Question

Suppose that it is known that the number of minutes that the daily 1 pm train from NJ Station arrives late to destination Linden is a random variable X with density function

f(x) = {c/15^2(15^2-x^2), -15<x<15;
0, elsewhere.

(a) Determine the value of c so that f(x) is a valid probability density function.

(b) Explain why c must be a particular value. That is, explain why probability density functions must have the specific property that you used to solve part (a).

(c) Use the value of c found in part (a). Find the probability that the train arrives between 0 (perfectly on time) and 10 minutes late.

(d) Plot the graph of the probability density function, shade in the area between x = 0 and x = 10, and explain what your result in (c) means with respect to your graph.

(e) Find the expected value of X, the variance of X, and the standard deviation of X. Note that, again, you should use the value found for c here.

Answer 1

Answer

It is known that the number of minutes that the daily 1 PM train from NJ Station arrives late to destination Linden is a random variable X with density function

(a) Since, f(x) is a probability density function, the f(x) must have the following property:

[since, f(- x) = f(x), therefore f(x) is an even function. By property of definite integration, , when f(x) is an even function]

Answer: The value of c is 1/20 so that f(x) is a valid probability density function.

(b) We know, for a continuous random variable X defined over ,

Since, is the total probability defined over the entire sample space, therefore

Hence,

(c) The probability that the train arrives between 0 and 10 minutes late is

Answer: The probability that the train arrives between 0 and 10 minutes late is 23/54.

(d) We plot the graph of the probability density function, shade in the area between x = 0 and x = 10.

From the shaded region of the graph, it is evident that the probability is slightly less than 0.5 or 1/2. In (c), we have found the result as 23/54 which is also slightly less than 1/2. This implies that our result in (c) matches with the graph drawn in here.

(e) The expected value of X is

Let,

Therefore,

which imples g(x) is an odd function. From the property of definite integration,

, if g(x) is an odd function.

Hence, E(X) = 0

The variance of X is

, [since, E(X) = 0]

Now,

, [since, the integral is an even function]

= 45

Therefore, Var (X) = 45

Hence, s.d.(X) =

Answer: The expected value of X is 0, the variance of X is 45 and the standard deviation of X is .