A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 students, what would be the probability that at least 6 students drive themselves to school? (CDF)

A) 0.139
B) 0.283
C) 0.717
D) 0.861

A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 students, what would be the probability class=

Answer :

Answer:

B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that [tex]p = 0.25[/tex]

Class of 18 students

This means that [tex]n = 18[/tex]

What would be the probability that at least 6 students drive themselves to school?

This is

[tex]P(X \geq 6) = 1 - P(X < 6)[/tex]

In which

[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006[/tex]

[tex]P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034[/tex]

[tex]P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096[/tex]

[tex]P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17[/tex]

[tex]P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213[/tex]

[tex]P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199[/tex]

[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718[/tex]

[tex]P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282[/tex]

Closest option is B, just a small rounding difference.