Question 476672: 1. ninety percent of licensed practical nurses are women (statistical abstract of the united states, 1992). suppose 10 licensed practical nurses are randomly selected.
A. what is the mean and standard devition?
B. what is the probability that three of the licensed practical nurses will be women?
C. what is the probability that at most 9 will be women?
D. If in fact only 3 are women, what conclusions could you draw?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. ninety percent of licensed practical nurses are women (statistical abstract of the united states, 1992). suppose 10 licensed practical nurses are randomly selected.
A. what is the mean and standard devition?
mean = np = 10*0.9 = 9
sqrt(npq) = sqrt(9*0.1) = 0.9487
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B. what is the probability that three of the licensed practical nurses will be women?:::P(x=3) = 10C3(0.9)^3*0.1^7 = 120*0.729*0.0000001 = 0.00000875
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C. what is the probability that at most 9 will be women?
P(0<= x <=9) = binomcdf(10,0.9,9) = 0.6513
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D. If in fact only 3 are women, what conclusions could you draw?
I'll leave that to you.
Cheers,
Stan H.
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