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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 107e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 107e

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# Forty percent of seeds from maize (modern-day corn) ears

ISBN: 9780321629111 32

## Solution for problem 107E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 107E

Forty percent of seeds from maize (modern-day corn) ears carry single spikelets, and the other 60% carry paired spikelets. A seed with single spikelets will produce an ear with single spikelets 29% of the time, whereas a seed with paired spikelets will produce an ear with single spikelets 26% of the time. Consider randomly selecting ten seeds. a. ?What is the probability that exactly five of these seeds carry a single spikelet and produce an ear with a single spikelet? b. ?What is the probability that exactly five of the ears produced by these seeds have single spikelets? What is the probability that at most five ears have single spikelets?

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Problem 107E Answer: Step1: We have 40% of seeds from maize (modern-day corn) ears carry single spikelets, and the other 60% carry paired spikelets. A seed with single spikelets will produce an ear with single spikelets 29% of the time, whereas a seed with paired spikelets will produce an ear with single spikelets 26% of the time. Consider randomly selecting 10 seeds. We need to find, a. What is the probability that exactly five of these seeds carry a single spikelet and produce an ear with a single spikelet b. What is the probability that exactly five of the ears produced by these seeds have single spikelets What is the probability that at most five ears have single spikelets Step2: a). Here, randomly selecting seeds n = 10 and The probability that each randomly selected seed carries a single spikelet and produce an ear with a single spikelet is p = 0.4 × 0.29 = 0.116. Since A random variable “X” follows binomial distribution with parameter “(n, p)” X ~B(n, p) X ~B(10, 0.116) The probability mass function of X is given by n x n-x p(x) = C x q , x = 0,1,2,...,n With mean E(x) = np and Var(x) = np(1 - p) Now, The probability that exactly five of these seeds carry a single spikelet and produce an ear with a single spikelet is given by P(X = 5) Consider, p(x) = C x q n-, x = 0,1,2,...,n 10 5 10-5 P(X = 5) = C50.116) (1-0.116) = 252(0.000021)(0.5398) = 0.002857. Therefore,The probability that exactly five of these seeds carry a single spikelet and produce an ear with a single spikelet is 0.002857. Step3: b). Here, randomly selecting seeds n = 10 and The probability for each ear to have single spikelet is therefore p = (0.4 × 0.29) + (0.6 × 0.26) = 0.272. Now, The probability that exactly five of the ears produced by these seeds have single spikelets is given by P(X = 5) Consider, p(x) = Cx q n-x, x = 0,1,2,...,n 10 5 10-5 P(X = 5) = C50.272) (1- 0.272) = 252(0.001488)(0.2044) = 0.0766. Therefore, The probability that exactly five of the ears produced by these seeds have single spikelets is 0.0766. Step4: Here, randomly selecting seeds n = 10 and The probability for each ear to have single spikelet is therefore p = (0.4 × 0.29) + (0.6 × 0.26) = 0.272. Now, The probability that at most five ears have single spikelets is given by P(X 5) o btained from Excel by using the formula “ =Binomdist(x,n,p,FALSE)” “=Binomdist(0 to 5,10,0.272,FALSE)” x P(X 5) 0 0.291422212 1 0.382409237 2 0.225811789 3 0.079017097 4 0.018145329 5 0.002857273 Sum P(X 5) = 0.999662938 Therefore, The probability that at most five ears have single spikelets is 0.9996.

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Forty percent of seeds from maize (modern-day corn) ears