# Statistics Practice Problems Help

• March 31st 2011, 12:48 PM
ChrisYee08
Statistics Practice Problems Help
I just need some help on these problems for my test tomorrow. I do not know how to even start this problem.

1. Suppose the random variable X represents the time until an event occurs and that the probability distribution of X is exponential with a mean of 50 years.

a. Find the median value of X (the median of the probability distribution).

b. Graph the probability distribution of X, with axis labels. Show where the mean and the median lie along the horizontal axis.

c. Find P(X ≤ mean) and compare with P(X ≤ median).

d. Find P(X > 1), P(X > 50) and P(X > 100), the probability that the waiting time is more than 1 year, 50 years and 100 years, respectively.

e. Find P(X > 101 years given that X > 100 years). Compare with P(X > 1).

• March 31st 2011, 12:59 PM
SpringFan25
(a)
suppose the median is k. The definition of a median means that

$P(X \leq K) = 0.5$

Use the cdf and set its value to 0.5.

$1-e^{-0.02k}=0.5$
solve for k.

(b) the graph looks like this
http://www.wolframalpha.com/input/?i=plot+0.02e^%28-0.02x%29%2C+x%3D0...100

(it extends to infinity, but i've only showed up to x=100)

(c) use the cdf to find $p(X\leq 50)$

post your attempt to at least part of (d) and (e) if you need help on those.
• March 31st 2011, 01:07 PM
ChrisYee08
wow ty so much, ill post my attempt in a bit after i finish making dinner. what does cdf mean anyways?
• March 31st 2011, 01:11 PM
SpringFan25
cdf = cumulative distribution function, which the probability that X is less than some value
• March 31st 2011, 01:12 PM
ChrisYee08
ook cool cool. ill put my answers here when i get them.
• March 31st 2011, 01:47 PM
ChrisYee08
for a) i got 34.66 as the median, sand for c i got .63 as the probability if less than the mean. are these right? and do i just use the cdf equation to find the other probabilities in d and e?
• March 31st 2011, 01:57 PM
SpringFan25
34.66 and 0.63 are correct.

You can just use the cdf for (d)

for (e) you need to use the conditional probability formula

$P(A|B) = \frac{P(A ~ and ~ B)}{P(B)}$
• March 31st 2011, 02:03 PM
ChrisYee08
so is just do 1-e^-.02k=x right? or would the - become a +?
• March 31st 2011, 02:13 PM
SpringFan25
• March 31st 2011, 02:24 PM
ChrisYee08
im asking about question D. should i still use the equation 1-e^-.02k=x ?
• March 31st 2011, 02:31 PM
SpringFan25
I think you have the right idea, but just to be sure:

$P(X>k) = 1-P(X \leq k)$

$P(X>k) = 1-\left[1-e^{-0.02k} \right]$

$P(X>k) = e^{-0.02k}$

for the first part of (d), we need k=1
$P(X>1) = e^{-0.02}$
• March 31st 2011, 02:34 PM
ChrisYee08
so the equation im supposed to use is e^-.02k = x ?
• March 31st 2011, 02:38 PM
SpringFan25
there is a difference between

e^-.02k = x

and

e^-.02k = P(x > k)

The red one is the correct thing to write and can be used to find P(x>1).
• March 31st 2011, 02:52 PM
ChrisYee08
kk ty, for d) i got P(x>1) = .98, P(x>50) = .37, P(x>100) = .14, and for e) i got, 0.14