Respuesta :
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.
