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Normal distribution

Normal Distribution Ppt

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Normal distribution

  1. 1. Normal Distribution
  2. 2. Definition •It is defined as a continuous frequency distribution of infinite range. •The normal distribution is a descriptive model that describes real world situations.
  3. 3. Importance • Many dependent variables are commonly assumed to be normally distributed in the population • If a variable is approximately normally distributed we can make inferences about values of that variable
  4. 4. Characteristics •Bell-Shaped
  5. 5. •Mean=Median=Mode
  6. 6. •Curve is asymptotic to the x- axis •Total area under the curve above the x-axis = 1 or 100%
  7. 7. Empirical Rule: 68% of the data 95% of the data 99.7% of the data
  8. 8. •Mathematical Formula Note constants: =3.14159 e=2.71828 The Normal Distribution: as mathematical function (pdf) 2 )( 2 1 2 1 )(       x exf This is a bell shaped curve with different centers and spreads depending on  and 
  9. 9. The Standard Normal Distribution (Z) •The mean (μ ) = 0 •Standard deviation (σ) =1 )1,0(~),(~ N x ZNX     
  10. 10. 11 33
  11. 11. Probabilities are depicted by areas under the curve • Total area under the curve is 1 • The area in red is equal to p(z > 1) • The area in blue is equal to p(-1< z <0)
  12. 12. Finding Areas under the Normal Curve •Example: ▫Find the area to the right of: 1. z= 1.25 2. z=-.34
  13. 13. •Example: ▫Find the area to the left of: 1. z=0.33 2. z=-0.21
  14. 14. •Example: ▫Find the area between: 1. z = -0.24 and z = 1.23 2. z = 0.15 and z = 2.04 3. z = -0.15 and z = -2.02
  15. 15. Applications of the Normal Distribution • Example: ▫ DGP University conducts placement examination to all incoming freshmen. The examination scores of the 1000 examinees last semester were approximately normally distributed with mean score of 80 and standard deviation of 5. What is the probability that randomly chosen student got a score below 70?above 82? Between 75 and 90?
  16. 16. Solution: a.below 70 •Given: ▫μ= 80 ▫σ =5 ▫x<70    x Z 5 8070  Z Z= -2
  17. 17. • P [x<70] = P[z<-2] = 0.5 – P[0<z<2] =0.5 – 0.4772 =0.0228 or 2.28% -2
  18. 18. Solution: above 82 5 8082  Z = 0.4
  19. 19. • P [x>82] = P[z>0.4] = 0.5 – P[0>z>0.4] =0.5 – 0.1554 =0.3446 or 34.46% 0.4
  20. 20. Exercise: (Using the same problem) Sketch of the curve and find: •Between 75 and 90 (Ronald and Jyx,) •How many got scores below 87? (Nikki,Dioniel ) above 77? (Frea,Gretchel) between 75 and 90 (Vhea and Sheryl)
  21. 21. Evaluation: (1/2) • X is a normally normally distributed variable with mean μ = 30 and standard deviation σ = 4. Find a) P(x < 40) b) P(x > 21) c) P(30 < x < 35)
  22. 22. • Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.)
  23. 23. Agreement: •A firms marketing manager believes that total sales for the firm next year can be modeled using a normal distribution, with a mean of P 2.5 million and a standard deviation of P300,000
  24. 24. a. What is the probability that the firm’s sales will exceed the P3 million? b. In order to cover fixed cost, the firm’s sales must exceed the break-even level of P 1.8 million. What is the probability that sales will exceed the break-even level?

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