harringtonrob16
21.02.2020 •
Mathematics
A dust mite allergen level that exceeds 2 micrograms per gram (mg/g) of dust has been associated with the development of allergies. Consider a random sample of four homes and let Y be the number of homes with a dust mite level that exceeds 2 mg/g. The probability distribution for Y=y, based on a study by the National Institute of Environmental Health Sciences, is shown in the following table.
y 0 1 2 3 4
p(y) .09 .30 .37 .20 .04
(a) Find the probability that three or four of the homes in the sample have a dust mite level that exceeds 2 µg/g.
(b) Find the probability that fewer than three homes in the sample have a dust mite level that exceeds 2 µg/g.
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Ответ:
(a) 0.24
(b) 0.76
Step-by-step explanation:
We are given that let Y be the number of homes with a dust mite level that exceeds 2 mg/g. The probability distribution for Y=y is shown below ;
y P(y)
0 0.09
1 0.30
2 0.37
3 0.20
4 0.04
(a) Probability that three or four of the homes in the sample have a dust mite level that exceeds 2 µg/g = P(y = 3) + P(y = 4)
= 0.20 + 0.04 = 0.24 .
(b) Probability that fewer than three homes in the sample have a dust mite level that exceeds 2 µg/g = P(y = 0) + P(y = 1) + P(y = 2)
= 0.09 + 0.30 + 0.37 = 0.76 .
Ответ:
The correct answer to this open question is the following.
Although there are no options attached, we can say the following.
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