 𝑃 𝑍 ˂ − 2.50
𝑃 𝑥 − 4 < 8 = 0.9332 − 0.0062
𝑃 𝑥 − 4 < 8 = 0.9270
(b)
𝑃 𝑥 − 4 > 8 = 1 − 𝑃 𝑥 − 4 < 8
𝑃 𝑥 − 4 > 8 = 1 − 𝑃(−4 ˂ 𝑥˂12)
𝑃 𝑥 − 4 > 8 = 1 − 𝑃(
−4−𝜇
𝜎
˂
𝑥−𝜇
𝜎
˂
12−𝜇
𝜎
)
𝑃 𝑥 − 4 > 8 = 1 − 𝑃(
−4−6
4
˂ 𝑍 ˂
12−6
4
)
𝑃 𝑥 − 4 > 8 = 1 − 𝑃(−2.50˂ 𝑍 ˂1.50)

16. 𝑃 𝑥 − 4 > 8 = 1 − {𝑃 𝑍 ˂ 1.50 − 𝑃 𝑍 ˂ − 2.50 }
𝑃 𝑥 − 4 > 8 = 1 − (0.9332 − 0.0062)
𝑃 𝑥 − 4 > 8 = 1 − 0.9270
𝑃 𝑥 − 4 > 8 = 0.0730
(c)
𝑃 𝑥 − 4 > 8 = 1 − 𝑃 𝑥 − 4 < 8
𝑃 𝑥 − 4 > 8 = 1 − 𝑃 −4 ˂ 𝑥˂12
𝑃 𝑥 − 4 > 8 = 1 − 0.9270 (from part a)
𝑃 𝑥 − 4 > 8 = 0.0730

17. Use of Normal Approximation to The Binomial Distribution
We can use the normal distribution as a close approximation
to the binomial distribution when n is large and
p is moderate (not to far from 0.5) .
OR
We can use the normal distribution as a close approximation
to the binomial distribution whenever n×p ≥ 5 and n×q ≥ 5.

18. As we know that the binomial distribution is a discrete distribution and
the normal distribution is a continuous distribution, When we use
normal approximation we should use the following continuity
correction rules.
(i) P(x1 ≤ X ≤ x2) = P(x1- 0.5 < X < x2+ 0.5)
(ii) P(x1 < X < x2) = P(x1+ 0.5 < X < x2- 0.5)
(iii) p(X > x) = p(X > x+0.5)
(iv) p(X ≥ x) = p(X > x – 0.5)
(v) p(X < x) = p(X < x – 0.5 )
(vi) p(X ≤ x) = p(X < x+0.5)
(vii) p(X = x) = P(x1- 0.5 < X < x2+ 0.5)

19. Example-4
In a sack of mixed grass seeds, the probability that a seed is
ryegrass is 0.35, find the probability that in a random
sample of 400 seeds from this sack,
(a). less than 120 are ryegrass seeds,
(b). more than 160 are ryegrass seeds,
(c). between 120 and 150 are ryegrass
(d). 150 are ryegrass.
Solution:
n = 400 ; p = 0.35
µ = np = 400 × 0.35 = 140
σ = 𝑛𝑝𝑞 = 400 × 0.35 × 0.65 = 9.54

20. (a). P 𝒙 < 𝟏𝟐𝟎 = 𝑷 𝒙 < 𝟏𝟏𝟗. 𝟓 = 𝑷
𝒙−𝝁
σ <
𝟏𝟏𝟗.𝟓−𝝁
σ
= 𝑷 𝒁 <
𝟏𝟏𝟗.𝟓−𝟏𝟒𝟎
𝟗.𝟓𝟒
= 𝑷 𝒁 < −𝟐. 𝟏𝟒𝟗
= 𝟎. 𝟎𝟏𝟓𝟖
(b). P 𝒙 > 𝟏𝟔𝟎 = 𝑷 𝒙 > 𝟏𝟔𝟎. 𝟓 = 𝑷
𝒙−𝝁
σ >
𝟏𝟔𝟎.𝟓−𝝁
σ
= 𝑷 𝒁 >
𝟏𝟔𝟎.𝟓−𝟏𝟒𝟎
𝟗.𝟓𝟒
= 𝑷 𝒁 > 𝟐. 𝟏𝟓
= 𝟏 − 𝑷 𝒁 < 𝟐. 𝟏𝟓
= 𝟏 − 𝟎. 𝟗𝟖𝟒𝟐
= 𝟎. 𝟎𝟏𝟓𝟖
120 140
140 160

21. (c).
P 120 < 𝑥 < 150 = 𝑃 120.5 < 𝑥 < 149.5
= 𝑃
120.5−140
9.54
< 𝑍 <
149.5−140
9.54
= 𝑃 −2.04 < 𝑍 < 0.99
= 𝑃 0 < 𝑍 < 0.99 − 𝑃 −2.04 < 𝑍 < 0
= 0.8389 − 0.0207
= 0.8182
120 140 150

22. (d) P x = 150 = P 149.5 < x < 150.5
P x = 150 = P
149.5−μ
σ
<
x−μ
σ
<
150.5−μ
σ
P x = 150 = P
149.5−140
9.54
< Z <
150.5−140
9.54
P x = 150 = P 0.99 < Z < 1.10
P x = 150 = P Z < 1.10 − P Z < 0.99
P x = 150 = 0.8643 − 0.8389
P x = 150 = 0.0254
140 149.5 150.5

23. Use of Normal Approximation to The Poisson Distribution
We can use the normal distribution as a close approximation to
the
Poisson distribution whenever Mean is large (more than 15).
As we know that the Poisson distribution is a discrete distribution
and the normal distribution is a continuous distribution, When we
use normal approximation we should use the following continuity
correction rules.
(i) P(x1 ≤ X ≤ x2) = P(x1- 0.5 < X < x2+ 0.5)
(ii) P(x1 < X < x2) = P(x1+ 0.5 < X < x2- 0.5)
(iii) p(X > x) = p(X > x+0.5)
(iv) p(X ≥ x) = p(X > x – 0.5)
(v) p(X < x) = p(X < x – 0.5 )
(vi) p(X ≤ x) = p(X < x+0.5)
(vii) p(X = x) = P(x1- 0.5 < X < x2+ 0.5)

24. Example-5
In a certain city area the number of accidents occurring in a month
follows a Poisson distribution with mean 3. Find the following
probability that there will be accidents during 10 months.
(a) at least 35 accidents
(b) Less 30 accidents
(c) Between 25 and 35
(d) 35 accidents.
Solution:
µ = 3×10 = 30 (in 10 months)
As we know in Poisson distribution mean is equal to variance
Therefore 𝛔 𝟐 = 𝟑𝟎
σ = 5.48
(a) 𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏 𝐱 > 𝟑𝟒. 𝟓
𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏
𝐱−𝛍
𝛔
>
𝟑𝟒.𝟓−𝛍
𝛔

25. 𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏 𝐙 >
𝟑𝟒. 𝟓 − 𝟑𝟎
𝟓. 𝟒𝟖
𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏 𝐙 > 𝟎. 𝟖𝟐
𝐏 𝐱 ≥ 𝟑𝟓 = 𝟏 − 𝐏 𝐙 < 𝟎. 𝟖𝟐
𝐏 𝐱 ≥ 𝟑𝟓 = 𝟏 − 𝟎. 𝟕𝟗𝟑𝟗
𝐏 𝐱 ≥ 𝟑𝟓 = 𝟎. 𝟐𝟎𝟔𝟏
(b)
𝐏 𝐱 < 𝟐𝟓 = (𝐱 < 𝟐𝟒. 𝟓)
𝐏 𝐱 < 𝟐𝟓 = (
𝐱−𝛍
𝛔
<
𝟐𝟒.𝟓−𝛍
𝛔
)
𝐏 𝐱 < 𝟐𝟓 = (𝐙 <
𝟐𝟒.𝟓−𝟑𝟎
𝟓.𝟒𝟖
)
𝐏 𝐱 < 𝟐𝟓 = (𝐙 < −𝟏. 𝟎)
𝐏 𝐱 < 𝟐𝟓 = 𝟎. 𝟏𝟓𝟖𝟕

26. (c) 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 𝟐𝟓. 𝟓 < 𝐱 < 𝟑𝟒. 𝟓
𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏
𝟐𝟓.𝟓−𝝁
𝝈
<
𝒙−𝝁
𝝈
<
𝟑𝟒.𝟓−𝝁
𝝈
𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏
𝟐𝟓.𝟓−𝟑𝟎
𝟓.𝟒𝟖
< 𝐙 <
𝟑𝟒.𝟓−𝟑𝟎
𝟓.𝟒𝟖
𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 −𝟎. 𝟖𝟐 < 𝐙 < 𝟎. 𝟖𝟐
𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 𝒁 < 𝟎. 𝟖𝟐 − 𝐏 𝒁 < −𝟎. 𝟖𝟐
𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝟎. 𝟕𝟗𝟑𝟗 − 𝟎. 𝟐𝟎𝟔𝟏
𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝟎. 𝟓𝟖𝟕𝟖

27. (d) P x = 35 = P 34.5 < x < 35.5
P x = 35 = P
34.5−μ
σ
<
x−μ
σ
<
35.5−μ
σ
P x = 35 = P
34.5−30
5.48
< Z <
35.5−30
5.48
P x = 35 = P 0.82 < Z < 1.00
P x = 35 = P Z < 1.00 − P Z < 0.82
P x = 35 = 0.8413 − 0.7939
P x = 35 = 0.0474 30 34.5 35.5

28. Using the Table in Reverse for any normal variable x
Example-6
The time taken by the milkman to deliver to the high street
is normally distributed with a mean of 12 minutes and a
standard deviation of 2 minutes. He delivers milk every
day. Estimate the time of day during the year.
(a) Longer than 0.0062 .
(b) Less than 0.1587.

29. Solution:
X is the time to delivers milk everyday. X ̴ N(12,22)
(a) Given p(x > k) = 0.0062
Standardising
𝑷
𝒙−𝝁
σ >
𝒌−𝝁
σ = 𝟎. 𝟎𝟎𝟔𝟐
𝑷 𝒛 >
𝒌−𝟏𝟐
𝟐
= 𝟎. 𝟎𝟎𝟔𝟐
Since
𝒌−𝟏𝟐
𝟐
= 𝟐. 𝟓𝟎
(The value of z at 0.0062 from normal table is 2.50)
k= 2.50×2+12 = 17 minute
Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
-2.6 .00466 .00453 .00440 .00427 .00415 .00402 .00391 .00379 .00368 .00357
-2.5 .00621 .00604 .00587 .00570 .00554 .00539 .00523 .00508 .00494 .00480
-2.4 .00820 .00798 .00776 .00755 .00734 .00714 .00695 .00676 .00657 .00639

30. (b)
Given p(x < k) = 0.1587
Standardrising
𝑷
𝒙−𝝁
σ <
𝒌−𝝁
σ = 𝟎. 𝟏𝟓𝟖𝟕
𝑷 𝒛 <
𝒌−𝟏𝟐
𝟐
= 𝟎. 𝟏𝟓𝟖𝟕
Since 𝒌−𝟏𝟐
𝟐
= −𝟏. 𝟎
(The value of z at 0.1587 from normal table is -1.0)
k= (-1.0)×2+12 = 10 minute
Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
-3.9 .00005 .00005 .00004 .00004 .00004 .00004 .00004 .00004 .00003 .00003
-1.0 .15866 .15625 .15386 .15151 .14917 .14686 .14457 .14231 .14007 .13786
-0.9 .18406 .18141 .17879 .17619 .17361 .17106 .16853 .16602 .16354 .16109

31. Example-7
The length of certain items follow a normal distribution with mean 
cm and standard deviation 6cm, it is known that 95% of the items have
length less than 82cm. Find the value of the mean .
Solution:
P(x  82) = 0.95
P
x−μ
σ
<
k−μ
σ
= 0.95
P Z <
82−μ
6
= 0.95
82−μ
6
= 1.65
82 - µ =(1.65)×6
µ = 82 – 9.9 = 72

32. Example-8
x~N(100, σ2
) and P(x ˂ 106) = 0.8849. find the standard deviation
σ.
Solution:
P(x  106) = 0.8849
P
x − μ
σ
<
106 − 100
σ
= 0.8849
P Z <
6
σ
= 0.8849
6
σ
= 1.20
6
1.20
= σ
σ = 5

33. Example-9
In a normal distribution, 5% of the items are up to 45 and 75% are
below 65.
(a)Find mean and standard deviation,
(b)Find P(45 ˂ x ˂ 65).
Solution:
(a) P 𝑥 > 45 = 0.05
P
𝑥−𝜇
𝜎
>
45−𝜇
𝜎
= 0.05
P 𝑍 >
45−𝜇
𝜎
= 0.05
45−𝜇
𝜎
= −1.65 (from Z- table)
45 - µ = -1.65 𝜎 ------(i)

34. 𝑷 𝒙 < 𝟔𝟓 = 𝟎. 𝟕𝟓
P
𝒙−𝝁
𝝈
<
𝟔𝟓−𝝁
𝝈
= 𝟎. 𝟕𝟓
P 𝒁 <
𝟔𝟓−𝝁
𝝈
= 𝟎. 𝟕𝟓
𝟔𝟓−𝝁
𝝈
= 𝟎. 𝟔𝟖 (from Z- table)
65 - µ = 0.68 𝝈 ------(ii)
Subtract (i) from (ii)
65 - µ = 0.68 𝝈
45 - µ = -1.65 𝝈
20 = 2.33 𝝈
𝝈 =
𝟐𝟎
𝟐. 𝟑𝟑
= 𝟖. 𝟓𝟖

35. Put 𝜎 = 8.58 in (ii)
65 - µ = 0.68 (8.58)
65 - µ = 5.83
µ = 59.17
(b)
P 45 < 𝑥 < 65 = 𝑷
𝟒𝟓−𝝁
𝝈
<
𝒙−𝝁
𝝈
<
𝟔𝟓−𝝁
𝝈
= 𝑷
𝟒𝟓−𝟓𝟗.𝟏𝟕
𝟖.𝟓𝟖
< 𝒁 <
𝟔𝟓−𝟓𝟗.𝟏𝟕
𝟖.𝟓𝟖
= 𝑷 −𝟏. 𝟔𝟓 < 𝒁 < 𝟎. 𝟔𝟖
= 𝑷 𝒁 < 𝟎. 𝟔𝟖 − 𝑷 𝒁 < −𝟏. 𝟔𝟓
= 0.7517 – 0.0495
= 0.7022

36. Example-10
In a normal distribution with µ = 100 and σ = 15 find
(a) 𝑄1 𝑎𝑛𝑑𝑄3
(b) 𝐷7
(c) 𝑃95
Solution:
(a)
𝑄1 = First quartile, (the points below which 25% area lies and above
which 75% area lies)
Thus P(X ≤ 𝑄1) = 0.25
𝑃
𝑥−𝜇
𝜎
≤
𝑄1−𝜇
𝜎
= 0.25
𝑃 𝑍 ≤
𝑄1−100
15
= 0.25
𝑃 𝑍 ≤ −0.67 = 0.25
(from Z- table)

37. Hence
𝑄1−100
15
= −0.67
𝑄1 = 100 − (0.67)(15)
𝑄1 = 89.95
and
𝑄3 = Third quartile, (the points below which 75% area lies and
above which 25% area lies)
Thus P(X ≤ 𝑄3) = 0.75
𝑃
𝑥−𝜇
𝜎
≤
𝑄3−𝜇
𝜎
= 0.75
𝑃 𝑍 ≤
𝑄3−100
15
= 0.75
𝑃 𝑍 ≤ 0.67 = 0.75
(from Z- table)

38. Hence
𝑄3−100
15
= 0.67
𝑄3 = 100 + (0.67)(15)
𝑄3 = 110.05
(b)
𝐷7 = 7th deciles, (the points below which 70% area lies and
above which 30% area lies)
Thus P(X ≤ 𝐷7) = 0.70
𝑃
𝑥−𝜇
𝜎
≤
𝐷7−𝜇
𝜎
= 0.70
𝑃 𝑍 ≤
𝐷7−100
15
= 0.70
𝑃 𝑍 ≤ 0.52 = 0.70
(from Z- table)

39. Hence
𝐷7−100
15
= 0.52
𝐷7 = 100 + (0.52)(15)
𝐷7 = 107.8
(c)
𝑃95 = 95th percentile, (the points below which 95% area lies and
above which 5% area lies)
Thus P(X ≤ 𝑃95) = 0.95
𝑃
𝑥−𝜇
𝜎
≤
𝑃95−𝜇
𝜎
= 0.95
𝑃 𝑍 ≤
𝑃95−100
15
= 0.95
𝑃 𝑍 ≤ 1.65 = 0.95
(from Z- table)

40. Hence
𝑃95−100
15
= 1.65
𝑃95 = 100 + 1.65 15
𝑃95 = 124.75