1)Use Markov’s inequality. P(X > 85) ≤ E[X]/85 = 75/85 = 15/17
2) Here we use Chebyshev. P(65 ≤ X ≤ 85) = P(|X − 75| ≤ 10) = 1 − P(|X − 75| > 10) ≥ 1 − var(X)/102 = 1 − 25/100 = 3/4
3) P(|X −75| ≤ 5) = 1−P(|X −75| > 5) ≥ 1−var(X)/25 = 0.9. So 1/n = 0.1 and so n = 10
4) P(|X − 75| ≤ 5) = P(|Z| ≤ "\\sqrt n" ) = 0.9 and so Φ("\\sqrt n" ) = 0.95. A normal table lookup gives "\\sqrt n" = 1.65 and so n ≈ 3. This is a situation where you should not trust the answer, since for n = 3 a CLT approximation is not valid anyway. However the Chebyshev is valid and gives a correct but possibly not tight answer
d)we assume that the data collected by the polling company is pretty representative of the population at large
Comments
Leave a comment