1 / 18

Normal Distribution

Normal Distribution. “Bell Curve”. Characteristics of a normal distribution…. Example: The heights of males and females in the UK. Few very short people. Few very tall people. Most people around the average height. Characteristics of a normal distribution…. Example:

hedwig-hubbard
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 “Bell Curve”

  2. Characteristics of a normal distribution… Example: The heights of males and females in the UK . Few very short people Few very tall people Most people around the average height

  3. Characteristics of a normal distribution… Example: X is the r.v. “The heights, in cm, of males and females in the UK” The probability distribution for the random variable X P(X = x) P(X = x) μ = 178 cm μ = 163 cm X Height is continuous, so X is a CONTINUOUS RANDOM VARIABLE

  4. Properties of a normal distribution For a continuous random variable, X: Symmetrical distribution Total area under curve = 1 Since P(X = x) = 1 Mean = mode = median asymptote -  < X < +

  5. Parameters of a normal distribution P(X = x) depends ONLY on two things: mean (μ) and variance (σ2) X ~ N (μ, σ2) Nearly ALL values of X lie within ± 3 standard deviations away from the mean μ – 3σ μ + 3 σ

  6. Shape of a normal distribution The values of the parameters (μ and σ2) change the shape of the curve. Example: Heights (cm) of adult males and females in the UK X ~ N (μ, σ2) XM ~ N (178, 252) XF ~ N (163, 252) μF μM X Different means  distributions have different location Equal variances  distributions have same dispersion

  7. Shape of a normal distribution The values of the parameters (μ and σ2) change the shape of the curve. Example: Length (cm) of hair in adult males and females in the 1970s X ~ N (μ, σ2) XF ~ N (30, 42) XM ~ N (30, 102) μF= μM X Equal means  distributions have same location Different variances  distributions have different dispersion

  8. P(W > b) P(V < a) b a V W P(c < X < d) X c d Calculating probabilities The probabilities are VERY difficult to calculate! All probabilities depend on the values of the mean and variance. Normal distributions for different variables will differ.

  9. The standard normal distribution Any variable can be transformed (coded) so it has a mean of zero and a variance of one:

  10. The standard normal distribution Any variable can be transformed (coded) so it has a mean of zero and a variance of one: Z ~ N(0, 12) Z is the standard normal distribution, and all its associated probabilities can be found using statistical tables.

  11. Transforming your variable Any variable can be transformed (coded) so it has a mean of zero and a variance of one. Example: X ~ N(100, 152) Z ~ N(0, 12) X Z Each value of X has a corresponding value of Z after standardising When X = 115,

  12. Probabilities 1 Example 1. X ~ N(10, 42) Z ~ N(0, 12) Find P(X < 12) Z Transform X using Find P(X < 14) Similarly…

  13. Z ~ N(0, 12) 0.5 Z Using tables Your statistical tables use a notation that you need to become familiar with… Φ(z) = P(Z < z) P(Z < 0.5) = Φ(0.5) = 0.6915 Φ(0.5) means P(Z < 0.5)

  14. Z ~ N(0, 12) Z Using tables Your statistical tables use a notation that you need to become familiar with… Φ(z) = P(Z < z) P(Z < 1) = Φ(0.5) = 0.8413 1 Φ(1) means P(Z < 1)

  15. Z ~ N(0, 12) Z Using tables Your statistical tables use a notation that you need to become familiar with… Φ(z) = P(Z < z) 0.5 1 P(0.5 < Z < 1) = Φ(1) – Φ(0.5) = 0.8413 – 0.6915 = 0.1498

  16. Z ~ N(0, 12) Z When z is negative Find P(Z < -1) Use SYMMETRY properties P(Z < -1) = P(Z > 1) -1 1 BUT tables only give you probabilities LESS THAN z! P(Z < -1) = 1 – P(Z < 1) Therefore Φ(-1) = 1 – Φ(1) = 1 – 0.8413 = 0.1587

  17. Find these probabilities… • 0.1587 • 0.0228 • 0.6247 Z

  18. Find these probabilities… • 0.1151 • 0.3159 • 0.2638 Y Z Classwork: Exercise 9A Q1 – 6

More Related