Presentation on theme: "4.3 NORMAL PROBABILITY DISTRIBUTIONS"— Presentation transcript:
1 4.3 NORMAL PROBABILITY DISTRIBUTIONS The Most Important Probability Distribution in Statistics
2 Normal DistributionsA random variable X with mean m and standard deviation s is normally distributed if its probability density function is given by
3 The Shape of Normal Distributions Normal distributions are bell shaped, and symmetrical around m.90m110Why symmetrical? Let m = Suppose x = 110.Now suppose x = 90
4 Normal Probability Distributions The expected value (also called the mean) E(X) (or ) can be any numberThe standard deviation can be any nonnegative numberThe total area under every normal curve is 1There are infinitely many normal distributions
6 The effects of m and sHow does the standard deviation affect the shape of f(x)?s= 2s =3s =4How does the expected value affect the location of f(x)?m = 10m = 11m = 12
7 µ = the mean of the distribution = the standard deviation X836912µ = 3 and = 1A family of bell-shaped curves that differ only in their means and standard deviations.µ = the mean of the distribution = the standard deviation
10 P(6 < X < 8)µ = 6 and = 2X36912XProbability = area under the density curveP(6 < X < 8) = area under the density curve between 6 and 8.ab
11 Probability = area under the density curve P(6 < X < 8)µ = 6 and = 2X369128XProbability = area under the density curveP(6 < X < 8) = area under the density curve between 6 and 8.ab6868
12 XProbability = area under the density curveP(6 < X < 8) = area under the density curve between 6 and 8.ab6868
13 Probabilities: area under graph of f(x) P(a < X < b) f(x) X P(a < X < b) = area under the density curve between a and b.P(X=a) = 0P(a < x < b) = P(a < x < b)
14 Standardizing Suppose X~N( Form a new random variable by subtracting the mean from X and dividing by the standard deviation :(XThis process is called standardizing the random variable X.
15 Standardizing (cont.)(X is also a normal random variable; we will denote it by Z:Z = (Xhas mean 0 and standard deviation 1: E(Z) = = 0; SD(Z) = = 1.The probability distribution of Z is called the standard normal distribution.
16 Standardizing (cont.)If X has mean and stand. dev. , standardizing a particular value of x tells how many standard deviations x is above or below the mean .Exam 1: =80, =10; exam 1 score: 92Exam 2: =80, =8; exam 2 score: 90Which score is better?
17 X 3 6 9 12 µ = 6 and = 2 Z 1 2 3 -1 -2 -3 .5 µ = 0 and = 1 (X-6)/2 836912µ = 6 and = 2(X-6)/2Z123-1-2-3.5µ = 0 and = 1
18 Pdf of a standard normal rv A normal random variable x has the following pdf:
19 Standard Normal Distribution Z123-1-2-22.214.171.124Z = standard normal random variable = 0 and = 1
20 Important Properties of Z #1. The standard normal curve is symmetric around the mean 0#2. The total area under the curve is 1;so (from #1) the area to the left of 0 is 1/2, and the area to the right of 0 is 1/2
21 Finding Normal Percentiles by Hand (cont.) Table Z is the standard Normal table. We have to convert our data to z-scores before using the table.The figure shows us how to find the area to the left when we have a z-score of 1.80:
22 Areas Under the Z Curve: Using the Table P(0 < Z < 1) = = .3413Z.1587.50.34131
23 Standard normal probabilities have been calculated and are provided in table Z.P(- <Z<z0)The tabulated probabilities correspondto the area between Z= - and some z0Z = z0z0.000.010.020.030.040.050.060.070.080.090.00.50000.50400.50800.51200.51600.51990.52390.52790.53190.53590.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.57530.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141…10.84130.84380.84610.84850.85080.85310.85540.85770.85990.86211.20.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015
24 Example – continued X~N(60, 8) 0.89440.89440.8944=0.8944P(z < 1.25)0.89440.8944In this example z0 = 1.25z0.000.010.020.030.040.050.060.070.080.090.00.50000.50400.50800.51200.51600.51990.52390.52790.53190.53590.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.57530.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141…10.84130.84380.84610.84850.85080.85310.85540.85770.85990.86211.20.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015
35 P( Z < 2.16) = .9846Z.1587.9846Area=.5.48462.16
36 ExampleRegulate blue dye for mixing paint; machine can be set to discharge an average of ml./can of paint.Amount discharged: N(, .4 ml). If more than 6 ml. discharged into paint can, shade of blue is unacceptable.Determine the setting so that only 1% of the cans of paint will be unacceptable