Skip to main content

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





Comments

Post a Comment

Popular posts from this blog

MCQ'S based on Basic Statistics (For B. Com. II Business Statistics)

    (MCQ Based on Probability, Index Number, Time Series   and Statistical Quality Control Sem - IV)                                                            1.The control chart were developed by ……         A) Karl Pearson B) R.A. fisher C) W.A. Shewhart D) B. Benjamin   2.the mean = 4 and variance = 2 for binomial r.v. x then value of n is….. A) 7 B) 10 C) 8 D)9   3.the mean = 3 and variance = 2 for binomial r.v. x then value of n is….. A) 7 B) 10 C) 8 D)9 4. If sampl...
Post a Comment
Read more

Measures of Central Tendency :Mean, Median and Mode

Changing Color Blog Name  Measures of Central Tendency  I. Introduction. II. Requirements of good measures. III. Mean Definition. IV . Properties  V. Merits and Demerits. VI. Examples VII.  Weighted Arithmetic Mean VIII. Median IX. Quartiles I. Introduction Everybody is familiar with the word Average. and everybody are used the word average in daily life as, average marks, average of bike, average speed etc. In real life the average is used to represent the whole data, or it is a single figure is represent the whole data. the average value is lies around the centre of the data. consider the example if we are interested to measure the height of the all student and remember the heights of all student, in that case there are 2700 students then it is not possible to remember the all 2700 students height so we find out the one value that represent the height of the all 2700 students in college. therefore the single value represent ...
1 comment
Read more

Business Statistics Notes ( Meaning, Scope, Limitations of statistics and sampling Methods)

  Business Statistics Paper I Notes. Welcome to our comprehensive collection of notes for the Business Statistics!  my aim is to provided you  with the knowledge you need as you begin your journey to comprehend the essential ideas of this subject. Statistics is a science of collecting, Presenting, analyzing, interpreting data to make informed business decisions. It forms the backbone of modern-day business practices, guiding organizations in optimizing processes, identifying trends, and predicting outcomes. I will explore several important topics through these notes, such as: 1. Introduction to Statistics. :  meaning definition and scope of  Statistics. 2. Data collection methods. 3. Sampling techniques. 4. Measures of  central tendency : Mean, Median, Mode. 5. Measures of Dispersion : Relative and Absolute Measures of dispersion,  Range, Q.D., Standard deviation, Variance. coefficient of variation.  6.Analysis of bivariate data: Correlation, Regr...
Post a Comment
Read more

Classification, Tabulation, Frequency Distribution, Diagrams & Graphical Presentation.

Business Statistics I    Classification, Tabulation, Frequency Distribution ,  Diagrams & Graphical Presentation. In this section we study the following point : i. Classification and it types. ii. Tabulation. iii. Frequency and Frequency Distribution. iv. Some important concepts. v. Diagrams & Graphical Presentation   I. Classification and it's types:        Classification:- The process of arranging data into different classes or groups according to their common  characteristics is called classification. e.g. we dividing students into age, gender and religion. It is a classification of students into age, gender and religion.  Or  Classification is a method used to categorize data into different groups based on the values of specific variable.  The purpose of classification is to condenses the data, simplifies complexities, it useful to comparison and helps to analysis. The following are some criteria to classi...
Post a Comment
Read more

Measures of Dispersion : Range , Quartile Deviation, Standard Deviation and Variance.

Measures of Dispersion :  I.  Introduction. II. Requirements of good measures. III. Uses of Measures of Dispersion. IV.  Methods Of Studying Dispersion:     i.  Absolute Measures of Dispersions :             i. Range (R)          ii. Quartile Deviation (Q.D.)          iii. Mean Deviation (M.D.)         iv. Standard Deviation (S. D.)         v. Variance    ii.   Relative Measures of Dispersions :              i. Coefficient of Range          ii. Coefficient of Quartile Deviation (Q.D.)          iii. Coefficient of Mean Deviation (M.D.)         iv. Coefficient of Standard Deviation (S. D.)         v. Coefficien...
Post a Comment
Read more

Basic Concepts of Probability and Binomial Distribution , Poisson Distribution.

 Probability:  Basic concepts of Probability:  Probability is a way to measure hoe likely something is to happen. Probability is number between 0 and 1, where probability is 0 means is not happen at all and probability is 1 means it will be definitely happen, e.g. if we tossed coin there is a 50% chance to get head and 50% chance to get tail, it can be represented in probability as 0.5 for each outcome to get head and tail. Probability is used to help us taking decision and predicting the likelihood of the event in many areas, that are science, finance and Statistics.  Now we learn the some basic concepts that used in Probability:  i) Random Experiment OR Trail: A Random Experiment is an process that get one or more possible outcomes. examples of random experiment include tossing a coin, rolling a die, drawing  a card from pack of card etc. using this we specify the possible outcomes known as sample pace.  ii)Outcome: An outcome is a result of experi...
1 comment
Read more

Statistical Inference I ( Theory of estimation : Efficiency)

🔖Statistical Inference I ( Theory of estimation : Efficiency)  In this article we see the  terms:  I. Efficiency. II. Mean Square Error. III. Consistency. 📚 Efficiency:  We know that  two unbiased estimator of parameter gives rise to infinitely many unbiased estimators of parameter. there if one of parameter have two estimators then the problem is to choose one of the best estimator among the class of unbiased estimators. in that case we need to some other criteria to to find out best estimator. therefore, that situation  we check the variability of that estimator, the measure of variability of estimator T around it mean is Var(T). hence If T is an Unbiased estimator of parameter then it's variance gives good precision. the variance is smaller then it give's greater precision. 📑 i. Efficient estimator: An estimator T is said to be an Efficient Estimator of 𝚹, if T is unbiased estimator of    𝛉. and it's variance is less than any other estima...
Post a Comment
Read more

The Power of Statistics: A Gateway to Exciting Opportunities

  My Blog The Power of Statistics: A Gateway to Exciting Opportunities     Hey there, future statistician! Ever wondered how Netflix seems to know exactly what shows you'll love, how sports teams break down player performance, or how businesses figure out their pricing strategies? The answer is statistics—a fascinating field that helps us make sense of data in our everyday lives. Let's dive into why choosing statistics for your B.Sc. Part First can lead you to some exciting opportunities.     Why Statistics Matters in Everyday Life     From predicting election outcomes and analyzing social media trends to understanding consumer behavior and optimizing public transport routes, statistics are crucial. It's the backbone of modern decision-making, helping us sift through complex data to uncover meaningful insights that drive innovation and progress.   The Role of Statistics in Future Opportunities ...
Post a Comment
Read more

Statistical Inference I ( Theory of Estimation) : Unbiased it's properties and examples

 📚Statistical Inference I Notes The theory of  estimation invented by Prof. R. A. Fisher in a series of fundamental papers in around 1930. Statistical inference is a process of drawing conclusions about a population based on the information gathered from a sample. It involves using statistical techniques to analyse data, estimate parameters, test hypotheses, and quantify uncertainty. In essence, it allows us to make inferences about a larger group (i.e. population) based on the characteristics observed in a smaller subset (i.e. sample) of that group. Notation of parameter: Let x be a random variable having distribution function F or f is a population distribution. the constant of  distribution function of F is known as Parameter. In general the parameter is denoted as any Greek Letters as θ.   now we see the some basic terms :  i. Population : in a statistics, The group of individual under study is called Population. the population is may be a group of obj...
Post a Comment
Read more