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Chapter 5. Continuous Probability Distributions Section 5.6: Normal Distributions Jiaping Wang Department of Mathematical Science 03/27/2013, Wednesday The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Outline Probability Density Function Mean and Variance More Examples Homework #9 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 1. Probability Density Function The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Probability Density Function In general, the normal density function is given by = 1 exp 2 − − 22 2 , −∞ < < ∞, where the parameters μ and σ are constants (σ >0) that determines the shape of the curve. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Standard Normal Distribution Let Z=(X-μ)/σ, then Z has a standard normal distribution 1 2 = exp − , −∞ < < ∞ 2 2 It has mean zero and variance 1, that is, E(Z)=0, V(Z)=1. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 2. Mean and Variance The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Mean and Variance ∞ ∞ = = − −∞ −∞ ∞ = ∙ − = . −∞ ∞ = −∞ − = ∞ / − = Γ / = . Then we have V(X)=E(X2)-E2(X)=1. As Z=(X-μ)/σX=Zσ+μE(X)=μ, V(X)=σ2. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Calculating Normal Probabilities 1 < < 2 = 2 1 2 exp − = 1 2 for z1<0<z2. 2 0 1 exp 1 2 2 − 2 + 2 1 exp 0 2 2 − 2 = 1 + 2 P(z1<Z<z2)=P(0<Z<z2)-P(0<Z<z1) =A2-A1 for 0<z1<z2 A property: P(Z<z)=1-P(Z>-z) for any z. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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)=10.3888=0.6112 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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)=0.5+0.4332=0.9332 2. P(Z≥1.5)=1-P(Z<1.5)=1-0.9332=0.0668 3. 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. 4. 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 5. z0=1.04 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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; 3. 99.7% of the values fall within 3 standard deviation of the mean in either direction. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 3. More Examples The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Homework #9 Page Page Page Page 223-224: 5.41, 5.42, 5.46 226: 5.60 (Optional) 232: 5.67 251: 5.82, 5.84. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL