6.2 Normal Distribution. Ziad Zahreddine

Transcription

1 6.2 Normal Distribution

2 Importance of Normal Distribution 1. Describes Many Random Processes or Continuous Phenomena 2. Can Be Used to Approximate Discrete Probability Distributions Example: Binomial 3. Basis for Classical Statistical Inference

3 Normal Distribution 1. Bell-Shaped & Symmetrical 2. Mean, Median, Mode Are Equal 3. Random Variable Has Infinite Range that cannot be listed. Mean Median Mode

4 Probability Density Function The Probability Density Function n (x; μ, σ) is given by n where ( ) ( 1 2) ( ) x; μ, σ = e x μ σ x = Value of Random Variable (- < x < ) μ = Mean of the random variable x σ = Population Standard Deviation π = ; e = π [ σ ] 2

5 Effect of Varying Parameters (μ & σ) The mean and standard deviation affect the flatness and center of the curve, but not the basic shape.

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7 6.3 Areas Under the Normal Curve Probability is area under curve! ( ) ( ) [ ] dx e d X c P d c x = < < ) ( σ μ π σ? ), ; ( ) ( = < < d c dx x n d X c P σ μ

8 Standardize the Normal Distribution

9 Standardize the Normal Distribution n 1 ( 1 2) ( x μ ) ( x; μ, σ ) = e σ 2π Normal Distribution [ σ ] 2 Infinitely Many Distributions

10 Standardize the Normal Distribution n 1 ( 1 2) ( x μ ) ( x; μ, σ ) = e σ 2π Normal Distribution [ σ ] 2 n 1 ( 1 2) z ( z;0,1) = e 2 2π Standard Normal Distribution Infinitely Many Distributions Only One Distribution

11 The Standard Normal Distribution Table for Standard Normal Distribution contains probability for the area between 0 and z. Partial table below shows components of table Value of z a combination of column and row Probability associated with a particular z value, in this case z=.13, p(0<z<.13) =.0517 Z

12 The Standard Normal Distribution What is P(-1.33 < z < 1.33)? Table gives us area A 1 Symmetry about the mean tell us that A 2 = A 1 P(-1.33 < z < 1.33) = P(-1.33 < z < 0) +P(0 < z < 1.33)= A 2 + A 1 = =.8164

13 The Standard Normal Distribution What is P(z > 1.64)? Table gives us area A 2 Symmetry about the mean tell us that A 2 + A 1 =.5 P(z > 1.64) = A 1 =.5 A 2 = =.0505

14 The Standard Normal Distribution What is P(z <.67)? Table gives us area A 1 Symmetry about the mean tell us that A 2 =.5 P(z <.67) = A 1 + A 2 = =.7486

15 The Standard Normal Distribution What is P( z > 1.96)? P( z > 1.96) = P(z < or z > 1.96) Table gives us area.5 - A 2 =.4750, so A 2 =.0250 Symmetry about the mean tell us that A 2 = A 1 P( z > 1.96) = A 1 + A 2 = =.05

16 Given a standard normal distribution, find the area under the curve that lies (a) to the right of z = 1.84 and (b) Between z = and z = Solution (a) The area to right of z = 1.84 is equal to 0.5 minus the area in the table. That is = (b) The area between 0 and 0.86 is By symmetry the area between and 0 is the same as the area between 0 and 1.97, which is the total area is =

17 Given a standard normal distribution, find the value of k such that (a) P(Z > k) = , and (b) P(k <Z< -0.18) = Solution (a) k lies to the right of 0. Why? The total area to the right of zero is = According to the table, the value of k such that P(0 < Z < k) = is k = (b) The area between and 0 is the same as the area between 0 and 0.18 which is The area between k and 0 becomes: = From the table, the symmetric value of k is Hence k = k k

18 The Normal Distribution What if values of interest were not standarized? We want to know P (8 < x < 12), with μ = 10 and σ = 1.5 Convert to standard normal using x = 8 z = (8 10)/1.5 = x = 12 z = (12 10)/1.5 = 1.33 z = x μ σ P(8 < x < 12) = P(-1.33 < z < 1.33) = 2(.4082) =.8164

19 The Normal Distribution We want to know P (45 < x < 62), with μ = 50 and σ = 10 Convert to standard normal using x = 45 z = (45 50)/10 = -0.5 x = 62 z = (62 50)/10 = 1.2 z = x μ σ P(45 < x < 62) = P(-0.5 < z < 1.2) = =

20 The Normal Distribution Steps for Finding a Probability Corresponding to a Normal Random Variable Sketch the distribution, locate mean, shade area of interest Convert to standard z values using z = x μ Add z values to the sketch σ Use tables to calculate probabilities, making use of symmetry property where necessary

21 The Normal Distribution Given that X has a normal distribution with mean of 27 and standard deviation of 3, find the probability that X assumes a value less than 20. Z value for x = 20 is P(x < 20) = P(z < -2.33) = =.0099 You could reasonably conclude that this is a rare event

22 The Normal Distribution You can also use the table in reverse to find a z-value that corresponds to a particular probability What is the value of z that will be exceeded only 10% of the time? Look in the body of the table for the value closest to.4, and read the corresponding z value Z = 1.28

23 The Normal Distribution Which values of z enclose the middle 95% of the standard normal z values? Using the symmetry property, z 0 must correspond with a probability of.475 From body of the table, we find that z 0 and z 0 are 1.96 and respectively.

24 The Normal Distribution Given a normally distributed variable x with mean 100,000 and standard deviation of 10,000, what value of x identifies the top 10% of the distribution? P ( 100,000 x x0 )= x0 μ x P 0 z = 0 100,000 P 0 z =. 40 σ 10,000 The z value corresponding with.40 is Solving for x 0 x 0 = 100, (10,000) = 100, ,800 = 112,800

25 6.4 Applications of the Normal Distribution Example A certain type of storage battery lasts, on average, 3 years, with a standard deviation of 0.5 year. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years. Solution z = = P( X < 2. 3)= P ( Z < 1. 4)= P( Z >1.4) = =

26 Example An electrical firm manufactures light bulbs that have a life, before burn-out, that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a bulb burns between 778 and 834 hours. Solution z1 = = 0.55 and z2 = = P ( 778 < X < 834) = P( 0.55 < Z < 0.85) = = σ = μ = 800

27 Example In an industrial process, the specifications on the diameter of a ball bearing is 3 ± 0.01 cm. The implication is that no part falling outside these specifications will be accepted. It is known that the diameter of a ball bearing has a normal distribution with mean 3 and standard deviation On the average, how many manufactured ball bearings will be scrapped? Solution z1 = = 2 and z2 = = P( 0 < X < 2) = By symmetry P( X > 2) = = P( X > 2 or X < 2) = = Therefore, on the average, 4.56% of manufactured ball bearings will be scrapped.

28 Example Gauges are used to reject all components where a certain dimension is not within the specification 1.50 ± d. It is known that this measurement is normally distributed with mean 1.50 and standard deviation 0.2. Determine the value of d such that the specifications cover 95% of the measurements. Solution From the table, the Z value corresponding to the area 0.95/2 = is Z = In other words P( 0 < Z < 1.96) = ( d) 1.50 Therefore 1.96 = 0.2 From which we obtain d = ( 0.2)(1.96) = d d

29 Example A machine makes electrical resistors having a mean resistance of 40 Ohms and a standard deviation of 2 Ohms. Assuming that the resistance follows a normal distribution and an be measured to any degree of accuracy, what percentage of resistors will have a resistance exceeding 43 Ohms? Solution The Z value corresponding to 43 is From the table P( 0 < Z < 1.5) = Therefore Z = = P (X > 43) = P (Z > 1.5) = 0.5 P (0 < Z < 1.5) = = % of the resistors will have a resistance exceeding 43 Ohms

30 Example The average grade for an exam is 74, and the standard deviation is 7. If 12% of the class are given A s, and the grades are curved to follow a normal distribution, what is the lowest possible A and the highest possible B? Solution = By examining the table, we take Z to be 0.12 the average of 1.17 and Z = = Therefore = X 7 Leading to X = ( 7)(1.175) + 74 = Therefore, the lowest A is 83 and the highest B is 82.

31 Sections 6.1, 6.2, 6.3, 6.4, P. 156 Assignment: 1, 2, 3, 4, 5, 7, 8.