Chapter 5. Continuous Probability Distributions

Transcription

1 Chapter 5. Continuous Probability Distributions Section 5.6: Normal Distributions Jiaping Wang Department of Mathematical Science 03/27/2013, Wednesday

2 Outline Probability Density Function Mean and Variance More Examples Homework #9

3 Part 1. Probability Density Function

4 Probability Density Function In general, the normal density function is given by f x = 1 2 x μ exp, < x <, where the σ 2π 2σ 2 parameters μ and σ are constants (σ >0) that determines the shape of the curve.

5 Standard Normal Distribution Let Z=(X-μ)/σ, then Z has a standard normal distribution f z = 1 2π exp z2 2, < z < It has mean zero and variance 1, that is, E(Z)=0, V(Z)=1.

6 Part 2. Mean and Variance

7 Mean and Variance E Z = zf z dd = = 1 22 z eee z2 2 z z2 eee 22 2 dd = 0. dd E Z 2 = z2 z2 eee 22 2 dz = 1 22 u1/ 2 eee u 2 du = 1 22 Γ / 2 0 = 1. Then we have V(X)=E(X 2 )-E 2 (X)=1. As Z=(X-μ)/σ X=Zσ+μ E(X)=μ, V(X)=σ 2.

8 Calculating Normal Probabilities P zz < Z < zz = zz 1 exp z2 z1 dd= 0 1 2π 2 z1 for z1<0<z2. 2π zz exp z2 dd π exp z2 2 dd = AA + AA A property: P(Z<z)=1-P(Z>-z) for any z. P(z1<Z<z2)=P(0<Z<z2)-P(0<Z<z1) =A2-A1 for 0<z1<z2

9 For example, P(-0.53<Z<1.0)=P(0<Z<1.0) +P(0<Z<0.53)= = P(0.53<Z<1.2)=P(0<Z<1.2)- P(0<Z<0.53)= = P(Z>1.2)=1-P(Z<1.22)= =0.6112

10 Example 5.13 If Z denotes a standard normal variable, find the following probabilities: 1. P(Z 1.5); 2. P(Z 1.5); 3. P(Z<-2); 4. P(-2 Z 1); 5. Also find a value of z say z0 such that P(0 Z z0)=0.35. Answer: 1. P(Z 1.5)=P(Z 0)+P(0<Z<1.5)= = P(Z 1.5)=1-P(Z<1.5)= = P(Z<-2)=1-P(Z -2)=1-P(-2 Z<0)-P(0<Z)=1-P(0<Z<2)-0.5= = P(-2 Z 1)=P(-2 Z<0)+P(0<Z 1)=P(0<Z 2)+P(0<Z 1)= = z0=1.04

11 Empirical Rule 1. 68% of the values fall within 1 standard deviation of the mean in either direction; 2. 95% of the values fall within 2 standard deviation of the mean in either direction; % of the values fall within 3 standard deviation of the mean in either direction.

12 Example 5.14 A firm that manufactures and bottles apple juice has a machine that automatically fills bottles with 16 ounces of juice. (The bottle can hold up to 17 ounces.) Over a long period, the average amount dispensed into the bottle has been 16 ounces. However, there is variability in how much juice is put in each bottle; the distribution of these amounts has a standard deviation of 1 ounces. If the ounces of fill per bottle can be assumed to be normally distributed, find the probability that the machine will overflow any one bottle. Answer: Let X denote the amount of liquid (in ounces) dispensed into one bottle by the Filling machine. Then X is following the normal distribution with mean 16 and standard Deviation 1. So we are interested in the probability that a bottle will overflow if the Machine attempts to put more than 17 ounces in it. P(X>17) = P((X-μ)/σ>(17-16)/1)=P(Z>1)=

13 Example 5.15 Suppose that another machine similar to the one described in Example 5.14 is operating in such a way that the ounces of fill have a mean value equal to the dial setting for amount of liquid but also has a standard deviation of 1.2 ounces. Find the proper setting for the dial so that the 17-ounce bottle will overflow only 5% of the time. Assume that the amount dispensed have a normal distribution. Answer: Let X denote the amount of liquid dispensed; we look for a value of μ so that P(X>17)=0.05, which is equivalent to P((X-μ)/1.2>(17- μ)/1.2)=0.05 or P(Z>z0)=0.05 with z0=(17- μ)/1.2. We know that when z0=1.645, P(Z>z0)=0.05, so (17- μ)/1.2=1.645 μ=

14 Part 3. More Examples

15 Additional Example 1 Let X be a normal random variable with mean 1 and variance 4. Find P(X 2-2X 8). Answer: P(X 2-2X 8)=P(X 2-2X +1 9)=P[(x-1) 2 9] = P(-3 (x-1) 3) =P(-3/2 (x-1)/2 3/2)=P(-1.5 Z 1.5)=2P(0 Z 1.5)=2(0.4332)=0.8664

16 Additional Example 2 Suppose that X is a normal random variable with parameters μ= 5, σ 2 = 49. Using the table of the normal distribution, compute: (a) P(X > 5.5); (b) P(4 < X < 6.5); (c) P(X < 8); (d) P( X-7 4). Answer: μ=5, σ=7. a). P(X>5.5)=P((X- μ)/ σ>(5.5-5)/7)=p(z>0.0714)=0.5-p(0<z<0.074)= = b). P(4<X<6.5)=P((4-5)/7<Z<(6.5-5)/7)=P( <Z<0.2143) =P(0<Z<0.2143)+P(0<Z<0.1429)= c). P(X<8)=P(Z<3/7)=P(Z<0.4286)=P(Z<0)+P(0<Z<0.4286)= = d). P( X-7 4)=P(X-7 4)+P(X-7-4)=P(X 11)+P(X 3)=P(Z 6/7)+P(Z -2/7) =P(Z 0.86)+P(Z -0.29)=0.5-P(0 Z 0.86)+0.5-P(0 Z 0.29) = =

17 Homework #9 Page : 5.41, 5.42, 5.46 Page 226: 5.60 (Optional) Page 232: 5.67 Page 251: 5.82, 5.84.