Chapter 5. Continuous Probability Distributionsmath.unt.edu/sites/ · 2013-03-27 · The UNIVERSITY...

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Chapter 5. Continuous Probability Distributions

Section 5.6: Normal Distributions

Jiaping Wang

Department of Mathematical Science

03/27/2013, Wednesday

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

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Part 1. Probability Density Function

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Probability Density Function

In general, the normal density function is given by 𝑓 𝑥 = 1

𝜎 2𝜋exp − 𝑥−𝜇 2

2𝜎2,−∞ < 𝑥 < ∞, where the

parameters μ and σ are constants (σ >0) that determines the shape of the curve.

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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 =12𝜋

exp −𝑧2

2 ,−∞ < 𝑧 < ∞

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Part 2. Mean and Variance

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Mean and Variance

𝑬 𝒁 = � 𝒛𝒇 𝒛 𝒅𝒅 = �𝒛𝟐𝟐

𝒆𝒅𝒆 −𝒛𝟐

𝟐𝒅𝒛

∞

−∞

∞

−∞

= 𝟏𝟐𝟐 ∫ 𝒛 ∙ 𝒆𝒅𝒆 − 𝒛𝟐

𝟐𝒅𝒛 = 𝟎.∞

−∞ 𝐄 𝐙𝟐

= �𝐳𝟐

𝟐𝟐𝐞𝐞𝐞 −

𝒛𝟐

𝟐𝐝𝐳 =

𝟏𝟐𝟐

� 𝒖𝟏/𝟐𝐞𝐞𝐞 −𝒖𝟐

𝐝𝐮 =𝟏𝟐𝟐

Γ𝟑𝟐

𝟐 𝟑/𝟐∞

𝟎= 𝟏.

∞

−∞

Then we have V(X)=E(X2)-E2(X)=1. As Z=(X-μ)/σX=Zσ+μE(X)=μ, V(X)=σ2.

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Calculating Normal Probabilities

𝑃 𝑧𝑧 < 𝑍 < 𝑧2 =∫ 1

2𝜋𝑧2𝑧1 exp −𝑧2

2𝑑𝑧=∫ 1

2𝜋0𝑧1 exp −𝑧2

2𝑑𝑧+∫ 1

2𝜋𝑧20 exp −𝑧2

2𝑑𝑧 = 𝐴𝑧 + 𝐴2

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

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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

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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

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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.

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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.

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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.

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Part 3. More Examples

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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

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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.

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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.