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Everyday a factory produces 5000 light bulbs, of which 2500 are type I and 2500 are type II. If a sample of 40 light bulbs is selected at random to be examined for defects, what is the probability that this sample contains at least 18 light bulbs of each type? Solution: Let X be the number of light bulbs of type I in the sample. We need to find P(18< X < 22). Since P(a randomly selected light bulb is type I) 1 we have 18 22 19 21 22 18 40 1 40 1 40 1 1 +...+ 22 There are two methods to get the numerical answer. Method 1: The normal distribution to approximate the answer. H = np = (40) = 20, o? (40) = 10. = npq = Let Y - Nu = 20,0 = 17.5 - 20 Y-u 22.5 - 20 P(18< X < 22)= P(17.5 < Y < 22.5)= P| V10 V10 = P(-0.79

Everyday a factory produces 5000 light bulbs, of which 2500 are type I and 2500 are type II. If a sample of 40 light bulbs is selected at random to be examined for defects, what is the probability that this sample contains at least 18 light bulbs of each type? Solution: Let X be the number of light bulbs of type I in the sample. We need to find P(18< X < 22). Since P(a randomly selected light bulb is type I) 1 we have 18 22 19 21 22 18 40 1 40 1 40 1 1 +...+ 22 There are two methods to get the numerical answer. Method 1: The normal distribution to approximate the answer. H = np = (40) = 20, o? (40) = 10. = npq = Let Y - Nu = 20,0 = 17.5 - 20 Y-u 22.5 - 20 P(18< X < 22)= P(17.5 < Y < 22.5)= P| V10 V10 = P(-0.79

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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I wonder how do we get the number 22 as I highlight? Please help me to explain. Thank you!

Everyday a factory produces 5000 light bulbs, of which 2500 are type I and 2500 are type II. If a sample of 40 light bulbs is selected at random to be examined for defects, what is the probability that this sample contains at least 18 light bulbs of each type? Solution: Let X be the number of light bulbs of type I in the sample. We need to find P(18 <X < 22). 1 we have 2 Since P(a randomly selected light bulb is type I) 22 1 18 22 19 21 18 40 1 40 1 1 40 P(18 < X <22)=|: +... + 2 2 2 18 19 22 There are two methods to get the numerical answer. Method 1: The normal distribution to approximate the answer. 1 = 10. 2 H = np = (40) = 20, oʻ = npq = (40) - (u = 20,0 = VT0). 17.5 – 20 Y - u 22.5 – 20 P(18< X < 22)= P(17.5 < Y < 22.5)= P %3D V10 V10 = P(-0.79 <Z <0.79)=0.5704
Transcribed Image Text:Everyday a factory produces 5000 light bulbs, of which 2500 are type I and 2500 are type II. If a sample of 40 light bulbs is selected at random to be examined for defects, what is the probability that this sample contains at least 18 light bulbs of each type? Solution: Let X be the number of light bulbs of type I in the sample. We need to find P(18 <X < 22). 1 we have 2 Since P(a randomly selected light bulb is type I) 22 1 18 22 19 21 18 40 1 40 1 1 40 P(18 < X <22)=|: +... + 2 2 2 18 19 22 There are two methods to get the numerical answer. Method 1: The normal distribution to approximate the answer. 1 = 10. 2 H = np = (40) = 20, oʻ = npq = (40) - (u = 20,0 = VT0). 17.5 – 20 Y - u 22.5 – 20 P(18< X < 22)= P(17.5 < Y < 22.5)= P %3D V10 V10 = P(-0.79 <Z <0.79)=0.5704
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