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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
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 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.
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 =12𝜋
exp −𝑧2
2 ,−∞ < 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
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.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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
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=0.9332 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 5.14
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
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
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 223-224: 5.41, 5.42, 5.46 Page 226: 5.60 (Optional) Page 232: 5.67 Page 251: 5.82, 5.84.
