Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, 5% of the slabs are classified as high, 85% as medium, and 10% as low. A group of 20 slabs that are independent regarding voids is selected for test ing. Let X, Y, and Z denote the number of slabs that are classified as high, medium, and low, respectively. (a) What are the name and values of the parameters of the joint probability distribution of X, Y , and
Z?
(b) What is the range of the joint probability distribution of X, Y, and Z?
(c) What are the name and the values of the parameters of the marginal probability distribution of X?
(d) Determine E[X] and Var(X).
Determine the following:
(e) P{X = 1, Y = 17, Z = 3}
(f) P{X ? 1, Y = 17, Z = 3}
(g) P{X ? 1}
(h) E[Y ]

(a) Name: Multinomial distribution

Parameters:      

(b) Range:

(c) Name: Binomial distribution

Parameters:      

Step-by-step explanation:

Given

High Slabs

Medium Slabs

Low Slabs

Solving (a): Names and values of joint pdf of X, Y and Z

Given that:

Number of voids considered as high slabs

Number of voids considered as medium slabs

Number of voids considered as low slabs

Since the variables are more than 2 (2 means binomial), then the name is multinomial distribution

The parameters are:

And the mass function is:

Solving (b): The range of the joint pdf of X, Y and Z

Given that:

The number of voids (x, y and z) cannot be negative and they must be integers; So:

Hence, the range is:

Solving (c): Names and values of marginal pdf of X

We have the following parameters attributed to X:

and

Hence, the name is: Binomial distribution

Solving (d): E(x) and Var(x)

In (c), we have:

and

In (b), we have:

However, the given values of x in this question implies that:

Hence:

This question implies that:

Because

--- for x

In (e), we have:

So:

In (a), we have:

So:

Expand

Using a calculator, we have:

So:

This implies that:

In (c), we established that X is a binomial distribution with the following parameters:

Such that:

So:

So:

Y has the following parameters

 and    

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

linear pair postulates