Question

the time that it takes for the next train to come follows a uniform distribution with f(x)=1/45 where x goes between 5 and 50 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 7 minutes.
f. find the probability that the time will be between 7 and 19 minutes.
g. find the 72nd percentile.
h. find the probability that the time is more than 25 minutes given (or knowing that) it is at least 7 minutes.
hint:
written hint
helpful videos: probability +, conditional probability + conditional probability + percentiles +

Explanation:

Step1: Identify distribution type

Given \(f(x)=\frac{1}{45}\) for \(x\in[5,50]\), it's a continuous uniform distribution. For a continuous uniform distribution on \([a,b]\), the probability - density function is \(f(x)=\frac{1}{b - a}\) where \(b - a=50 - 5 = 45\).

Step2: Calculate the mean

The formula for the mean \(\mu\) of a continuous uniform distribution on \([a,b]\) is \(\mu=\frac{a + b}{2}\). Here \(a = 5\) and \(b = 50\), so \(\mu=\frac{5+50}{2}=\frac{55}{2}=27.5\).

Step3: Calculate the standard - deviation

The formula for the standard deviation \(\sigma\) of a continuous uniform distribution on \([a,b]\) is \(\sigma=\sqrt{\frac{(b - a)^2}{12}}\). Substituting \(a = 5\) and \(b = 50\), we have \(b - a=45\), then \(\sigma=\sqrt{\frac{45^2}{12}}=\sqrt{\frac{2025}{12}}\approx13.2288\).

Step4: Calculate \(P(X\leq7)\)

The cumulative - distribution function of a continuous uniform distribution on \([a,b]\) is \(F(x)=\frac{x - a}{b - a}\) for \(a\leq x\leq b\). So \(P(X\leq7)=\frac{7 - 5}{50 - 5}=\frac{2}{45}\approx0.0444\).

Step5: Calculate \(P(7\lt X\lt19)\)

\(P(7\lt X\lt19)=\frac{19 - 7}{50 - 5}=\frac{12}{45}\approx0.2667\).

Step6: Calculate the 72nd percentile

Let \(k\) be the 72nd percentile. We know that \(F(k)=0.72\). Using \(F(k)=\frac{k - a}{b - a}\), substituting \(a = 5\), \(b = 50\), and \(F(k)=0.72\), we get \(0.72=\frac{k - 5}{45}\). Solving for \(k\): \(k-5=0.72\times45\), \(k - 5 = 32.4\), \(k=32.4 + 5=37.4\).

Step7: Calculate \(P(X\gt25|X\geq7)\)

By the formula for conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here \(A=\{X\gt25\}\) and \(B = \{X\geq7\}\). \(P(A\cap B)=P(X\gt25)=\frac{50 - 25}{50 - 5}=\frac{25}{45}\), \(P(B)=\frac{50 - 7}{50 - 5}=\frac{43}{45}\). So \(P(X\gt25|X\geq7)=\frac{\frac{25}{45}}{\frac{43}{45}}=\frac{25}{43}\approx0.5814\).

Answer:

a. continuous
b. uniform
c. 27.5000
d. 13.2288
e. 0.0444
f. 0.2667
g. 37.4000
h. 0.5814