Random variable and probability distributions ?

1. A continuous random variable X that can assume values between x = 1 and x=3 has a density function given by f(x) = 1/2. a.) Show that the area under the curve is equal to 1 b.) Find P(2%26lt;X%26lt;2.5). c.) P(X%26lt;1.6)





thanks for the hlp...i'm really confuse





2. From a box containing 4 black balls and 2 green balls,3 balls are drawn in succession, each ball being replaced in the box before the next draw is made.Find the probability distribution for the number of green balls ?





3. The proportion of people who respond to a certain mail-order solicitation is a continuous radom variable X that has the denssity function. f(x) = { 2(x+2)/5, 0%26lt;x%26lt;1


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a.) show that P(0-X-1)=1 b.) Find the probabi;ity that more than 1/4 but fewer than 1/2 of the people contacted will respond to this type of solicitation.

Random variable and probability distributions ?
1a. Integrate 1/2 dx from 1 to 3.. it should equal one


1b. Integrate 1/2 dx from 2 to 2.5 should be 0.25


1c. Integrate 1/2 dx from 1 (lower bound) to 1.6





2. This question is ambiguous. The number of green balls assuming what?





3a. Integrate f(x) dx from 0 to 1 to prove it's equal to one


3b. Integrate f(x) dx from 0.25 to 0.50
Reply:1)





for any continuous distribution you need to verify that the probability density function (pdf) is really a pdf by making sure the function integrates to 1 over the defined region.





∫f(x)dx from x = 1 to x = 3


= 1/2 x for x from 1 to 3


= 1/2 * (3 - 1) = 1/2 * 2 = 1





b) to find the probability of any continuous random variable for a given range you just need to integrate the pdf over the range for the probability.





= 1/2 * (2.5 - 2) = 1/4





c) integrate f(x) from 1 to 1.6 and you get a prob of 0.3





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2)


Let X be the number of green balls drawn


X ~ Binomial ( 3, 1/3 )





binomial with three trials (the three balls being drawn with replacement) and a success probability for each draw of 1/3





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3)


same logic as from 1.


a) integrate over the whole region to show the pdf will integrate to 1


b) integrate f(x) from x = 1/4 to x = 1/2 and you'll get 0.2375