Lesson 5: Continuous Random Variables 5.2 The Normal Ran

Transcription

1 Lesson 5: Continuous Random Variables 5.2 The Normal Random Variable September 18, 2013

2 Goals will be discusses, extensively. are ubiquitous in statistics. Work with two types of problems Compute Probability Compute cut-off values.

3 Fundamentals of The most commonly encountered random variable in nature and real life is the normal random variable. A Normal random variable X is characterized by the fact that the graph of its pdf y = f(x) has a perfect bell shape. The graph is symmetric around the vertical line x = µ, through the mean µ. In fact, the pdf f(x) is given by f(x) = 1 e (x µ) π < x <

4 Continued Preview We will not use the formula. However, note that f(x) is fully determined, if we know µ and. Using this formula and that of mean and variance, one can prove that µ is the mean and 2 is the variance of X. Review Animations Observe, in Animation 5.2.1, mean µ is the x coordinate of the line of symmetry. Also, the standard deviation is the distance from the line of symmetry (x = µ) to the point on the bell curve, where the concavity changes. We also write X N(µ,) or say X is N(µ,) random variable to mean that X is a normal random variable, with mean µ and standard deviation.

5 Standard Definition. A normal random variable Z with mean µ = 0 and standard deviation = 1 (i.e. a N(0, 1) random variable) is called a Random Variable. The notation Z is used fairly consistently to denote a standard normal random variable.

6 Continued Preview Following are some comments about importance, properties and usage of N(0, 1) random variables, which reinterpret the above properties of N(µ, ) random variables. The graph of the pdf y = f(x) of the standard random variable Z is symmetric around the y axis. Total area under the graph and above the x axis is 1. On each side of the y axis the corresponding area is.5. Review Animations 5.2.2, for the standard normal random variable and to computation of probability.

7 Continued Important comments: The next theorem reduces any problem to compute proability for any X N(µ,) to that for standard normal (Z N(0,1)). The process is called standardization. The proability tables of standard random variable, used to play a big role in probability computations. This became outdated due to the advent of simple and sophisticated tools (like TI-84, Excel, SAS, Minitab). The normalcdf functions of TI-84 will be used to compute probability for standard normal.

8 Standardization Preview Theorem: Standardization: Let X be a N(µ, ) random variable. Then Z = X µ is a standard random varable. So, ( a µ P(a X b) = P Z b µ ) To compute the right hand side, the normalcdf function, under the Distr -menu of TI-84, will be used. We will be working on two types of problems. First, we deal with probability computation. Reading Assignment: Exercise

9 Exercise (edited) Let X be a normal random variable with mean µ = 4.5 and standard deviation = 1.7. (1) Compute P(3 X 6.5) (2) Compute P(X 3.5) (3) Compute P(1 X 3) (4) Compute P(2.5 X)

10 Solution: Preview Solution: Here the mean µ = 4.5 and standard deviation = 1.7 (1) ( 3 µ P(3 X 6.5) = P X µ 6.5 µ ) ( = P Z ) = P (.8824 Z ) = normalcdf(.8824, ) =.6915

11 Solution: Preview (2) ( X µ P(X 3.5) = P 3.5 µ ) ( = P Z ) = P (Z.5882) 1.7 = normalcdf( 5,.5882) =.2782

12 Solution: Preview (3) ( 1 µ P(1 X 3) = P X µ 3 µ ) ( = P Z ) = P ( Z.8824) = normalcdf( ,.8824,) =.1690

13 (4) ( 2.5 µ P(2.5 X) = P ( = P 1.7 X µ ) ) Z = P ( Z) = normalcdf( , 5) =.8803

14 Exercise Preview The diameter of the pumpkins in my patch has normal distribution with mean 13 inches and standard deviation 4.5 inches. WWhat proportion (i.e., probability) of pumpkins is above 22 inches? Solution: Here the mean µ = 13 and standard deviation = 4.5. Let X = diameter of the pumpkins. Then X N(13,4.5). ( 22 µ P(22 X) = P X µ ) ( ) 22 µ = P Z ( ) = P Z = P(2 Z) = normalcdf(2,5) =

15 Exercise Preview The weight X at birth weight of babies is normally distributed with mean µ = 114 oz and standard deviation = 18 oz. What percent of babies will have birth weight below 141 oz? Solution: Here the mean µ = and standard deviation = 18. Let X = birth weight of babies. Then X N(114,18). ( X µ P(X 141) = P 141 µ ) ( = P Z 141 µ ) ( = P Z ) = P (Z 1.5) 18 = normalcdf( 5,1.5) =.9332 Which is percent.

16 Inverse Probability and Cut-Off Values In certain situation, the probability is known and the variable values would have to be determined. This concept is called inverse probability or Cut-Off Values. Following is an example of such a situation.

17 Example Most common examples of inverse probability or cut-off values would be percentiles. Suppose X is the birth weight of babies in a county. The 75th percentile x 75 of the birth weight (or for an age group), is given by the equation P(X x 75 ) =.75. Remark. Pediatricians have charts of percentile weights. These charts have age on the horizontal axis and weight on the vertical axis. Several graphs are given in the chart, representing various percentiles (I guess, one for each 5 percent). So, there will be one graph for each of 5th, 10th and up to 95th or 100th percentile.

18 Example Preview The annual income X in a county is normally distributed with mean µ = $37,000 and standard deviation = $15,000. Your annual income is $75,000. Do you think that your annual income is above 90 percentile? Solution: Here the mean µ = 37,000 and standard deviation = 15,000. We have X N(37000,15000). Denote x 90 = 90 th percentile of the annual income. So, ( X µ P(X x 90 ) =.90 OR P x ) 90 µ =.90

19 P ( Z x ) 90 µ =.90 Also P(Z < ) =.90 The last equality follows: TI-84, invnorm(.90) = By comparing the above two equations x 90 µ = So, x 90 = µ = = That means that the 90th percentile of the annual income is x 90 = $ We conclude your annual income $75,000 is above 90th percentile.

20 Exercise Preview The length X of a fish in a lake has normal distribution with mean 67 cm and standard deviation 21 cm. On a fishing trip to the lake, you are instructed to keep only those in the upper 33 percent in length. What is the cut-off length, above which you are permitted keep? Solution: Here the mean µ = 67 and standard deviation = 21 Let X = the length of fish in the lake. Then X N(67,21). Let L = the cut-off length. Then, we are given P(L < X) =.33 The way invnormal is used, we like to work with P(X L) = 1 P(L < X) = 1.33 =.67

21 So, Ss, P ( X µ P L µ ) =.67 ( Z L µ ) =.67 Also P(Z.4399) =.67 The last equality follows: TI-84, invnorm(.67) =.4399 Comparing two equations L µ =.4399 So, L = µ = = cm.

22 Exercise Preview The weight X of babies (of a fixed age) is normally distributed with with mean µ = 212 oz and standard deviation = 25 oz. Doctors would be concerned (not necessarily alarmed) if a baby is among the lower 5 percent in weight. Find the cut-off weight L below which the doctors will be concerned. Solution: Here the mean µ = 212 and standard deviation = 25 Let X = the weight of the babies of this age group. Then X N(212,25) Let U = the cut-off length below which doctors will be concerned. Then, we have ( X µ P(X U) =.05 OR P U µ ) =.05

23 So, ( P Z U µ ) =.05 Also P(Z < ) =.05 Last equality, From TI-84, invnorm(.05) = Comparing two equations, So, U µ = U = µ+( ) = 212+( ) 25 = oz.

24 Exercise Preview Monthly water consumption X per household, in a subdivision in Kansas City, has normal distribution with mean gallons and standard deviation 3000 gallons. It has been decided that a surcharge will be imposed for those in the top 25 percent. Find the cut-off consumption L in gallons. Solution: Here the mean µ = and standard deviation = 3000 Let X = Monthly water consumption by households Then X N(15000,3000). Let L = the cut-off. Then, we are given P(L < X) =.25 The way invnormal is used, we like to work with P(X L) = 1 P(L < X) = 1.25 =.75

25 So, Ss, P ( X µ P L µ ) =.75 ( Z L µ ) =.75 Also P(Z.6745) =.75 The last equality follows: TI-84, invnorm(.75) =.6745 Comparing two equations L µ =.6745 So, L = µ = =