Page 312, Exercise 50

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1 Millersville University Name Answer Key Department of Mathematics MATH 130, Elements of Statistics I, Homework 4 November 5, 2009 Page 312, Exercise 50 Simulation According to the U.S. National Center for Health Statistics, there is a 98% probability that a 20-year-old male will survive to age 30. (a) Using statistical software, simulate taking 100 random samples of size 30 from this population. Your results may differ (b) Using the results of the simulation, compute the probability that exactly 29 of the 30 male survive to age 30. After creating a frequency table of the results, Age Frequency Total 100 the probability P(X 29) 31/ (c) Compute the probability that exactly 29 of the 30 males survive to age 30, using the binomial probability distribution. Compare the results with part (b).

2 The results are close. P(X 29) ( 30 C 29 )(0.98) 29 (1 0.98) (30)( )() (d) Using the results of the simulation, compute the probability that at most 27 of the 30 male survive to age P(X 27) P(X k) k0 (e) Compute the probability that at most 27 of the 30 males survive to age 30, using the binomial probability distribution. Compare the results with part (d). P(X 27) These results are again, close. 27 k0 27 P(X k) ( 30 C k ) (0.98) k (1 0.98) 30 k k0 2 (f) Compute the mean number of male survivors in the 100 simulations of the probability experiment. Is it close to the expected value? x X (X P(X)) 30 x26 (x P(X x)) (26)(0.01) + (27)(0.01) + (28)(0.06) + (29)(0.31) + (30)(0.62) 29.5 The expected value of the mean number of survivors is µ X np (30)(0.98) 29.4, so the results are close.

3 (g) Compute the standard deviation of the number of male survivors in the 100 simulations of the probability experiment. Compare the result to the theoretical standard deviation of the probability distribution. s 2 X (X 2 P(X)) (x) 2 30 x26 (x 2 P(X x)) (29.5) 2 (26 2 )(0.01) + (27 2 )(0.01) + (28 2 )(0.06) + (29 2 )(0.31) + (30 2 )(0.61) (29.5) s The theoretical standard deviation is The results are close. X np(1 p) (30)(0.98)(1 0.98) (h) Did the simulation yield any unusual results? Not in this simulation. Section 7.3, Exercise 22 Light Bulbs General Electric manufactures a decorative Crystal Clear 60-watt light bulb that it advertises will last 1500 hours. Suppose the lifetimes of the bulbs are approximately normally distributed with a mean of 1550 hours and a standard deviation of 57 hours. (a) What proportion of the light bulbs will last less than the advertised time? The Z-score associated with a lifetime of 1500 hours is P(X < 1500) P(Z < 0.88) (b) What proportion of the light bulbs will last more than 1650 hours? The Z-score associated with a lifetime of 1650 hours is P(X > 1650) P(Z > 1.75) 1 P(Z < 1.75)

4 (c) What is the probability that a randomly selected GE Crystal Clear 60-watt bulb lasts between 1625 and 1725 hours? The Z-score associated with a lifetime of 1625 hours is , while the Z-score associated with a lifetime of 1725 hours is , P(1625 X 1725) P(1.32 Z 3.07) (d) What is the probability that a randomly selected GE Crystal Clear 60-watt bulb lasts longer than 1400 hours? The Z-score associated with a lifetime of 1400 hours is P(X > 1400) P(Z > 2.63) 1 P(Z < 2.63) Section 7.3, Exercise 26 Manufacturing Ball bearings are manufactured with a mean diameter of 5 millimeter (mm). Because of variability in the manufacturing process, the diameters of the ball bearings are approximately normally distributed with a standard deviation of mm. (a) What proportion of ball bearings has a diameter more than 5.03 mm? The Z-score associated with a diameter of 5.03 mm is P(X > 5.03) P(Z > 1.50) 1 P(Z < 1.50)

5 (b) Any ball bearings that have diameter less than 4.95 mm or greater than 5.05 mm are discarded. What proportion of ball bearings will be discarded? The Z-score associated with a diameter of 4.95 mm is , while the Z-score associated with a diameter of 5.05 mm is , P(X < 4.95 or X > 5.05) 1 P(4.95 < X < 5.05) 1 P( 2.50 < Z < 2.50) 1 ( ) (c) Using the results of part (b), if 30,000 ball bearings are manufactured in a day, how many should the plant manager expect to discard? number of discards (30000)(0.0124) 372 (d) If an order comes in for 50,000 ball bearings, how many bearings should the plant manager manufacture if the order states that all ball bearings must be between 4.97 mm and 5.03 mm? The Z-score associated with a diameter of 4.97 mm is , while the Z-score associated with a diameter of 5.03 mm is , the proportion of ball bearings meeting the diameter requirements of the customer are P(4.97 < X < 5.03) P( 1.50 < Z < 1.50) Therefore if 50,000 acceptable ball bearings are needed the plant should manufacture a total of ball bearings

6 Section 7.3, Exercise 30 Earthquakes The magnitude of earthquakes since 1900 that measure 0.1 or higher on the Richter scale in California is approximately normally distributed with µ 6.2 and 0.5, according to data obtained from the U.S. Geological Survey. (a) Determine the 40th percentile of the magnitude of earthquakes in California. The 40th percentile is the magnitude of earthquake separating the bottom 40% of magnitudes from the top 60% of magnitudes. If P(Z < z) 0.40 then z X µ X X (b) Determine the magnitude of earthquakes that make up the middle 85% of magnitudes. If the area in the middle of the bell-shaped curve is 0.85 then the area in both tails combined is 0.15 and the area in the right tail is If P(Z < z) then z the middle 85% of the area under the standard normal curve lies between z 1.44 and z The magnitudes of the earthquakes associated with these Z-scores are 1.44 X 1 µ X X X 2 µ X X Section 7.3, Exercise 32 Speedy Lube The time required for Speedy Lube to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 2.5 minutes. (a) Speedy Lube guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half price. What percent of customers receives the service for half price? The Z-score associated with a service time of 20 minutes is

7 and thus P(X > 20) P(Z > 1.20) 1 P(Z < 1.20) Therefore 11.51% of customers receive the service for half price. (b) If Speedy Lube does not want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? If the area under the standard normal curve to the left of z is 0.97, then z The waiting time for service associated with this Z-score is 1.88 X µ X X The guaranteed service time should be increased to 22 minutes.

STAT509: Continuous Random Variable

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