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Suppose that f(x) is a probability density function defined as follows:

f(x) = (3/8) x², for 0 ≤ x ≤ g
f(x) = 0, otherwise

a) What is the value of "g" in this probability density function?

b) Calculate the probability that a random variable following this distribution falls in the interval [1, 3].

User Kkurni
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1 Answer

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Final answer:

To find the value of "g", the total area under the pdf must equal 1, leading to an integral that will solve for g. To find the probability in the interval [1, 3], integrate the function between those limits.

Step-by-step explanation:

Finding the Value of "g" and Probability Interval

To find the value of "g" in the given probability density function (pdf), we utilize the property that the total area under the curve of a pdf must equal 1, representing the total probability of all outcomes in a continuous probability distribution. The function provided is f(x) = (3/8) x² for the interval from 0 to g. To find g, we solve the following integral:

\[\int_{0}^{g} (3/8) x² dx = 1\]

This requires integrating the function and then solving for g to ensure the area under the curve adds up to 1.

Next, to calculate the probability that a random variable falls within the interval [1, 3], we evaluate the integral of f(x) from 1 to 3:

\[P(1 \leq x \leq 3) = \int_{1}^{3} (3/8) x² dx\]

This yields the desired probability which is the area under the function between x=1 and x=3.

User Piyush
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