Answer:
(a) 0.24
(b) 0.76
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 .