Question

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. Calculate the mean and variance for the defective transistors. Mean = Variance = (Please round answers to 4 decimal places.)

Answer 1

Answer:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X         0           1           2         3

P(X)    0.92     0.03    0.03     0.02

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

Answer 2

The required  value of mean 0.330 and standard deviation 0.5545 for the defective transistors.

Given that,

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.

We have to find,

Calculate the mean and variance for the defective transistors. Mean = Variance.

According to the question,

Let,  X the random variable who represent the number of defective transistors.

The following probability distribution for X,

X         0           1           2         3

P(X)   0.92     0.03    0.03     0.02

To calculate the expected value with the following formula:

$E(X) = \sum^{n}_{i=1} X_i. P.(X_i)$

On substitute all the values in the formula,

$E(X) = 0\times0.92 + 1\times0.03+2\times0.03 + 3\times0.02\\\\E(X) = 0.15$

To find the variance first the second moment given by:

$E(X^2) = \sum^{n}_{i=1} X_i^2. P.(X_i)$

On substitute all the values in the formula;

$E(X) = 0^2\times0.92 + 1^2\times0.03+2^2\times0.03 + 3^2\times0.02\\\\E(X) = 0.330$

Therefore, The variance is calculated with this formula:

$Var(X) = E(X)^2 - (E(X))^2 = 0.33 - (0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance,

$Standard \ deviation = \sqrt{0.0375} = 0.5545$

Hence, The required  value of mean 0.330 and standard deviation 0.5545 for the defective transistors.

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