Respuesta :
Answer:
Step-by-step explanation:
Let X be the time waiting for an individual.
X is U (0,10) and continuous
Hence pdf of X is
[tex]f(x) =0.1[/tex]
a) the probability that the individual waits
more than 7 minutes=[tex]P(X>7) = \int\limits^10_7 {\frac{1}{10} } \, dx \\=0.1(10-7)\\=0.3[/tex]
b) the probability that the individual waits
between 2 and 7 minutes
[tex]=P(2<x<7) =\int\limits^7_2 {0.1} \, dx \\=0.1(7-2)\\=0.5[/tex]
Uniform distribution has probability density distributed uniformly over its defined interval. The needed probabilities for the given case are:
- Probability that the individual waits more than 7 minutes = 0.3
- Probability that the individual waits between 2 and 7 minutes = 0.5
How to calculate the probability of an interval in uniform distribution?
Suppose that a random variable X has uniform distribution, such that its range is from X = a to X = b, then its pdf is given by:
[tex]f(x) = \dfrac{1}{b-a}; a < x < b\\\\f(x) = 0; x < a , b < x[/tex]
For the given context, lets assume that the random variable X tracks the waiting time of an individual for bus on the station.
Then X can have value from 0 to 10 minutes, and thus, a = 0, b= 10
Thus, the needed probabilities are calculated as:
a) P(X > 7) = P(An individual waits more than 7 minutes)
[tex]P(X > 7) = 1 - P(X \leq 7) = 1 - \int_0^{7}\dfrac{dx}{10-0} = 1 - [\dfrac{x}{10}]^{7}_0 \\\\P(X > 7) = 1 - 7/10 = 0.3[/tex]
b) P(waiting time is between 2 and 7) = P(2 < X < 7)
It is calculated by integrating the probability density from X = 2 to X = 7.
or
[tex]P(2 < X < 7) = \int_2^{7}\dfrac{dx}{10-0} = [\dfrac{x}{10}]^{7}_2 \\\\P(2 < X < 7) = 7/10 - 2/10 = 0.5[/tex]
Thus, the needed probabilities for the given case are:
- Probability that the individual waits more than 7 minutes = 0.3
- Probability that the individual waits between 2 and 7 minutes = 0.5
Learn more about uniform distribution here:
https://brainly.com/question/6083006
