Question

The pal.f. of a continuous random variable is \(f(x)=k / x^{3}, 5 \leq x \leq 10 ; 0\) elsewhere, then the value of \(k\) is (a) 1 (b) 50 (c) \(200 / 6\) (d) 200 .

Short answer

The value of \(k\) is \(\frac{200}{3}\).

Step-by-step solution

Understand the Density Function

The given probability density function (p.d.f.) is \(f(x) = \frac{k}{x^3}\) for values \(x\) between 5 and 10. Outside this range, \(f(x) = 0\). The task is to find the constant \(k\) such that the total probability over its domain, from 5 to 10, equals 1.

Set Up the Integral for Total Probability

We need the integral of the function from 5 to 10 to equal 1. This means solving: \[\int_{5}^{10} \frac{k}{x^3} \, dx = 1\]

Compute the Integral

First, find the antiderivative of \(\frac{1}{x^3}\), which is \(-\frac{1}{2x^2}\). Using the Fundamental Theorem of Calculus, evaluate: \[\int_{5}^{10} \frac{1}{x^3} \, dx = \left[-\frac{1}{2x^2}\right]_5^{10} = -\frac{1}{2 \times 10^2} + \frac{1}{2 \times 5^2}\]This simplifies to:\[-\frac{1}{200} + \frac{1}{50}\]

Simplify the Result of the Integral

Perform the arithmetic operation: \[\frac{1}{50} - \frac{1}{200} = \frac{4}{200} - \frac{1}{200} = \frac{3}{200}\]So, the integral simplifies to \(\frac{3}{200}\).

Solve for the Constant \(k\)

Set the computed integral equal to \(1\) since it must represent the total probability: \[k \times \frac{3}{200} = 1\]Solving for \(k\), multiply both sides by \(\frac{200}{3}\):\[k = \frac{200}{3}\]

Validate the Solution

Ensure that the solution makes sense by re-checking the calculations for the integral and confirming that the value satisfies the equation. The total probability from 5 to 10 should indeed sum up to 1 with this \(k\).

Key concepts

Continuous Random Variable

A continuous random variable can take any value within a specified range. Unlike discrete random variables, which stick to distinct separate values, continuous variables fill the gaps in between.
For instance, think about measuring the exact time it takes for a coffee cup to cool from boiling to room temperature. It doesn't just snap to 1 or 2 minutes. It could be any millisecond in between.
Now, in probability, we describe how likely these values are to occur using a probability density function (p.d.f.). This function helps us understand how the probabilities for a continuous random variable are distributed over its range.
In our exercise, the continuous random variable is defined over the range from 5 to 10. Outside this boundary, any probability is zero. This restriction forms our valid calculation domain for integrating the density function to find the constant \(k\).

Integral Calculus

Integral calculus plays a crucial role in finding probabilities for continuous random variables. It helps us add up probabilities over a continuous range to find totals. The process of integration enables us to determine the total probability as it accumulates.
In our exercise, we need to calculate the integral of the function \(f(x) = \frac{k}{x^3}\) from 5 to 10. This integral represents the total probability over this interval, which must sum to 1.
To do this, we find the antiderivative of \(\frac{1}{x^3}\), which is achieved by reversing the process of differentiation. The antiderivative \(-\frac{1}{2x^2}\) allows us to apply limits in the function to calculate total area under the curve, aligning our work with probability rules for continuous functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the two main operations in calculus: differentiation and integration. It's our key tool for linking functions to their antiderivatives in a meaningful way.
Simply put, the theorem allows us to evaluate the integral of a function by using its antiderivative. By evaluating this antiderivative at the boundaries of our interval, we can effectively calculate the accumulation of a quantity—like our total probability here.
In the exercise, we determine another layer of our solution by computing \(\left[-\frac{1}{2x^2}\right]_5^{10}\). This is essentially doing a final subtraction of computed values at the defined limits of integration (10 and 5).

• The result: \(-\frac{1}{200} + \frac{1}{50}\), simplifies to \(\frac{3}{200}\).
• We then solve \(k \times \frac{3}{200} = 1\) to find \(k\).

This process exemplifies the essence of how integral calculus, guided by the Fundamental Theorem, provides clear solutions to probability-based questions.