SOLUTION: Help--how do i figure this one out? Driving me crazy !!!
Twenty bearded men gather. Ten have eight-inch beards, six have ten-inch beards, and four have twelve-inch beards. Wh
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-> SOLUTION: Help--how do i figure this one out? Driving me crazy !!!
Twenty bearded men gather. Ten have eight-inch beards, six have ten-inch beards, and four have twelve-inch beards. Wh
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Question 5081: Help--how do i figure this one out? Driving me crazy !!!
Twenty bearded men gather. Ten have eight-inch beards, six have ten-inch beards, and four have twelve-inch beards. What is the expected length of a randomly selected man from this group?
Tried 10/20 x 6/20 x 4/20 =12inches, that doesn't seem right. Please help! Found 3 solutions by rapaljer, apeokukuy2k, JWG:Answer by rapaljer(4671) (Show Source):
You can put this solution on YOUR website! This is sort of like a weighted mean. Multiply number of beards time length of beard, sum the products, and divide by the number of men with beards. I get this:
= 9.4 inches
You can put this solution on YOUR website! i think that u calculate the probability of selecting a man with eigth -inch beared from the 20 people gathered,then u do likewise for the rest.The one with the highest probability is the most likely to be picked at random.
prob of selecting eight -inch beared from 20 people is 10 combination 1 /20 combination 1.that for 10 inch man is 6 combination 1 / 20 combination 1 and that for a twelve inch beared man is 4 combination 1 / 20 combination 1.the one with the highest probability is the most likely to be selected at random
You can put this solution on YOUR website! They want to know who is the most likely to be picked.
8 inch: 10
10 inch: 6
12 inch: 4
There are 10+6+4=20 bearded men in the group. Immediately one can see that 8-inch bearded men make up most of the group. 10/20 = .5 or 50%.