Question

16. From past experience, a company has found that in cartons of transistors, 92% contain no defective transistors, 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Find the mean, variance, and standard deviation for the defective transistors.

Answer 1

Answer:

E(x)=0.15

V(x)=0.3075

S(x)=0.5545

Step-by-step explanation:

The mean of a discrete variable is calculated as:

$E(x)=x_1*p(x_1)+x_2*p(x_2)+...+x_n*p(n)$

where $x_1,x_2,...,x_n$ are the values that the variable can take and $p(x_1),p(x_2),...,p(x_n)$ are their respective probabilities.

So, if we call x the number of defective transistors in cartons, we can calculate the mean E(x) as:

$E(x)=(0*0.92)+(1*0.03)+(2*0.03)+(3*0.02)=0.15$

Because there are 0 defective transistor with a probability of 0.92, 1 defective transistor with a probability of 0.03, 2 defective transistors with a probability of 0.03 and 3 defective transistors with a probability of 0.01.

At the same way, the variance V(x) is calculated as:

$V(x)=E(x^2)-(E(x))^2$

Where $E(x^2)=x_1^2*p(x_1)+x_2^2*p(x_2)+...+x_n^2*p(n)$

So, the variance V(x) is equal to:

$E(x^2)=(0^2*0.92)+(1^2*0.03)+(2^2*0.03)+(3^2*0.02)=0.33\\V(x)=0.33-(0.15)^2\\V(x)=0.3075$

Finally, the standard deviation is calculated as:

$S(x)=\sqrt{V(x)} \\S(x)=\sqrt{0.3075} \\S(x)=0.5545$

Answer 2

Answer:

mean = 0.15

variance = 0.3075

standard deviation = 0.5707

Step-by-step explanation:

- See the attachment an fill in the cells with "bold" marked columns.

- Where mean = E(X)

                     E(X) = Sum ( Xi*P(Xi) )

                             = 0 + 0.03 + 0.06 + 0.06

                             = 0.15

- The variance = Var (X)

                     Var (X) = Sum ( X2*P(X) ) - E(X)^2

                                 = (0 + 0.03 + 0.12 + 0.18) - 0.15^2

                                 = 0.3075

- The standard deviation = S(X)

                     S(X) = √Var(X) = √0.3075 = 0.5707