Answer :
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
The Chebyshev's Theorem also known as Chebyshev's Inequality can be applied to several probability distributions.
The given parameters are:
- [tex]\mu = 10[/tex] --- the mean
- [tex]\sigma^2 = 4[/tex] --- the variance
Start by calculating the standard deviation using
[tex]\sigma = \sqrt{\sigma^2}[/tex]
So, we have:
[tex]\sigma = \sqrt{4}[/tex]
[tex]\sigma =2[/tex]
Using Chebyshev’s theorem, we have:
[tex]P(|X - \mu| \ge k\sigma) = \frac{1}{k\²}[/tex]
(a) Calculate P(|X −10|≥3)
We have:
[tex]P(|X - \mu| \ge k\sigma) = \frac{1}{k\²}[/tex]
By comparison P(|X −10|≥3) with [tex]P(|X - \mu| \ge k\sigma) = \frac{1}{k\²}[/tex];
[tex]k\sigma = 3[/tex]
Substitute [tex]\sigma =2[/tex]
[tex]k\times 2= 3[/tex]
Divide both sides by 2
[tex]k= \frac 32[/tex]
[tex]P(|X - \mu| \ge k\sigma) = \frac{1}{k\²}[/tex] becomes
[tex]P(|X - 10| \ge 3) = \frac{1}{(3/2)^2}[/tex]
Evaluate the exponent
[tex]P(|X - 10| \ge 3) = \frac{1}{9/4}[/tex]
Rewrite the above fraction as
[tex]P(|X - 10| \ge 3) = \frac{4}{9}[/tex]
Hence, the value of the required probability is [tex]P(|X - 10| \ge 3) = \frac{4}{9}[/tex]
(b) Calculate P(|X −10|<3)
Using the complement rule of probability, we have:
[tex]P(|X - 10|<3) + P(|X - 10|\ge3) = 1[/tex]
Make P(|X - 10|<3) the subject of formula
[tex]P(|X - 10|<3) = 1 - P(|X - 10|\ge3)[/tex]
In (a), we have: [tex]P(|X - 10| \ge 3) = \frac{4}{9}[/tex]
So, the equation becomes
[tex]P(|X - 10|<3) = 1 - \frac 49[/tex]
Take LCM
[tex]P(|X - 10|<3) = \frac{9 -4}9[/tex]
Simplify the numerator
[tex]P(|X - 10|<3) = \frac{5}9[/tex]
Hence, the value of the required probability is [tex]P(|X - 10|<3) = \frac{5}9[/tex]
Read more about Chebyshev's theorem at:
https://brainly.com/question/5179184