This document discusses solving probability problems using the normal distribution. It provides an example of test scores that are normally distributed with a mean of 14 and standard deviation of 2.5. It then calculates: 1) The number of students who scored between 12 and 15, which is 444. 2) The number of students who scored above 18, which is 55. 3) The number of students who scored below 8, which is 8. The calculations are shown by converting the data values to standard normal variables and looking them up in a z-table.

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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