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# Ch 2 - 102E

## Problem 102E Chapter 2

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Probability and Statistics for Engineering and the Sciences | 9th Edition

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Problem 102E

102E

Step-by-Step Solution:

Answer : Step 1 : The accompanying table categorizing each student in a sample according to gender and eye color appeared in the article. The table is given below. Eye Colour Gender Blue Brown Green Hazel Total Male 370 352 198 187 1107 Female 359 290 110 160 919 Total 729 642 308 347 2026 From the above table we consider A = blue B = Brown C=Green and D=Harzel. Let F denote the selected individual is a female. a). We have to calculate both P(F) and P(C). Here the total number of females is 919 and The total number of eye colour is 2026 Number of females So P(F) = Total number Number of females P(F) = Total number P(F) = 919 2026 P(F) = 0.4536 Here green = C So P(C) = Number of green Total number Number of green P(C) = Total number 308 P(C) = 2026 P(C) = 0.152 Hence P(F) = 0.4536 and P(C) = 0.152 Step 2 : b). Now we have to calculate P(F C). Number of females with green P(F C) = Total number From the above table number of females with green is 110. 110 P(F C) = 2026 P(F C) = 0.0543 Now we have to find P(F) * P(C). If they were independent then P(F C) would be equal to P(F) * P(C). P(F) * P(C) = (0.4536) (0.152) P(F) * P(C) = 0.006895 So we can clearly see they are not independent. Step 3 : c). If the selected individual has green eyes. Let the probability that he or she is a female. From the above table number of females with green is 110. Number of green is 308. Number of females with green P(F/C) = Number of green P(F/C) = 110 308 P(F/C) = 0.3571 Therefore the probability that he or she is a female is 0.3571.

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