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Normal distribution , Properties and Examples.

Probability Distribution.  

Normal Distribution:

Normal distribution is the maximally used Probability distribution in the theory of statistics. 

 Definition: A Random variable X is used said to be follow a normal distribution with mean m and variance  σ2  then its probability density function is 


f(x) = 
the variable x is said to be normally distributed with mean m and variance  σ2   then it is denoted as 
X ~ N(m ,σ2 ).and if in normal distribution the values of parameter changes as m= 0 and σ2 = 1 then distribution of x is said to be standard normal distribution and it is denoted as X ~ N(0 ,1 ).

  if X ~ N(m ,σ2 ). and we take transformation as z = (x-m)/σ then the distribution of z is standard normal distribution. and it is denoted as Z ~ N(0 ,1 ). the pdf of f(z) is called standard normal distribution and defined as 

note: 
1 mean of the distribution is m.
2. the variance of the distribution is σ2 
3. the shape of normal distribution is bell-shaped and it is symmetrical about the x=m
4. normal curve is uni-modal.
5.the quartile deviation is Q.D. = (2/3) σ.
6. the distribution has mean = median = mode.
7. the skewness is 0 and kurtosis is 3.
8. the mean deviation is (4/5)σ Approximately.

Examples

Example1. For Normal distribution the lower quartile is 83 and the standard deviation is 15. find 
i. Median,  ii. Mean Deviation.

Solution: 
i. we know that for normal distribution lower quartile means Q1 
 therefore Q1  = mean - (2/3) σ
the given value are σ = 15,  Q1  = 83
using the property of normal distribution  Q1 = mean - (2/3)  σ
83 = mean - (2/3) x 15
83 = mean - 10
mean = 83 +10 = 93
therefore the mean of normal distribution is 93.
 we known the property normal distribution for mean, median, mode is 
mean = median= mode
for the normal distribution mean = median 
therefore mean = median = 93

ii. mean deviation
the mean deviation of normal distribution is (4/5) σ
therefore the mean deviation = (4/5) σ   = (4/5) x 15 = 4 x 3 = 12
therefore the mean deviation is 12.
the median and mean deviation of normal distribution is 93 and 12 respectively. 

Example 2.  The mean and standard deviation of normal distribution are 60 and 15. find the Q1 , Q3  and mean deviation.
Solution: 
Given value are mean = 60 and standard deviation = 15 
i.e. m= 60 and σ = 15
 we known that  
Q1  = mean - (2/3) σ
Q1  =  60 - (2/3) x 15 
Q1  = 60 -  10 
Q1  = 50

Q3  = mean + (2/3) σ
Q3  = 60 + (2/3) x 15
Q3  = 60 + 10
Q3  = 70

Mean Deviation =  (4/5) σ  = (4/5) x 15  =  4 x 3 = 12

the Q1 = 50 , Q3 = 70  and mean deviation = 12.

Example 3. The mean and standard deviation normal distribution is 92 and 25 respectively. find the three quartiles and mean deviation. 

Solution : 
Given     mean = 92 and standard deviation = 25
 i.e. m = 92 and σ = 25.
we know that 
three quartiles are Q1  = mean - (2/3) σ,Q2 = median,   Q3  = mean + (2/3) σ 
and Mean Deviation = (4/5) σ 
therefore 
Q1  = mean - (2/3) σ
Q1  =  92 - (2/3) x 25 
Q1  = 92 - 16.6666
Q1  = 75.3334

Q2 = median
we known the property normal distribution for mean, median, mode is 
mean = median= mode
but for the normal distribution mean = median 
therefore mean = median = 92
Q2 = 92

Q3  = mean + (2/3) σ
Q3  = 92 + (2/3) x 25
Q3  = 92 + 16.6666
Q3  = 108.6666

Mean Deviation =  (4/5) σ  = (4/5) x 25  =  4 x 5 = 20

The three quartiles Q1  = 75.3334,  Q2 = median = 92, Q3  = 108.6666 and mean deviation = 20

 Normal Distribution:

Properties of Normal Distribution Or Normal Curve.

Following are some important properties of Normal Distribution.

1.The  shape of normal distribution curve is bell-shaped. And symmetric. it is symmetric at maximum value at X = m (i.e. m = mean).

2. The curve has maximum height at X =m ( and maximum value is model value) therefore the m is mode of the distribution and at point m the curve divided into two equal parts ( we known median divide the data into two parts) hence m is median of normal distribution. hence if the curve is symmetric then mean = median =mode of normal distribution.

3.The mean and variance of normal distribution is m and σ^2 respectively.

4.The Kurtosis is 3 and Skewness is zero of normal distribution.

5.The curve is symmetric about mean  then first and 3rd quartiles are same distance from mean.

i.e. first quartile = Q1 = m - (2/3)σ and 3rd Quartile = Q3 = m+(2/3)σ

6. the total area under curve is equal to 1.

7.The Normal probability curve follow the rule in which approximately 68% of data falls within limits m-σ and m+σ, 95% of  data falls within limits m-2σ and m+2σ and 99.7% of the data falls within limits m-3σ and m+3σ.

Example

Example 1. The weight of 4000 Students are found to be normally distributed with mean 50 kg and standard deviation is 5kg. Find the number of students having weight i) less than 45 kg, ii) between 45 and 60 kg. 

(for a standard normal variable z the area under the curve z=0 to z= 1 is 0.3413 and z= 0 to z= 2 is 0.4772).

Solution:-

 Here X- is weight of the students and it has normal distribution with mean 50 kg and standard deviation is 5kg.

i) the number of students having weight less than 45kg.  i.e. x < 45

therefore firstly we finding the percentage of number of students having weight less then 45kg using standard normal distribution. i.e. we find the P(z < 45) form this we get percentage of students having weight is less than 45kg.

we known that if  x has  normal distribution then z = (x-mean of x)/(standard deviation) , z has standard normal distribution. 

therefore P( x < 45) = P{(x-mean of x)/(standard deviation) < (45-mean of x)/(standard deviation)}

P(z < (45-50)/5) = P( z < -5/5) = P(z < -1) = P(-∞ < z < - 1 ) 

i.e. area of curve is -∞ to - 1 is obtained difference between as area of z =-∞ to z= 0  and z= -1 to z= 0 in symbolic form 

 P(z < -1) = P(-∞ < z  <  0) - P( -1 < z < 0)

 P(z < -1) = 0.5 - 0.3413

    P(z < -1)    = 0.1587

 i.e. P( x < 45)  = 0.1587 

there 15.87 % of the students having weight is less than 45 kg 

then the number of students having weight less than 45 kg is = total number of students into probability of weight less than 45 kg.

the number of students having weight less than 45 kg is = 4000 x 0.1587 = 634.8

it is approximately 634 students having weight less than 45 kg.

the number of students having weight less than 45 kg is 634.

similarly 

ii)   the number of students having weight between 45 and 60 kg i.e. 45 < x  < 60

therefore P( 45 < x  < 60) = 

P{(45-mean of x)/(standard deviation) < (x-mean of x)/(standard deviation) < (60-mean of x)/(standard deviation)}

 P( 45 < x  < 60)  =  P{((45-50)/5) < z < (60-50)/5) } = P{ (-5/5) < z < (10/5)} = P(-1 < z < 2)

P(-1 < z < 2) = P( -1 < z  <  0) + P( 0 < z < 2)

P(-1 < z < 2) = 0.3413 + 0.4772  = 0.8185

81.85 % of the students having weight is between 45 to 60 kg.

the number of students having weight between 45 to 60 kg = total number of students into probability of weight between 45 to 60. 

the number of students having weight between 45 to 60 kg = 4000 x 0.8185  = 3274

it is approximately 3274 students having weight between 45 to 60 kg.


The number of students having weight between 45 to 60 kg is 3274.

The number of students having weight less than 45 kg is 634.


Normal Distribution Example:

Example 1. The weight of 1000 Students are found to be normally distributed with mean 50 kg and standard deviation is 5kg. Find the number of students having weight i) less than 45 kg, ii)between 45 and 60 kg. 

(for a standard normal variable z the area under the curve z=0 to z= 1 is 0.3413 and z= 0 to z= 2 is 0.4772).

Solution:-

 Here X- is weight of the students and it has normal distribution with mean 50 kg and standard deviation is 5kg.

i) the number of students having weight less than 45kg.  i.e. x < 45

therefore firstly we finding the percentage of number of students having weight less then 45kg using standard normal distribution. i.e. we find the P(z < 45) form this we get percentage of students having weight is less than 45kg.

we known that if  x has  normal distribution then z = (x-mean of x)/(standard deviation) , z has standard normal distribution. 

therefore P( x < 45) = P{(x-mean of x)/(standard deviation) < (45-mean of x)/(standard deviation)}

P(z < (45-50)/5) = P( z < -5/5) = P(z < -1) = P(-∞ < z < - 1 ) 

i.e. area of curve is -∞ to - 1 is obtained difference between as area of z =-∞ to z= 0  and z= -1 to z= 0 in symbolic form 

 P(z < -1) = P(-∞ < z  <  0) - P( -1 < z < 0)

 P(z < -1) = 0.5 - 0.3413

    P(z < -1)    = 0.1587

 i.e. P( x < 45)  = 0.1587 

there 15.87 % of the students having weight is less than 45 kg 

then the number of students having weight less than 45 kg is = total number of students into probability of weight less than 45 kg.

the number of students having weight less than 45 kg is = 1000 x 0.1587 = 158.7

it is approximately 159 students having weight less than 45 kg.

the number of students having weight less than 45 kg is 159

similarly 

ii)   the number of students having weight between 45 and 60 kg i.e. 45 < x  < 60

therefore P( 45 < x  < 60) = 

P{(45-mean of x)/(standard deviation) < (x-mean of x)/(standard deviation) < (60-mean of x)/(standard deviation)}

 P( 45 < x  < 60)  =  P{((45-50)/5) < z < (60-50)/5) } = P{ (-5/5) < z < (10/5)} = P(-1 < z < 2)

P(-1 < z < 2) = P( -1 < z  <  0) + P( 0 < z < 2)

P(-1 < z < 2) = 0.3413 + 0.4772  = 0.8185

81.85 % of the students having weight is between 45 to 60 kg.

the number of students having weight between 45 to 60 kg = total number of students into probability of weight between 45 to 60. 

the number of students having weight between 45 to 60 kg = 1000 x 0.8185  = 818.5 

it is approximately 819 students having weight between 45 to 60 kg.


The number of students having weight between 45 to 60 kg is 819.

The number of students having weight less than 45 kg is 159.


Example 2. For a normal variable x with mean 25 and S.D. 10. find the area between x = 25 to x =35.

Solution: 

we known that if  x has  normal distribution then z = (x-mean of x)/(standard deviation) , z has standard normal distribution. 

therefore

 P( 25 < x < 35) = P{(25-mean of x)/(standard deviation) < (x-mean of x)/(standard deviation) < (35-mean of x)/(standard deviation)}

P( 25 < x < 35) = P{(25-25)/(10) < (x-mean of x)/(standard deviation) < (35- 25)/10)}

P( 25 < x < 35) = P( 0 < z  < 1)

therefore area between z= 0 to z = 1 is 0.3413

hence

P( 25 < x < 35) = P( 0 < z  < 1) = 0.3413

The area between x = 25 to x =35 is 0.3413





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