homework_5_ans - Course MTH 1210 Homework #5 (answers) (The...

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Course MTH 1210 Homework #5 ( answers ) ( The Normal Distribution ) Mar 24’09 Use the MS Excel program Three_distributions.xls, sheet 3 1. A certain type of light bulb has an average life of 500 hr , with a standard deviation of 100 hr . The length of life of the bulb can be closely approximated by a normal curve. An amusement park buys and installs 10,000 such bulbs. Find the total number that can be expected to last for each of the following periods of time: (a) less than 500 hr, (b) between 290 and 540 hr, (c) more than 300 hr. Solution. In this problem we are given that the mean (average) µ = 500 hr and the standard deviation σ = 100 hr. Run the program Three_distributions.xls, sheet 3 and enter the values of 500 and 100 in the cells G58 and G59 respectively (type in each value and press [ENTER] key). We’ll work with the rectangular block of cells E53 – K62 . (a) To answer this question, we should evaluate the product 10,000*f(X ≤ 500) where f(X ≤ 500) is the cumulative probability that the length of life of a bulb is less than or equal to 500 hr. Enter the value of 500 in the cell G58 for x . We find that f(X ≤ x) = 0.5 (see the cell I59 ). Thus, 10,000*f(X ≤ 500) = 10,000*0.5 = 5000 . x = 500 f(x) = 0.003989 µ = 500 σ = 100 f(X ≤ x) = 0.5 (b) The number in question is 10,000*f(290 ≤ X ≤ 540) = 10,000*[ f(X ≤ 540) – f(X ≤ 290) ]. To evaluate the cumulative probability f(X ≤ 540) that the length of life of a bulb is less than or equal to 540 hr, we enter the value of 540 in the cell G58 for x and find in the cell I59 that f(X ≤ x) = 0.655422 . x =
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This note was uploaded on 11/24/2011 for the course MATH 3000 taught by Professor Kzaer during the Spring '05 term at St. Johns College MD.

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homework_5_ans - Course MTH 1210 Homework #5 (answers) (The...

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