A random variable X has a mean μ = 10 and a variance σ2 = 4. Using Chebyshev’s theorem, find (a) P(|X −10|≥3); (b) P(|X −10| < 3); (c) P(5

c) P(5<X<15); (d) the value of the constant c such that P(|X −10|≥c) ≤ 0.04

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

d bonding propertys

explanation:

carbon can form so many different compounds not just with oxygen

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