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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 ]

Respuesta :

MrRoyal

Answer:

(a) Name: Multinomial distribution

Parameters: [tex]p_1 = 5\%[/tex]   [tex]p_2 = 85\%[/tex]   [tex]p_3 = 10\%[/tex]  [tex]n = 20[/tex]

(b) Range: [tex]\{(x,y,z)| x + y + z=20\}[/tex]

(c) Name: Binomial distribution

Parameters: [tex]p_1 = 5\%[/tex]      [tex]n = 20[/tex]

[tex](d)\ E(x) = 1[/tex]   [tex]Var(x) = 0.95[/tex]

[tex](e)\ P(X = 1, Y = 17, Z = 3) = 0[/tex]

[tex](f)\ P(X \le 1, Y = 17, Z = 3) =0.07195[/tex]

[tex](g)\ P(X \le 1) = 0.7359[/tex]

[tex](h)\ E(Y) = 17[/tex]

Step-by-step explanation:

Given

[tex]p_1 = 5\%[/tex]

[tex]p_2 = 85\%[/tex]

[tex]p_3 = 10\%[/tex]

[tex]n = 20[/tex]

[tex]X \to[/tex] High Slabs

[tex]Y \to[/tex] Medium Slabs

[tex]Z \to[/tex] Low Slabs

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

Given that:

[tex]X \to[/tex] Number of voids considered as high slabs

[tex]Y \to[/tex] Number of voids considered as medium slabs

[tex]Z \to[/tex] 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:

[tex]p_1 = 5\%[/tex]   [tex]p_2 = 85\%[/tex]   [tex]p_3 = 10\%[/tex]  [tex]n = 20[/tex]

And the mass function is:

[tex]f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z[/tex]

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

Given that:

[tex]n = 20[/tex]

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

[tex]x + y + z = n[/tex]

[tex]x + y + z = 20[/tex]

Hence, the range is:

[tex]\{(x,y,z)| x + y + z=20\}[/tex]

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

We have the following parameters attributed to X:

[tex]p_1 = 5\%[/tex] and [tex]n = 20[/tex]

Hence, the name is: Binomial distribution

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

In (c), we have:

[tex]p_1 = 5\%[/tex] and [tex]n = 20[/tex]

[tex]E(x) = p_1* n[/tex]

[tex]E(x) = 5\% * 20[/tex]

[tex]E(x) = 1[/tex]

[tex]Var(x) = E(x) * (1 - p_1)[/tex]

[tex]Var(x) = 1 * (1 - 5\%)[/tex]

[tex]Var(x) = 1 * 0.95[/tex]

[tex]Var(x) = 0.95[/tex]

[tex](e)\ P(X = 1, Y = 17, Z = 3)[/tex]

In (b), we have: [tex]x + y + z = 20[/tex]

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

[tex]x + y + z = 1 + 17 + 3[/tex]

[tex]x + y + z = 21[/tex]

Hence:

[tex]P(X = 1, Y = 17, Z = 3) = 0[/tex]

[tex](f)\ P{X \le 1, Y = 17, Z = 3)[/tex]

This question implies that:

[tex]P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) + P(X = 1, Y = 17, Z = 3)[/tex]

Because

[tex]0, 1 \le 1[/tex] --- for x

In (e), we have:

[tex]P(X = 1, Y = 17, Z = 3) = 0[/tex]

So:

[tex]P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) +0[/tex]

[tex]P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)[/tex]

In (a), we have:

[tex]f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z[/tex]

So:

[tex]P(X=0; Y=17; Z = 3) = \frac{20!}{0! * 17! * 3!} * (5\%)^0 * (85\%)^{17} * (10\%)^{3}[/tex]

[tex]P(X=0; Y=17; Z = 3) = \frac{20!}{1 * 17! * 3!} * 1 * (85\%)^{17} * (10\%)^{3}[/tex]

[tex]P(X=0; Y=17; Z = 3) = \frac{20!}{17! * 3!} * (85\%)^{17} * (10\%)^{3}[/tex]

Expand

[tex]P(X=0; Y=17; Z = 3) = \frac{20*19*18*17!}{17! * 3*2*1} * (85\%)^{17} * (10\%)^{3}[/tex]

[tex]P(X=0; Y=17; Z = 3) = \frac{20*19*18}{6} * (85\%)^{17} * (10\%)^{3}[/tex]

[tex]P(X=0; Y=17; Z = 3) = 20*19*3 * (85\%)^{17} * (10\%)^{3}[/tex]

Using a calculator, we have:

[tex]P(X=0; Y=17; Z = 3) = 0.07195[/tex]

So:

[tex]P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)[/tex]

[tex]P(X \le 1, Y = 17, Z = 3) =0.07195[/tex]

[tex](g)\ P(X \le 1)[/tex]

This implies that:

[tex]P(X \le 1) = P(X = 0) + P(X = 1)[/tex]

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

[tex]p_1 = 5\%[/tex]      [tex]n = 20[/tex]

Such that:

[tex]P(X=x) = ^nC_x * p_1^x * (1 - p_1)^{n - x}[/tex]

So:

[tex]P(X=0) = ^{20}C_0 * (5\%)^0 * (1 - 5\%)^{20 - 0}[/tex]

[tex]P(X=0) = ^{20}C_0 * 1 * (1 - 5\%)^{20}[/tex]

[tex]P(X=0) = 1 * 1 * (95\%)^{20}[/tex]

[tex]P(X=0) = 0.3585[/tex]

[tex]P(X=1) = ^{20}C_1 * (5\%)^1 * (1 - 5\%)^{20 - 1}[/tex]

[tex]P(X=1) = 20 * (5\%)* (1 - 5\%)^{19}[/tex]

[tex]P(X=1) = 0.3774[/tex]

So:

[tex]P(X \le 1) = P(X = 0) + P(X = 1)[/tex]

[tex]P(X \le 1) = 0.3585 + 0.3774[/tex]

[tex]P(X \le 1) = 0.7359[/tex]

[tex](h)\ E(Y)[/tex]

Y has the following parameters

[tex]p_2 = 85\%[/tex]  and    [tex]n = 20[/tex]

[tex]E(Y) = p_2 * n[/tex]

[tex]E(Y) = 85\% * 20[/tex]

[tex]E(Y) = 17[/tex]

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