c) P(5<X<15); (d) the value of the constant c such that P(|X −10|≥c) ≤ 0.04
Answer:
a) 0.4444
b) 0.5556
c) 0.8400
d) 10
Explanation:
Chebyshev’s theorem states that
P(|X - μ| ≥ kσ) = 1/k²
μ = Mean = 10
σ = standard deviation = √variance = √4 = 2
a) P(|X −10|≥3)
Comparing this with P(|X - μ| ≥ kσ)
It is evident that kσ = 3
2k = 3
k = 3/2
k² = 2.25
1/k² = 0.444
P(|X −10|≥3) = 0.444
b) P(|X −10| < 3) = 1 - P(|X −10|≥3)
And P(|X −10|≥3) was found in (a) to be 0.444
So,
P(|X −10| < 3) = 1 - P(|X −10|≥3) = 1 - 0.4444 = 0.5556
c) P(5<X<15) = P(|X-10|<5) = 1 - P(|X-10|≥5)
P(|X-10|≥5) = 1/k²
Comparing with P(|X - μ| ≥ kσ) = 1/k²
where 2k = 5
k = 5/2
k² = 6.25
1/k² = 0.16
P(|X-10|≥5) = 0.16
P(5<X<15) = P(|X-10|<5) = 1 - P(|X-10|≥5) = 1 - 0.16 = 0.8400
d) P(|X −10|≥c) ≤ 0.04
1/k² = 0.04
k = 5
c = kσ = 5×2 = 10
