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www.sanjivanimba.org.in
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
www.sanjivanimba.org.in
302-DECISION SCIENCE
Unit No.5. Probability
5.2.7 Case 1: 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
www.sanjivanimba.org.in
NETWORK ANALYSIS
 At the End of the Session Student will be able to
understand-
A. Case 1: Normal Probability Distribution
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Probability Distribution
The weekly wages of 1000 workers are normally
distributed around a mean of Rs. 70 and Standard
Deviation of Rs. 5. Estimate the number of workers whose
weekly wages will be-
i. Between Rs. 70 and Rs. 72
ii. Between Rs. 69 and Rs. 72
iii. More than Rs. 75
iv. Less than Rs. 63
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Case 1: Normal Distribution
Let x denote the weekly wages in Rupees.
Thus, x is a normal variable with mean, m=70 and 𝜎=5
Standard Normal variate corresponding to x is-
z =
𝑥 − 𝑚
𝜎
=
𝑥 − 70
5
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Case 1: Normal Distribution
i. Let us find P(wages between Rs. 70 and Rs. 72) i.e. P(70≤x≤72)
Now, when x1=70,
z1 =
70 − 70
5
= 0
And x2= 72,
z2 =
72 − 70
5
= 0.4
P(70≤x≤72)= P(0≤z≤0.4)=0.1554 -(From the z Table)
Probability that the workers will have weekly wages between Rs.
70 and Rs. 72 is 0.1554
Hence, the number of such workers among 1000 workers will be-
f(70≤x≤72)=1000* P(70≤x≤72)=1000*0.1554=155.4 i.e. 155
www.sanjivanimba.org.in
Case 1: Normal Distribution
ii. Let us find P(wages between Rs. 69 and Rs. 72) i.e. P(69≤x≤72)
Now, when x1=69,
z1 =
69 − 70
5
= −0.2
And x2= 72,
z2 =
72 − 70
5
= 0.4
P(69≤x≤72)= P(-0.2≤z≤0.4)
= P(-0.2≤z≤0)+P(0≤z≤0.4)
= P(0≤z≤0.2)+P(0≤z≤0.4) - (By Symmetry)
= 0.0793+0.1554=0.2347 -(From the z Table)
Probability that the workers will have weekly wages between Rs. 69 and Rs. 72 is
0.2347
Hence, the number of such workers among 1000 workers will be-
f(69≤x≤72)=1000* P(69≤x≤72)=1000*0.2347=234.7 i.e. 235
www.sanjivanimba.org.in
Case 1: Normal Distribution
iii. Let us find P(wages more than Rs. 75) i.e. P(x≥75)
Now, when x=75,
z1 =
75 − 70
5
= 1
P(x≥75) = P(z ≥ 1) = P(0≤x≤∞) - P(0≤x≤1)
= 0.5 - 0.3413 = 0.1587 -(From the z Table)
Probability that the workers will have weekly wages more
than Rs. 75 is 0.1587
Hence, the number of such workers among 1000 workers
will be-
f(x≥75)=1000* P(x≥75)=1000*0.1587=158.7 i.e. 159
www.sanjivanimba.org.in
Case 1: Normal Distribution
iv. Let us find P(wages less than Rs. 63) i.e. P(x≤63)
Now, when x=63,
z1 =
63 − 70
5
= −1.4
P(x ≤ 63) = P(z ≤ -1.4) = P(z ≥ 1.4) = P(0≤z≤∞) - P(0≤z≤1.4)
= 0.5 - 0.4192 = 0.0808 -(From the z Table)
Probability that the workers will have weekly wages less
than Rs. 63 is 0.0808
Hence, the number of such workers among 1000 workers
will be-
f(x ≤ 63)=1000* P(x ≤ 63)=1000*0.0808=80.8 i.e. 81
www.sanjivanimba.org.in
EXERCISE
 Explain case 1: Normal Probability Distribution
www.sanjivanimba.org.in
For More Details Contact
Dr. V M Tidake
tidkevishal@gmail.com
tidkevishalmba@sanjivani.org.in

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5.2.7 case 1 normal probability distribution

  • 1. www.sanjivanimba.org.in 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
  • 2. www.sanjivanimba.org.in 302-DECISION SCIENCE Unit No.5. Probability 5.2.7 Case 1: 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
  • 3. www.sanjivanimba.org.in NETWORK ANALYSIS  At the End of the Session Student will be able to understand- A. Case 1: Normal Probability Distribution
  • 4. www.sanjivanimba.org.in Probability Distribution The weekly wages of 1000 workers are normally distributed around a mean of Rs. 70 and Standard Deviation of Rs. 5. Estimate the number of workers whose weekly wages will be- i. Between Rs. 70 and Rs. 72 ii. Between Rs. 69 and Rs. 72 iii. More than Rs. 75 iv. Less than Rs. 63
  • 5. www.sanjivanimba.org.in Case 1: Normal Distribution Let x denote the weekly wages in Rupees. Thus, x is a normal variable with mean, m=70 and 𝜎=5 Standard Normal variate corresponding to x is- z = 𝑥 − 𝑚 𝜎 = 𝑥 − 70 5
  • 6. www.sanjivanimba.org.in Case 1: Normal Distribution i. Let us find P(wages between Rs. 70 and Rs. 72) i.e. P(70≤x≤72) Now, when x1=70, z1 = 70 − 70 5 = 0 And x2= 72, z2 = 72 − 70 5 = 0.4 P(70≤x≤72)= P(0≤z≤0.4)=0.1554 -(From the z Table) Probability that the workers will have weekly wages between Rs. 70 and Rs. 72 is 0.1554 Hence, the number of such workers among 1000 workers will be- f(70≤x≤72)=1000* P(70≤x≤72)=1000*0.1554=155.4 i.e. 155
  • 7. www.sanjivanimba.org.in Case 1: Normal Distribution ii. Let us find P(wages between Rs. 69 and Rs. 72) i.e. P(69≤x≤72) Now, when x1=69, z1 = 69 − 70 5 = −0.2 And x2= 72, z2 = 72 − 70 5 = 0.4 P(69≤x≤72)= P(-0.2≤z≤0.4) = P(-0.2≤z≤0)+P(0≤z≤0.4) = P(0≤z≤0.2)+P(0≤z≤0.4) - (By Symmetry) = 0.0793+0.1554=0.2347 -(From the z Table) Probability that the workers will have weekly wages between Rs. 69 and Rs. 72 is 0.2347 Hence, the number of such workers among 1000 workers will be- f(69≤x≤72)=1000* P(69≤x≤72)=1000*0.2347=234.7 i.e. 235
  • 8. www.sanjivanimba.org.in Case 1: Normal Distribution iii. Let us find P(wages more than Rs. 75) i.e. P(x≥75) Now, when x=75, z1 = 75 − 70 5 = 1 P(x≥75) = P(z ≥ 1) = P(0≤x≤∞) - P(0≤x≤1) = 0.5 - 0.3413 = 0.1587 -(From the z Table) Probability that the workers will have weekly wages more than Rs. 75 is 0.1587 Hence, the number of such workers among 1000 workers will be- f(x≥75)=1000* P(x≥75)=1000*0.1587=158.7 i.e. 159
  • 9. www.sanjivanimba.org.in Case 1: Normal Distribution iv. Let us find P(wages less than Rs. 63) i.e. P(x≤63) Now, when x=63, z1 = 63 − 70 5 = −1.4 P(x ≤ 63) = P(z ≤ -1.4) = P(z ≥ 1.4) = P(0≤z≤∞) - P(0≤z≤1.4) = 0.5 - 0.4192 = 0.0808 -(From the z Table) Probability that the workers will have weekly wages less than Rs. 63 is 0.0808 Hence, the number of such workers among 1000 workers will be- f(x ≤ 63)=1000* P(x ≤ 63)=1000*0.0808=80.8 i.e. 81
  • 10. www.sanjivanimba.org.in EXERCISE  Explain case 1: Normal Probability Distribution
  • 11. www.sanjivanimba.org.in For More Details Contact Dr. V M Tidake tidkevishal@gmail.com tidkevishalmba@sanjivani.org.in