# Thread: Some probability questions

1. ## Some probability questions

Anyone good with probability? I'm so stuck. Here are the questions I'm asked for the latest homework... If someone could help me through it that would be great.

1. A psychologist asks the subjects of her experiment to learn a particular task. The probability density function is determined to be f(x)= (32)/(3x^3) minutes on [2,4]

A. Find the probability it will take less than 3 minutes for the subject to learn the task.

B. Find the probability it will take more than 2.5 minutes for the subject to learn the task.

C. Find the expected number of minutes it will take to learn the particular task. (i.e find the expected value of the PDF)

D. Find the variance of the number of minutes it will take to learn the particular task. (Hint: remember to subtract the square of the expected value!)

2. Find the median for the probability density function f(x)=(x^2)/21 on [1,4]

3. The distance (in meters) that seeds are dispersed from a certain kind of plant is an exponential random variable density function f(x)=0.1e^(-0.1x) for x in [0,infinity).

A. Find the probability the seeds are less than 3 meters from the plant.

B. Bonus! Find the probability the seeds are more than 4 meters from the plant.

2. ## Re: Some probability questions

For question number 2 is this correct:
integrate (x^2)/21 which results in (x^3)/63 then you plug 4 in and 1 in and and take the difference. After that you divide by 3 for an answer of 1/3 or .3333?

3. ## Re: Some probability questions

No. Let $F(x)$ be the (cumulative) distribution function. Then the median is the "smallest" $x$ with $F(x)\geq 1/2$.
For your question, $$F(x)={x^3-27\over 63}$$
So the median is the smallest $x$ with $x^3-27\geq 63/2$; i.e. $x^3\geq 117/2$. So the median is $$\sqrt[3]{117/2}$$ or about 3.88.