Question

the time that it takes for the next train to come follows a uniform distribution with f(x) = 1/15 where x goes between 7 and 22 minutes. round answers to 4 decimal places when possible.
a. this is a select an answer distribution.
b. it is a select an answer distribution.
c. the mean of this distribution is
d. the standard deviation is
e. find the probability that the time will be at most 20 minutes.
f. find the probability that the time will be between 8 and 10 minutes.
g. find the 32nd percentile.
h. find the probability that the time is more than 15 minutes given (or knowing that) it is at least 11 minutes.

Explanation:

Step1: Identify distribution type

It is a continuous uniform distribution since \(f(x)=\frac{1}{15}\) for \(7\leq x\leq22\).

Step2: Calculate the mean

The formula for the mean of a uniform distribution \(U(a,b)\) is \(\mu=\frac{a + b}{2}\), where \(a = 7\) and \(b=22\). So \(\mu=\frac{7 + 22}{2}=\frac{29}{2}=14.5\).

Step3: Calculate the standard - deviation

The formula for the standard deviation of a uniform distribution \(U(a,b)\) is \(\sigma=\sqrt{\frac{(b - a)^2}{12}}\). Substituting \(a = 7\) and \(b = 22\), we have \(\sigma=\sqrt{\frac{(22 - 7)^2}{12}}=\sqrt{\frac{15^2}{12}}=\sqrt{\frac{225}{12}}\approx4.3301\).

Step4: Calculate \(P(X\leq20)\)

The formula for the cumulative - distribution function of a uniform distribution \(U(a,b)\) is \(P(X\leq x)=\frac{x - a}{b - a}\) for \(a\leq x\leq b\). Here, \(P(X\leq20)=\frac{20 - 7}{22 - 7}=\frac{13}{15}\approx0.8667\).

Step5: Calculate \(P(8\leq X\leq10)\)

\(P(8\leq X\leq10)=\frac{10 - 8}{22 - 7}=\frac{2}{15}\approx0.1333\).

Step6: Calculate the 32nd percentile

Let the 32nd percentile be \(k\). Using the formula \(P(X\leq k)=\frac{k - a}{b - a}=0.32\). Substituting \(a = 7\) and \(b = 22\), we get \(\frac{k - 7}{22 - 7}=0.32\), then \(k-7=0.32\times15\), and \(k = 7+4.8 = 11.8\).

Step7: Calculate \(P(X > 15|X\geq11)\)

By the formula for conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here, \(A=\{X > 15\}\) and \(B = \{X\geq11\}\). \(P(A\cap B)=P(X > 15)=\frac{22 - 15}{22 - 7}=\frac{7}{15}\), \(P(B)=\frac{22 - 11}{22 - 7}=\frac{11}{15}\). So \(P(X > 15|X\geq11)=\frac{P(X > 15)}{P(X\geq11)}=\frac{\frac{22 - 15}{22 - 7}}{\frac{22 - 11}{22 - 7}}=\frac{7}{11}\approx0.6364\).

Answer:

c. 14.5
d. 4.3301
e. 0.8667
f. 0.1333
g. 11.8
h. 0.6364