1 / 20

Normal Distribution

Normal Distribution. A random variable X having a probability density function given by the formula. is said to have a Normal Distribution with parameters  and  2 . Symbolically, X ~ N(,  2 ). Properties of Normal Distribution.

bryce
Download Presentation

Normal Distribution

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Normal Distribution A random variable X having a probability density function given by the formula is said to have a Normal Distribution with parameters  and 2. Symbolically, X ~ N(, 2).

  2. Properties of Normal Distribution • The curve extends indefinitely to the left and to the right, approaching the x-axis as x increases in magnitude, i.e. as x  , f(x)  0. • The mode occurs atx=. • The curve is symmetric about a vertical axis through the mean  • The total area under the curve and above the horizontal axis is equal to 1. i.e.

  3. Empirical Rule (Golden Rule) • The following diagram illustrates relevant areas and associated probabilities of the Normal Distribution. Approximate 68.3% of the area lies within ±, 95.5% of the area lies within ±2, and 99.7% of the area lies within ±3.

  4. For normal curves with the same , they are identical in shapes but the means  are centered at different positions along the horizontal axis.

  5. For normal curves with the same mean , the curves are centered at exactly the same position on the horizontal axis, but with different standard deviations , the curves are in different shapes, i.e. the curve with the larger standard deviation is lower and spreads out farther, and the curve with lower standard deviation and the dispersion is smaller.

  6. Normal Table If the random variable X ~ N(, 2), then we can transform all the values of X to the standardized values Z with the mean 0 and variance 1, i.e. Z ~ N(0, 1), on letting

  7. Standardizing Process This can be done by means of the transformation. The mean of Z is zero and the variance is respectively,

  8. Diagrammatic of the Standardizing Process Transforms X ~ N(, 2) to Z ~ N(0, 1). Whenever X is between the values x=x1 and x=x2, Z will fall between the corresponding values z=z1 and z=z2, we have P(x1 < X < x2) = P(z1 < Z < z2). It illustrates by the following diagram:

  9. The normal table can be used to find values like P(Z > a), P(Z < b) and P(a  Z  b). We illustrate with the following examples. Example 1: P(-1.28 < Z < 0) = ? Solution: P(-1.28 < Z < 0) = P(0 < Z < 1.28) = 0.3997

  10. Example 2: P(Z < -1.28) = ? Solution: P(Z < -1.28) = P(Z > 1.28) = 0.5 – 0.3997 =0.1003

  11. Example 3: P(Z > -1.28) = ? Solution: P(Z > -1.28) = P(Z < 1.28) = 0.5 + 0.3997 = 0.8997

  12. Example 4: P(-2.28 < Z < -1.28) = ? Solution: P(-2.28 < Z < -1.28) = P(1.28 < Z < 2.28) = 0.4887 – 0.3997 = 0.0890

  13. Example 5: P(-1.28 < Z < 2.28) = ? Solution: P(-1.28 < Z < 2.28) = 0.3997 + 0.4887 = 0.8884

  14. Example 6: If P(Z > a) = 0.8, find the value of a? Solution: From the Normal Table A(0.84)  0.3  a  - 0.84

  15. Example 7: If P(Z < b) = 0.32, find the value of b? Solution: P(Z < b) = 0.32 P(b < Z < 0) = 0.5 – 0.32 = 0.18 From table, A(0.47)  0.18  b  -0.47

  16. Example 8: If P(|Z > c) = 0.1, fin the values of c? Solution: P(|Z > c) = 0.1  P(Z > c) = 0.05  P( c > Z > 0) = 0.5 – 0.05 = 0.45 From table, A(1.645)  0.45  c  1.645

  17. Transformation Example 9: If X ~ N(10, 4), find • P(X  12); • P(9.5  X  11); • P(8.5  X  9) ?

  18. Solution: (a) For the distribution of X with =10, =2

  19. Solution: (b) For the distribution of X with =10, =2 P(9.5  X  11) = P(- 0.25 Z  0.5) = 0.0987 + 0.1915 = 0.2902

  20. Solution: (c) For the distribution of X with =10, =2 P(8.5  X  9) = P(- 0.75 Z  - 0.5) = 0.2734 – 0.1915 = 0.0819

More Related