Respuesta :
Answer:
0.3968
Step-by-step explanation:
For exponential distribution :
P(A < x) = 1 - e^-x/m
Where m = mean of exponential distribution
Mean of distribution = 4
x = 3
P(A < 3) = 1 - e^-3/4
P(A < 3) = 1 - (0.4723665)
P(A < 3) = 0.5276335 =
Hence, p = 0.5276
(1 - p) = 1 - 0.5276 = 0.4724
Binomial distribution :
P(n, x) = nCx * p^x * (1 - p)^(n-x)
Served on atleast 4 of the next 6 days ;
P(4) + P(5) + P(6):
(6C4 * 0.5276^4 * 0.4724^2) + (6C5 * 0.5276^5 * 0.4724^1) + (6C6 * 0.5276^6 * 0.4724^0)
= (15 * 0.5276^4 * 0.4724^2) + (6 * 0.5276^5 * 0.4724^1) + (1 * 0.5276^6 * 0.4724^0)
= 0.3968
Using the exponential and the binomial distribution, it is found that there is a 0.3969 = 39.69% probability that a person is served in less than 3 minutes on at least 4 of the next 6 days.
First, we find the probability that a person is served in less than 3 minutes on a single day, using the exponential distribution.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation: Â
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by: Â
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
In this problem, mean of 4 minutes, hence [tex]m = 4, \mu = \frac{1}{4} = 0.25[/tex].
The probability that a person is served in less than 3 minutes on a single day is:
[tex]P(X \leq 3) = 1 - e^{-0.25(3)} = 0.5267[/tex]
Now, for the 6 days, we use the binomial distribution.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 6 days, hence [tex]n = 6[/tex]
- On each day, a 0.5276 probability, hence [tex]p = 0.5276[/tex].
The probability is:
[tex]P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)[/tex]
Hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{6,4}.(0.5276)^{4}.(0.4724)^{2} = 0.2594[/tex]
[tex]P(X = 5) = C_{6,5}.(0.5276)^{5}.(0.4724)^{1} = 0.1159[/tex]
[tex]P(X = 6) = C_{6,6}.(0.5276)^{6}.(0.4724)^{0} = 0.0216[/tex]
Then:
[tex]P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) = 0.2594 + 0.1159 + 0.0216 = 0.3969[/tex]
0.3969 = 39.69% probability that a person is served in less than 3 minutes on at least 4 of the next 6 days.
A similar problem is given at https://brainly.com/question/24863377
