It is found that on average, 9% of the population recycles its garbage, with a standard deviation of 3% over the country. If a random sample of 49 households is taken, what is the probability that more than 4 households recycle their garbage?
2006-12-01 09:37:00 · 2 answers · asked by Moose 2 in Science & Mathematics ➔ Mathematics
Ok, I'm going to try and do this without going to get my statistics text book.
4/49 is 8.16%
9%-8.16%=0.84%
0.84%/3%=.28 Standard Deviations
The probablility that a value is more than .28 standard deviations less than the mean is the same that a value is less than .28 standard deviations more than the mean.
Use a Z value table for the value z=.28
Answer: 61.0%
2006-12-02 04:12:52 · answer #1 · answered by hobbes84k 3 · 0⤊ 0⤋
This seems binomial where you'd say (where X = # of households recycling) P( X >4 ) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3) - P(X=4)
where P(X=x) = 49 nCr x (.09)^x (.91)^49-x
2006-12-02 12:04:13 · answer #2 · answered by Modus Operandi 6 · 0⤊ 0⤋