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The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Chapter 5. Continuous Probability Distributions Section 5.6: Normal Distributions Jiaping Wang Department of Mathematical Science 03/27/2013, Wednesday
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Chapter 5. Continuous Probability Distributions Section 5.6: Normal Distributions . Jiaping Wang Department of Mathematical Science 03/27/2013, Wednesday. Outline. Probability Density Function Mean and Variance More Examples Homework #9. Part 1 . Probability Density Function. - PowerPoint PPT Presentation

### Transcript of Chapter 5. Continuous Probability Distributions Section 5.6: Normal Distributions The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

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 byhere 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

It has mean zero and variance 1, that is, E(Z)=0, V(Z)=1.

𝑓 (𝑧 )= 1√2𝜋

exp (− 𝑧 22 ) ,−∞< 𝑧<∞ The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Part 2. Mean and Variance The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Mean and Variance

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

=+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 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)=1-0.3888=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=0.93322. P(Z≥1.5)=1-P(Z<1.5)=1-0.9332=0.06683. 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=0.81855. 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 toP((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 theFilling 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 theMachine 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

Let X be a normal random variable with mean 1 and variance 4. FindP(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

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=0.4721b). 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+0.1389c). P(X<8)=P(Z<3/7)=P(Z<0.4286)=P(Z<0)+P(0<Z<0.4286)=0.5+0.1664=0.6664d). 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 223-224: 5.41, 5.42, 5.46Page 226: 5.60 (Optional)Page 232: 5.67Page 251: 5.82, 5.84.