Question

1. from past experience, a company has found that in carton of transistors, 92% contain no defective transistors, 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors.

a. construct a probability distribution below.
x
p(x)
b. calculate the mean, variance, and standard deviation for the defective transistors.
e(x) μ=
σ²=
σ=

Explanation:

Step1: Construct the probability distribution

Let \(X\) be the number of defective transistors.

• When \(X = 0\), \(P(X)=0.92\)
• When \(X = 1\), \(P(X)=0.03\)
• When \(X = 2\), \(P(X)=0.03\)
• When \(X = 3\), \(P(X)=0.02\)

Step2: Calculate the mean \(\mu=E(X)\)

$$
LATEXBLOCK0
$$

Step3: Calculate the variance \(\sigma^{2}\)

$$
LATEXBLOCK1
$$

Step4: Calculate the standard deviation \(\sigma\)

$$ \sigma=\sqrt{\sigma^{2}}=\sqrt{0.3075}\approx0.5545 $$

Answer:

a. The probability distribution is:

\(X\)	\(P(X)\)
\(1\)	\(0.03\)
\(2\)	\(0.03\)
\(3\)	\(0.02\)

b.

• Mean \(\mu = 0.15\)
• Variance \(\sigma^{2}=0.3075\)
• Standard deviation \(\sigma\approx0.5545\)