Answer:
a
[tex]\mu_x =7.5 \ minutes [/tex]
b
[tex]\sigma_{x} = 4.33 \ minutes [/tex]
c
[tex]P(5 < X < 11 ) = 0.4 [/tex]
d
[tex]P(W = 4 ) = 0.228[/tex]
Step-by-step explanation:
From the question we are told that
The interval is [tex](w , y) = (0 , 15)[/tex]
Considering question a
Generally the mean waiting time is mathematically represented as
[tex]\mu_x = \frac{w + y}{2}[/tex]
=> [tex]\mu_x = \frac{0 + 15}{2}[/tex]
=> [tex]\mu_x =7.5 \ minutes [/tex]
Considering question b
Generally the standard deviation is mathematically represented as
[tex]\sigma_{x} = \frac{y - w}{\sqrt{\sqrt{12} } }[/tex]
=> [tex]\sigma_{x} = \frac{15 - 0}{\sqrt{\sqrt{12} } }[/tex]
=> [tex]\sigma_{x} = 4.33 \ minutes [/tex]
Considering question c
Generally the probability density function for a random variable X is
[tex]f(x) = \left \{ {{ \frac{1}{y} \ \ \ \ \ 0 < X < y } \atop {0} \\ \ \ \ Otherwise} \right.[/tex]
=> [tex]f(x) = \left \{ {{ \frac{1}{15} \ \ \ \ \ 0 < X < 15 } \atop {0} \\ \ \ \ Otherwise} \right.[/tex]
Generally the probability that the waiting time is between 5 and 11 minutes is mathematically represented as
[tex]P(5 < X < 11 )= \int\limits^{11}_{5} {f(x)} \, dx[/tex]
=> [tex]P(5 < X < 11 )= \int\limits^{11}_{5} {\frac{1}{15}} \, dx[/tex]
=> [tex]P(5 < X < 11 ) = \frac{1}{15} x | \left \ 11 } \atop {5}} \right.[/tex]
=> [tex]P(5 < X < 11 ) = \frac{1}{15} [ 11 - 5][/tex]
=> [tex]P(5 < X < 11 ) = 0.4 [/tex]
Considering question d
Generally the probability the waiting time is less than 5 minutes is mathematically represented as
[tex]P(X < 5) = \int\limits^5_{-\infty} {\frac{1}{15} } \, dx[/tex]
Given that time cannot be negetive
[tex]P(X < 5) = \int\limits^5_{0} {\frac{1}{15} } \, dx[/tex]
[tex]P(X < 5) = \frac{1}{15} * x | \left \ 5 } \atop {0}} \right.[/tex]
[tex]P(X < 5) =0.33.[/tex]
Generally the probability that the waiting time is less than 5 minutes on exactly 4 of 10 mornings follows a binomial distribution because the number of trial is finite i.e 10 the probability of success is probability that the waiting time is less than 5 minutes and probability of failure is probability that the waiting time is not less than 5 minutes(i.e there is only two outcome )
Hence for a random variable K representing the number of times in the total of 10 mornings that the waiting is less than 5 minutes then
The probability the random variable(W) is exactly 4 is mathematically represented as
[tex]P(W = 4 ) = ^{10}C_4 * (0.33)^4 * (1- p)^{10 -4}[/tex]
Here C denotes combination
So
[tex]P(W = 4 ) = 0.228[/tex]
