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Announcements Finite Probability Friday, November 18th I MyMathLab 9 is Monday Nov 28 I 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”): I Center is µ. I Width determined by σ I 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