Respuesta :
Answer:
(a) The value of P (X ≤ 2) is 0.8729.
(b) The value of P (X ≥ 5) is 0.0072.
(c) The value of P (1 ≤ X ≤ 4) is 0.7154.
(d) The probability that none of the 25 boards is defective is 0.2774.
(e) The expected value and standard deviation of X are 1.25 and 1.09 respectively.
Step-by-step explanation:
The random variable X is defined as the number of defective boards.
The probability that a circuit board is defective is, p = 0.05.
The sample of boards selected is of size, n = 25.
The random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of X is:
[tex]P(X=x)={25\choose x}0.05^{x}(1-0.05)^{25-x};\ x=0,1,2,3...[/tex]
(a)
Compute the value of P (X ≤ 2) as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
[tex]P(X\leq =x)=\sum\limits^{2}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.2774+0.3650+0.2305\\=0.8729[/tex]
Thus, the value of P (X ≤ 2) is 0.8729.
(b)
Compute the value of P (X ≥ 5) as follows:
P (X ≥ 5) = 1 - P (X < 5)
[tex]=1-\sum\limits^{4}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=1-0.9928\\=0.0072[/tex]
Thus, the value of P (X ≥ 5) is 0.0072.
(c)
Compute the value of P (1 ≤ X ≤ 4) as follows:
P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
[tex]=\sum\limits^{4}_{x=1}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.3650+0.2305+0.0930+0.0269\\=0.7154[/tex]
Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.
(d)
Compute the value of P (X = 0) as follows:
[tex]P(X=0)={25\choose 0}0.05^{0}(1-0.05)^{25-0}=1\times 1\times 0.277389=0.2774[/tex]
Thus, the probability that none of the 25 boards is defective is 0.2774.
(e)
Compute the expected value of X as follows:
[tex]E(X)=np=25\times 0.05=1.25[/tex]
Compute the standard deviation of X as follows:
[tex]SD(X)=\sqrt{np(1-p)}=\sqrt{25\times 0.05\times (1-0.05)}=1.09[/tex]
Thus, the expected value and standard deviation of X are 1.25 and 1.09 respectively.
