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Chapter 5. Continuous Probability Distributions Section 5.6: Normal Distributions

This chapter explores normal distributions, including probability density functions, mean and variance calculations, and examples of calculating normal probabilities.

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Chapter 5. Continuous Probability Distributions Section 5.6: Normal Distributions

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  1. Chapter 5. Continuous Probability DistributionsSection 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. Part1. Probability Density Function

  4. Probability Density Function In general, the normal density function is given by here the 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 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 Then we have V(X)=E(X2)-E2(X)=1. As Z=(X-μ)/σX=Zσ+μE(X)=μ, V(X)=σ2.

  8. Calculating Normal Probabilities =+ for z1<0<z2. 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)=0.3159+0.2019=0.5178 P(0.53<Z<1.2)=P(0<Z<1.2)-P(0<Z<0.53)=0.3849-0.2019 =0.1830 P(Z>1.2)=1-P(Z<1.22)=1-0.3888=0.6112

  10. Example 5.13 If Z denotes a standard normal variable, find the following probabilities: P(Z≤1.5); P(Z≥1.5); P(Z<-2); P(-2≤Z≤1); Also find a value of z – say z0 – such that P(0≤Z≤z0)=0.35. • Answer: • P(Z≤1.5)=P(Z≤0)+P(0<Z<1.5)=0.5+0.4332=0.9332 • P(Z≥1.5)=1-P(Z<1.5)=1-0.9332=0.0668 • P(Z<-2)=1-P(Z≥-2)=1-P(-2≤Z<0)-P(0<Z)=1-P(0<Z<2)-0.5=0.5-0.4772=0.228. • P(-2≤Z≤1)=P(-2≤Z<0)+P(0<Z≤1)=P(0<Z≤2)+P(0<Z≤1)=0.4772+0.3413=0.8185 • z0=1.04

  11. Empirical Rule • 68% of the values fall within 1 standard deviation of the mean in either direction; • 95% of the values fall within 2 standard deviation of the mean in either direction; • 99.7% of the values fall within 3 standard deviation of the mean in either direction.

  12. 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=15.026.

  13. 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)=0.1587.

  14. Part 3. More Examples

  15. Additional Example 1 Let X be a normal random variable with mean 1 and variance 4. Find P(X2-2X ≤ 8). Answer: P(X2-2X ≤ 8)=P(X2-2X +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, σ2= 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)=0.5-0.0279=0.4721 b). P(4<X<6.5)=P((4-5)/7<Z<(6.5-5)/7)=P(-0.1429<Z<0.2143) =P(0<Z<0.2143)+P(0<Z<0.1429)=0.0832+0.0557+0.1389 c). P(X<8)=P(Z<3/7)=P(Z<0.4286)=P(Z<0)+P(0<Z<0.4286)=0.5+0.1664=0.6664 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) =1- 0.3054 – 0.1141= 0.5805.

  17. Homework #9 Page 223-224: 5.41, 5.42, 5.46 Page 226: 5.60 (Optional) Page 232: 5.67 Page 251: 5.82, 5.84.

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