CASE STUDY ON NORMAL PROBABILITY DISTRIBUTION

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Unit No.5.
PROBABILITY DISTRIBUTION
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean, EDP & Associate Professor MBA
1
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in

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302-DECISION SCIENCE
Unit No.5. Probability
5.2.8 Case 2: Normal
Probability Distribution
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean EDP & Associate Professor MBA
2
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in

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NETWORK ANALYSIS
 At the End of the Session Student will be able to
understand-
A. Case 2: Normal Probability Distribution

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Probability Distribution
In a sample of 1000 scores, the mean of a certain test is 14
and the Standard Deviation is 2.5 Assuming the
distribution to be normal, Find-
i. How many students have scored between 12 and 15
ii. How many scored above 18
iii. How many scored below 8

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Case 1: Normal Distribution
Let x denote the marks scored by student in the given test.
Thus, x is a normal variable with mean, m=14 and 𝜎=2.5
Standard Normal variate corresponding to x is-
z =
𝑥 − 𝑚
𝜎
=
𝑥 − 14
2.5

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Case 1: Normal Distribution
i. Let us find P(student scores between 12 and 15) i.e. P(12≤x≤15)
Now, when x1=12,
z1 =
12 − 14
2.5
= −0.8
And x2= 15,
z2 =
15 − 14
2.5
= 0.4
P(12≤x≤15)= P(-0.8≤z≤0.4)
= P(-0.8≤z≤0) + P(0≤z≤0.4)
= P(0.8≤z≤0) + P(0≤z≤0.4)
= 0.2881 + 0.1554 = 0.4435 -(From the z Table)
Hence, the number of students between 12 and 15 among 1000 workers
will be-
f(12≤x≤15)=1000* P(12≤x≤15)=1000*0.4435=443.5 i.e. 444

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Case 1: Normal Distribution
ii. Let us find P(student scored above 18) i.e. P(x≥18)
Now, when x=18,
z1 =
18 − 14
2.5
= 1.6
P(x≥18) = P(z ≥ 1.6) = P(0≤x≤∞) - P(0≤x≤1.6)
= 0.5 - 0.4452 = 0.0548 -(From the z Table)
Probability that the student scores more than 18 is 0.0548
Hence, the number of such students among 1000 students
will be-
f(x≥18)=1000* P(x≥18)=1000*0.0548=54.8 i.e. 55

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Case 1: Normal Distribution
iii. Let us find P(student scores less than 8) i.e. P(x≤8)
Now, when x=8,
z1 =
8 − 14
2.5
= −2.4
P(x ≤ 8) = P(z ≤ -2.4) = P(z ≥ 2.4) = P(0≤z≤∞) - P(0≤z≤2.4)
= 0.5 - 0.4918 = 0.0082 -(From the z Table)
Probability that the students scores less than 8 is 0.0082
Hence, the number of such students among 1000 students
will be-
f(x ≤ 8)=1000* P(x ≤ 8)=1000*0.0082 = 8.2 i.e. 8

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EXERCISE
 Explain case 2: Normal Probability Distribution

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For More Details Contact
Dr. V M Tidake
tidkevishal@gmail.com
tidkevishalmba@sanjivani.org.in