### ppt - Department of Mathematics

```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
=
1
exp
2
−
−
22
2
, −∞ <  < ∞, 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
1
2
=
exp −
, −∞ <  < ∞
2
2
It has mean zero and variance 1,
that is, E(Z)=0, V(Z)=1.
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Part 2. Mean and Variance
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Mean and Variance
∞
∞

=
=
−

−∞
−∞
∞

=
∙  −  = .
−∞

∞
=
−∞

−
=

∞

/

−  =
Γ

/
= .
Then we have V(X)=E(X2)-E2(X)=1.
As Z=(X-μ)/σX=Zσ+μE(X)=μ, V(X)=σ2.
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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.
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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)=10.3888=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.
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 5.15
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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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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Part 3. More Examples
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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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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).
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
Page
Page
Page
223-224: 5.41, 5.42, 5.46
226: 5.60 (Optional)
232: 5.67
251: 5.82, 5.84.
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```