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Chapter 23: Problem 3

A p.d.f., \(f(x)\), for a continuous variable \(X\) is given by $$ f(x)=\frac{3}{10}\left(x^{2}+1\right), \quad 11.25)\).

Short Answer

Expert verified
Question: Verify if the function \(f(x) = \frac{3}{10}(x^2+1)\) for \(1 \leq x \leq 2\) can be a probability density function (pdf) for a continuous random variable X. If it is a valid pdf, find the following probabilities: (a) \(P(1.7 < X < 2)\) (b) \(P(X < 1.5)\) (c) \(P(X > 1.25)\) Answer: (a) Yes, the function \(f(x) = \frac{3}{10}(x^2+1)\) for \(1 \leq x \leq 2\) is a valid probability density function. (b) \(P(1.7 < X < 2) = 0.445\) (c) \(P(X < 1.5) = 0.295\) (d) \(P(X > 1.25) = 0.553\)

Step by step solution

01

(a) Verify that f(x) can be a p.d.f

To verify if the given function \(f(x)\) can be a p.d.f, it must satisfy the following two conditions: 1. \(f(x) \geq 0\) for all \(x\) 2. \(\int_{-\infty}^{\infty} f(x) dx = 1\) First, we can see that the function is non-negative because \(x^2 + 1\) is always non-negative and multiplying by \(\frac{3}{10}\) won't change its sign. Now, we will evaluate the integral of the function over its domain and check if it equals to 1: $$ \int_{1}^{2} \frac{3}{10}(x^2+1) dx $$
02

Evaluate the integral

Next, we will evaluate the integral by finding its antiderivative and applying the fundamental theorem of calculus: $$ \int_{1}^{2} \frac{3}{10}(x^2+1) dx = \frac{3}{10}\left[\frac{1}{3}x^3 + x\right]_{1}^{2} = \frac{1}{10}(8 - \frac{1}{3}) + \frac{3}{10}(2 - 1). $$
03

Check if integral is 1

Now, we will check if the evaluated integral equals to 1: $$ \frac{1}{10}(8 - \frac{1}{3}) + \frac{3}{10}(2 - 1) = \frac{1}{10}(7\frac{2}{3}) +\frac{3}{10} = \frac{23}{30} +\frac{3}{10} = 1. $$ Since the integral of \(f(x)\) over its domain is 1, we can conclude that it is a valid p.d.f.
04

(b) Find \(P(1.7

To find the probability of the event \(1.7 < X < 2\), we will integrate \(f(x)\) over the interval \((1.7, 2)\): $$ P(1.7 < X < 2) = \int_{1.7}^{2} \frac{3}{10}(x^2+1) dx. $$ Now, we will use the antiderivative we found earlier and apply the fundamental theorem of calculus: $$ P(1.7 < X < 2) = \frac{3}{10}\left[\frac{1}{3}x^3 + x\right]_{1.7}^{2} = \frac{1}{10}\left[(8 - \frac{1}{3}) - (4.913 - \frac{1}{3})\right] + \frac{3}{10}(2 - 1.7). $$ Calculating the above expression, we get: $$ P(1.7 < X < 2) = \frac{1}{10}\left(7\frac{2}{3} - 4\frac{56}{100}\right) + \frac{3}{10}(0.3) = 0.445. $$
05

(c) Find \(P(X

Similarly, to find the probability of the event \(X < 1.5\), we will integrate \(f(x)\) over the interval \((1, 1.5)\): $$ P(X < 1.5) = \int_{1}^{1.5} \frac{3}{10}(x^2+1) dx. $$ We will again use the antiderivative we found earlier and apply the fundamental theorem of calculus: $$ P(X < 1.5) = \frac{3}{10}\left[\frac{1}{3}x^3 + x\right]_{1}^{1.5} = \frac{1}{10}\left[(3.375 - \frac{1}{3}) - (1 - \frac{1}{3})\right] + \frac{3}{10}(1.5 - 1). $$ Calculating the above expression, we get: $$ P(X < 1.5) = \frac{1}{10}\left(3\frac{8}{27} - \frac{2}{3}\right) + \frac{3}{10}(0.5) = 0.295. $$
06

(d) Find \(P(X>1.25)\)

Finally, to find the probability of the event \(X > 1.25\), we will integrate \(f(x)\) over the interval \((1.25, 2)\): $$ P(X > 1.25) = \int_{1.25}^{2} \frac{3}{10}(x^2+1) dx. $$ We will again use the antiderivative we found earlier and apply the fundamental theorem of calculus: $$ P(X > 1.25) = \frac{3}{10}\left[\frac{1}{3}x^3 + x\right]_{1.25}^{2} = \frac{1}{10}\left[(8 - \frac{1}{3}) - (2.44125 - \frac{1}{3})\right] + \frac{3}{10}(2 - 1.25). $$ Calculating the above expression, we get: $$ P(X > 1.25) = \frac{1}{10}\left(7\frac{2}{3} - 2\frac{5775}{10000}\right) + \frac{3}{10}(0.75) = 0.553. $$

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Most popular questions from this chapter

Silicon chips are manufactured by four machines, A, B, C and D. Machines A, B, C and D manufacture \(20 \%, 25 \%, 35 \%\) and \(20 \%\) of the components respectively. Of those silicon chips manufactured by machine A, \(2.1 \%\) are faulty. The respective figures for machines \(\mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) are \(3 \%, 1.6 \%\) and \(2.5 \%\). A silicon chip is selected at random. Calculate the probability that it is (a) made by machine \(\mathrm{C}\) and is faulty (b) made by machine \(\mathrm{A}\) and is not faulty (c) faulty.

The data set \(A=\left\\{x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right\\}\) has a mean of \(\bar{x}\) and a standard deviation of \(\sigma .\) The data set \(B\) is \(\left\\{k x_{1}, k x_{2}, k x_{3}, \ldots, k x_{n}\right\\}\), the data set \(C\) is \(\left\\{x_{1}+k, x_{2}+k, x_{3}+k, \ldots, x_{n}+k\right\\}\) where \(k\) is a constant. (a) State the mean of set \(B\). (b) State the mean of set \(C\). (c) State the standard deviation of set \(B\). (d) State the standard deviation of set \(C\).

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