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Consider randomly selecting a student who is among the 12,000 registered for the current semester in a college. Let X be the number of courses the selected student is taking, and suppose that X has the following probability distribution: x 1 2 3 4 5 6 7 f(x) 0.01 0.02 0.12 0.25 0.42 0.16 0.02 a. Find the cdf of X. b. Find the expected number of courses a student is taking this semester. c. Find the variance of X. d. Find the third quartile of this distribution.

Sagot :

Answer:

a)

[tex]P(X \leq 1) = 0.01[/tex]

[tex]P(X \leq 2) = 0.03[/tex]

[tex]P(X \leq 3) = 0.15[/tex]

[tex]P(X \leq 4) = 0.40[/tex]

[tex]P(X \leq 5) = 0.82[/tex]

[tex]P(X \leq 6) = 0.98[/tex]

[tex]P(X \leq 7) = 1[/tex]

b)

4.61

c) 1.1579

d) 5 is the third quartile of this distribution.

Step-by-step explanation:

We are given the following discrete distribution:

[tex]P(X = 1) = 0.01[/tex]

[tex]P(X = 2) = 0.02[/tex]

[tex]P(X = 3) = 0.12[/tex]

[tex]P(X = 4) = 0.25[/tex]

[tex]P(X = 5) = 0.42[/tex]

[tex]P(X = 6) = 0.16[/tex]

[tex]P(X = 7) = 0.02[/tex]

a. Find the cdf of X.

Probability of X being less of equal to x. Than

[tex]P(X \leq 1) = P(X = 1) = 0.01[/tex]

[tex]P(X \leq 2) = P(X = 1) + P(X = 2) = 0.01 + 0.02 = 0.03[/tex]

[tex]P(X \leq 3) = P(X \leq 2) + P(X = 3) = 0.03 + 0.12 = 0.15[/tex]

[tex]P(X \leq 4) = P(X \leq 3) + P(X = 4) = 0.15 + 0.25 = 0.40[/tex]

[tex]P(X \leq 5) = P(X \leq 4) + P(X = 5) = 0.4 + 0.42 = 0.82[/tex]

[tex]P(X \leq 6) = P(X \leq 5) + P(X = 6) = 0.82 + 0.16 = 0.98[/tex]

[tex]P(X \leq 7) = P(X \leq 6) + P(X = 7) = 0.98 + 0.02 = 1[/tex]

b. Find the expected number of courses a student is taking this semester.

Multiplication of every outcome by its probability.

[tex]E = 0.01*1 + 0.02*2 + 0.12*3 + 0.25*4 + 0.42*5 + 0.16*6 + 0.02*7 = 4.61[/tex]

c. Find the variance of X.

Multiplication of the squared subtraction of each value and the mean, multiplied by its probability. So

[tex]V = 0.01*(1-4.61)^2 + 0.02*(2-4.61)^2 + 0.12*(3-4.61)^2 + 0.25*(4-4.61)^2 + 0.42*(5-4.61)^2 + 0.16*(6-4.61)^2 + 0.02*(7-4.61)^2 = 1.1579[/tex]

d. Find the third quartile of this distribution.

100*(3/4) = 75th percentile

The cdf passes 0.75 when X = 5, so 5 is the third quartile of this distribution.

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