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Let X be the number of the cars being repaired at a repair shop. We have the following information:
At any time, there are at most 3 cars being repaired.
The probability of having 2 cars at the shop is the same as the probability of having one car.
The probability of having no car at the shop is the same as the probability of having 3 cars.
The probability of having 1 or 2 cars is half of the probability of having 0 or 3 cars.
(a) What is the sample space?
(b) Find the pmf of X and draw the histogram of pmf.
(c) Find the cdf of X and draw the histogram of cdf.

Sagot :

MrRoyal

Answer:

(a) Sample Space

[tex]S = \{0,1,2,3\}[/tex]

(b) PMF

[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}[/tex]

(c) CDF

[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}[/tex]

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

[tex]S = \{0,1,2,3\}[/tex]

Solving (b): The PMF

From the question, we have:

[tex]P(2) = P(1)[/tex]

[tex]P(0) = P(3)[/tex]

[tex]P(1\ or\ 2) = 0.5 * P(0\ or\ 3)[/tex]

[tex]P(1\ or\ 2) = 0.5 * P(0\ or\ 3)[/tex] can be represented as:

[tex]P(1) + P(2) = 0.5[P(0) + P(3)][/tex]

Substitute [tex]P(2) = P(1)[/tex] and [tex]P(0) = P(3)[/tex]

[tex]P(1) + P(1) = 0.5[P(0) + P(0)][/tex]

[tex]2P(1) = 0.5[2P(0)][/tex]

[tex]2P(1) = P(0)[/tex]

[tex]P(0)= 2P(1)[/tex]

Also note that:

[tex]P(0) + P(1) + P(2) + P(3) = 1[/tex]

Substitute [tex]P(2) = P(1)[/tex] and [tex]P(0) = P(3)[/tex]

[tex]P(0) + P(1) + P(1) + P(0) = 1[/tex]

[tex]2P(1) + 2P(0) = 1[/tex]

Substitute [tex]P(0)= 2P(1)[/tex]

[tex]2P(1) + 2*2P(1) = 1[/tex]

[tex]2P(1) + 4P(1) = 1[/tex]

[tex]6P(1) = 1[/tex]

Solve for P(1)

[tex]P(1) = \frac{1}{6}[/tex]

To calculate others, we have:

[tex]P(2) = P(1)[/tex]

[tex]P(2) = P(1) = \frac{1}{6}[/tex]

[tex]P(0)= 2P(1)[/tex]

[tex]P(0) =2 * \frac{1}{6}[/tex][tex]P(0) =\frac{1}{3}[/tex]

[tex]P(3) = P(0) =\frac{1}{3}[/tex]

Hence, the PMF is:

[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}[/tex]

See attachment (1) for histogram

Solving (c): The CDF ; F(x)

This is calculated as:

[tex]F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)[/tex]

For x = 0;

We have:

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

[tex]P(X \le 0) = 1/3[/tex]

For x = 1

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

[tex]P(X \le 1) = 1/3 + 1/6[/tex]

[tex]P(X \le 1) = 1/2[/tex]

For x = 2

[tex]P(X \le 2) = P(0) + P(1) + P(2)[/tex]

[tex]P(X \le 2) = 1/3 + 1/6 + 1/6[/tex]

[tex]P(X \le 2) = 2/3[/tex]

For x = 3

[tex]P(X \le 3) = P(0) + P(1) + P(2) + P(3)[/tex]

[tex]P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3[/tex]

[tex]P(X \le 3) = 1[/tex]

Hence, the CDF is:

[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}[/tex]

See attachment (2) for histogram

View image MrRoyal
View image MrRoyal
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