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 <em>X</em> are 1.25 and 1.09 respectively.
Step-by-step explanation:
The random variable <em>X</em> is defined as the number of defective boards.
The probability that a circuit board is defective is, <em>p</em> = 0.05.
The sample of boards selected is of size, <em>n</em> = 25.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is:
![P(X=x)={25\choose x}0.05^{x}(1-0.05)^{25-x};\ x=0,1,2,3...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7B25%5Cchoose%20x%7D0.05%5E%7Bx%7D%281-0.05%29%5E%7B25-x%7D%3B%5C%20x%3D0%2C1%2C2%2C3...)
(a)
Compute the value of P (X ≤ 2) as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
![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](https://tex.z-dn.net/?f=P%28X%5Cleq%20%3Dx%29%3D%5Csum%5Climits%5E%7B2%7D_%7Bx%3D0%7D%7B%7B25%5Cchoose%20x%7D0.05%5E%7Bx%7D%281-0.05%29%5E%7B25-x%7D%7D%5C%5C%3D0.2774%2B0.3650%2B0.2305%5C%5C%3D0.8729)
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)
![=1-\sum\limits^{4}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=1-0.9928\\=0.0072](https://tex.z-dn.net/?f=%3D1-%5Csum%5Climits%5E%7B4%7D_%7Bx%3D0%7D%7B%7B25%5Cchoose%20x%7D0.05%5E%7Bx%7D%281-0.05%29%5E%7B25-x%7D%7D%5C%5C%3D1-0.9928%5C%5C%3D0.0072)
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)
![=\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](https://tex.z-dn.net/?f=%3D%5Csum%5Climits%5E%7B4%7D_%7Bx%3D1%7D%7B%7B25%5Cchoose%20x%7D0.05%5E%7Bx%7D%281-0.05%29%5E%7B25-x%7D%7D%5C%5C%3D0.3650%2B0.2305%2B0.0930%2B0.0269%5C%5C%3D0.7154)
Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.
(d)
Compute the value of P (X = 0) as follows:
![P(X=0)={25\choose 0}0.05^{0}(1-0.05)^{25-0}=1\times 1\times 0.277389=0.2774](https://tex.z-dn.net/?f=P%28X%3D0%29%3D%7B25%5Cchoose%200%7D0.05%5E%7B0%7D%281-0.05%29%5E%7B25-0%7D%3D1%5Ctimes%201%5Ctimes%200.277389%3D0.2774)
Thus, the probability that none of the 25 boards is defective is 0.2774.
(e)
Compute the expected value of <em>X</em> as follows:
![E(X)=np=25\times 0.05=1.25](https://tex.z-dn.net/?f=E%28X%29%3Dnp%3D25%5Ctimes%200.05%3D1.25)
Compute the standard deviation of <em>X</em> as follows:
![SD(X)=\sqrt{np(1-p)}=\sqrt{25\times 0.05\times (1-0.05)}=1.09](https://tex.z-dn.net/?f=SD%28X%29%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B25%5Ctimes%200.05%5Ctimes%20%281-0.05%29%7D%3D1.09)
Thus, the expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.