Respuesta :
Answer:
a) 91.33% probability that at most 6 will come to a complete stop
b) 10.91% probability that exactly 6 will come to a complete stop.
c) 19.58% probability that at least 6 will come to a complete stop
d) 4 of the next 20 drivers do you expect to come to a complete stop
Step-by-step explanation:
For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.
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.
20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.
This means that [tex]p = 0.2[/tex]
20 drivers
This means that [tex]n = 20[/tex]
a. at most 6 will come to a complete stop?
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115[/tex]
[tex]P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576[/tex]
[tex]P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369[/tex]
[tex]P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054[/tex]
[tex]P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182[/tex]
[tex]P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746[/tex]
[tex]P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091[/tex]
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133[/tex]
91.33% probability that at most 6 will come to a complete stop
b. Exactly 6 will come to a complete stop?
[tex]P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091[/tex]
10.91% probability that exactly 6 will come to a complete stop.
c. At least 6 will come to a complete stop?
Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 6) + P(X \geq 6) = 1[/tex]
We want [tex]P(X \geq 6)[/tex]. So
[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) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042[/tex]
[tex]P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958[/tex]
19.58% probability that at least 6 will come to a complete stop
d. How many of the next 20 drivers do you expect to come to a complete stop?
The expected value of the binomial distribution is
[tex]E(X) = np = 20*0.2 = 4[/tex]
4 of the next 20 drivers do you expect to come to a complete stop
