Respuesta :
Answer:
(a) The value of f (0) is 0.3487.
(b) The value of f (2) is 0.1937.
(c) The value of P (X ≤ 2) is 0.9298.
(d) The value of P (X ≥ 1) is 0.6513.
(e) The value of E (X) is 1.
(f) The value of V (X) is 0.9 and σ is 0.9487.
Step-by-step explanation:
The random variable X follows a Binomial distribution with parameter n = 10 and p = 0.10.
The probability mass function of X is:
[tex]P(X=x)={10\choose x}0.10^{x}(1-0.10)^{10-x};\ x=0,1,2,3...[/tex]
(a)
Compute the value of f (0) as follows:
f (0) = P (X = 0)
[tex]={10\choose 0}0.10^{0}(1-0.10)^{10-0}\\=1\times 1\times 0.348678\\=0.348678\\\approx0.3487[/tex]
Thus, the value of f (0) is 0.3487.
(b)
Compute the value of f (2) as follows:
f (2) = P (X = 2)
[tex]={10\choose 2}0.10^{2}(1-0.10)^{10-2}\\=45\times 0.01\times 0.43047\\=0.1937115\\\approx0.1937[/tex]
Thus, the value of f (2) is 0.1937.
(c)
Compute the value of P (X ≤ 2) as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
[tex]=\sum\limits^{0}_{2} {{10\choose x}0.10^{x}(1-0.10)^{10-x}}\\=0.3487+0.3874+0.1937\\=0.9298[/tex]
Thus, the value of P (X ≤ 2) is 0.9298.
(d)
Compute the value of P (X ≥ 1) as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
= 1 - 0.3487
= 0.6513
Thus, the value of P (X ≥ 1) is 0.6513.
(e)
Compute the expected value of X as follows:
[tex]E(X)=n\times p[/tex]
[tex]=10\times 0.10\\=1[/tex]
Thus, the value of E (X) is 1.
(f)
Compute the variance of X as follows:
[tex]V(X)=np(1-p)[/tex]
[tex]=10\times 0.10\times (1-0.10)\\=0.90[/tex]
Compute the standard deviation of X as follows:
[tex]SD(X)=\sqrt{V(X)}[/tex]
[tex]=\sqrt{0.90}\\=0.94868\\\approx0.9487[/tex]
Thus, the value of V (X) is 0.9 and σ is 0.9487.
