Download Normal Distributions II

Announcements
Finite Probability
Friday, November 18th
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MyMathLab 9 is Monday Nov 28
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Problem Set 9 is due Monday Nov 28
Today: Sec. 7.6: The Normal Distribution II
Transform regions on a normal curve to the standard
normal distribution.
Solve word problems involving normally distributed
random variables.
Next Class: Sec. 6.7 Activity
Cherveny
Nov 18
Math 1004: Probability
Review: Normal Curve
Key properties of the normal curve (a.k.a. the “bell curve”):
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Center is µ.
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Width determined by σ
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Total area under a normal curve is 1
Cherveny
Nov 18
Math 1004: Probability
Review: Finding Area Under Normal Curves
Translation to the standard normal curve:
In terms of probability,
P(a ≤ X ≤ b) = P
a−µ
b−µ
≤Z ≤
σ
σ
We can use Appendix A to calculate the righthand side!
Cherveny
Nov 18
Math 1004: Probability
Normal Distribution Example
Example
Suppose that the height (at the shoulder) of adult African elephant
is normally distributed with µ = 3.3 meters and σ = .2 meter. The
elephant on display at the Smithsonian Institution has height 4
meters and is the largest elephant on record.
(a) What is the probability that an adult African elephant has
height 4 meters or more?
(b) What is the probability that in a herd of five adult elephants,
exactly 3 of them will be between 3 and 3.5 meters? (Assume
their heights are independent)
Answer: (a) .0002 (b) C (5, 3)(.7745)3 (.2255)2 ≈ .2362
Cherveny
Nov 18
Math 1004: Probability
Practice
1. The amount of gas sold weekly by a gas station is normally
distributed with µ = 30, 000 gallons and σ = 4000. If the
station has 39,000 gallons at the beginning of the week, what
is the probability of its running out of gas that week?
2. Bolts produced by a machine are acceptable if their length is
within the range from 5.95 to 6.05 cm. Suppose the lengths
of bolts produced are normally distributed with µ = 6 cm and
σ = .02.
(a) What is the probability that all bolts in a ten bolt package will
be acceptable?
(b) What is the probability that a ten bolt package has at least two
unacceptable bolts?
3. The price of a stock next week is estimated to be normal with
mean $120 and standard deviation $5. An option will be
worth $10 if the price is over $132, $5 if the price is $125 to
$132 and zero otherwise. How much would you be willing to
pay for this option?
Cherveny
Nov 18
Math 1004: Probability
Practice Answers
1. P(X ≥ 39000) = P(Z ≥ 2.25) = .0122
2. (a) Probability a single bolt is acceptable:
P(5.95 ≤ X ≤ 6.05) = P(−2.5 ≤ Z ≤ 2.5) = 0.9876.
Probability all ten acceptable: (.9876)10 = .8827.
(b) 1 − (.9876)10 − C (10, 9)(.9876)9 (.0124) = .0065
3. Let X = price of the stock next week.
Probability
Value of Option
0
P(X < 125) = .8413
5
P(125 < X < 132) = .1505
10
P(X ≥ 132) = .0082
The expected value of the option is
0(.8413) + 5(.1505) + 10(.0082) = $0.8345. You’d be willing
to pay up to 83 cents for the option.
Cherveny
Nov 18
Math 1004: Probability