SOLUTION
a. The mean of this distribution
Mean of a uniform distribution is given as
[tex]\begin{gathered} \frac{b+a}{2} \\ =\frac{9+0}{2} \\ =\frac{9}{2} \\ =4.5 \end{gathered}[/tex]
Hence, the mean is 4.5
b. The standard deviation
This is given as
[tex]\begin{gathered} \frac{b-a}{\sqrt[]{12}} \\ =\frac{9-0}{\sqrt[]{12}} \\ =\frac{9}{\sqrt[]{12}} \\ =2.5981 \end{gathered}[/tex]
Hence the standard deviation is 2.5981
c. The probability that the person will wait more than 6 minutes
This is
[tex]\begin{gathered} P(x>6)=\frac{9-6}{9-0} \\ =\frac{3}{9} \\ =0.3333 \end{gathered}[/tex]
Hence, the answer is 0.3333
d. Suppose that the person has already been waiting for 2.4 minutes. Find the probability that the person's total waiting time will be between 4.8 and 6.8 minutes.
This can be written as
[tex]P(4.82.4)[/tex]
From here,
[tex]f(x)=\frac{1}{9-2.4}=\frac{1}{6.6}[/tex]
So,
[tex]\begin{gathered} P(4.82.4)=(6.8-4.8)\times\frac{1}{6.6} \\ =2\times\frac{1}{6.6} \\ =0.3030 \end{gathered}[/tex]
Hence the answer is 0.3030
e. 37% of all customers wait at least how long for the train?
This becomes
100 - 37 = 63
[tex]k(\frac{1}{9})=0.63[/tex]
Then we find k
[tex]\begin{gathered} k(\frac{1}{9})=0.63 \\ k=\frac{0.63}{\frac{1}{9}} \\ k=5.67\text{ minutes } \end{gathered}[/tex]
Hence, the answer is 5.67 minutes
