Distribution Function,Continuous Distribution,Probability Density Function
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Subjective Questions
Q1:
Rationalize the denominator.
\(\frac{1}{\sqrt{3} -1}\)
Type: Very_short Difficulty: Easy
<p>\(\frac{1}{\sqrt{3} -1}\)</p>
<p>= \(\frac{1}{\sqrt{3} -1}\) \(\times\) \(\frac{\sqrt{3} +1}{\sqrt{3}+1}\)</p>
<p>= \(\frac{\sqrt{3} +1}{(\sqrt{3})-(1)^2}\)</p>
<p>= \(\frac{\sqrt{3} +1}{3-1}\)</p>
<p>= \(\frac{\sqrt{3} +1}{2}\)</p>
<p> </p>
Q2:
Rationalize the denominator.
\(\frac{1}{\sqrt{5} -1}\)
Type: Very_short Difficulty: Easy
<p>\(\frac{1}{\sqrt{5} -1}\)</p>
<p>= \(\frac{1}{\sqrt{5} -1}\) \(\times\) \(\frac{\sqrt{5} +1}{\sqrt{5}+1}\)</p>
<p>= \(\frac{\sqrt{5} +1}{(\sqrt{5})-(1)^2}\)</p>
<p>= \(\frac{\sqrt{5} +1}{5 -1}\)</p>
<p>= \(\frac{\sqrt{5} +1}{4}\)</p>
<p> </p>
Q3:
Solve \(\sqrt{2x + 3}\) = 5
Type: Very_short Difficulty: Easy
<p>\(\sqrt{2x + 3}\) = 5</p>
<p>Squaring both sides of the equation we get,</p>
<p>2x + 3 = 25</p>
<p>or, 2x = 25 -3</p>
<p>or, 2x = 22</p>
<p>x = 11</p>
<p>Checking for x = 11,</p>
<p>\(\sqrt{2\times11 + 3}\) = 5</p>
<p>or, \(\sqrt{22 + 3}\) = 5</p>
<p>or, \(\sqrt{25}\) = 5</p>
<p> 5 = 5 (True)</p>
<p>∴ x = 11</p>
<p> </p>
Q4:
Solve \(\sqrt x - 2\) = 3
Type: Short Difficulty: Easy
<p>\(\sqrt{ x}\)-2 = 5</p>
<p>or, \(\sqrt{x}\) = 3+2</p>
<p>or, \(\sqrt{x}\) = 5</p>
<p>Squaring on both sides,</p>
<p>or, \(\sqrt{x}\)<sup>2</sup> =(5)<sup>2</sup></p>
<p>or, x = 25</p>
<p>Checking for x = 25,</p>
<p>\(\sqrt{25}\) - 2 = 3</p>
<p>or, 5 - 2 = 3</p>
<p>or, 3 = 3 (true)</p>
<p>Hence, x = 25.</p>
Q5:
Find the square root of 13 + 2\(\sqrt{30}\)
Type: Very_short Difficulty: Easy
<p>Here, 13 + 2\(\sqrt{30}\) can be written as</p>
<p> 10 + 3 + 2\(\sqrt{30}\)</p>
<p>= 10 + 2 \(\sqrt{30}\) + 3</p>
<p>= (\(\sqrt{10}^2)\) + 2 \(\sqrt{10}\) \(\times\) \(\sqrt{3}\) + ( \(\sqrt{3}^2)\)</p>
<p>= (\(\sqrt{10}\) + \(\sqrt{3})^2\)</p>
<p>Square root of 13 + 2\(\sqrt{30}\) = \(\sqrt{10} + \sqrt{3}\)</p>
<p> </p>
<p> </p>
<p> </p>
Q6:
If a and b are rational numbers, Find the value of a and b in
\(\frac{3+ \sqrt{6}}{\sqrt{3 + \sqrt{2}}}\) = a + b \(\sqrt{3}\)
Type: Short Difficulty: Easy
<p>Here, \(\frac{3+ \sqrt{6}}{\sqrt{3 + \sqrt{2}}}\) =a + b \(\sqrt{3}\)</p>
<p>or, \(\frac{3+ \sqrt{6}}{\sqrt{3 + \sqrt{2}}}\) \(\times\) \(\frac{\sqrt3- \sqrt{2}}{\sqrt{3 - \sqrt{2}}}\)= a+b\(\sqrt{3}\)</p>
<p>or, \(\frac{3 \sqrt{3} - 3\sqrt{2} + \sqrt{18} - \sqrt{12}} {3-2}\) = a + b \(\sqrt{3}\)</p>
<p>or, 3\(\sqrt{3} - 3 \sqrt{2} + 3\sqrt{2} - 2 \sqrt{3}\) = a + b \(\sqrt{3}\)</p>
<p>or, \(\sqrt{3}\) = a + b \(\sqrt{3}\)</p>
<p>or, 0 + 1. \(\sqrt{3}\) = a + b \(\sqrt{3}\)</p>
<p>Comparing the coefficients, we get a =0 and b=1</p>
<p>∴ a = 0 and b= 1</p>
<p> </p>
Q7:
Simplify:
\(\frac{\sqrt{a + 1} - \sqrt{a^2 -1}} {\sqrt{a^2 + 1} + \sqrt{a^2 -1}}\)
Type: Short Difficulty: Easy
<p>\(\frac{\sqrt{a + 1} - \sqrt{a^2 -1}} {\sqrt{a^2 + 1} + \sqrt{a^2 -1}}\) \(\times\) \(\frac{\sqrt{a + 1} - \sqrt{a^2 -1}} {\sqrt{a^2 + 1} - \sqrt{a^2 -1}}\)</p>
<p>= \(\frac{(\sqrt{a^2 + 1} - \sqrt{a^2 -1)^2}} {(\sqrt{a^2 + 1})^2 - (\sqrt{a^2 -1})^2}\)</p>
<p>=\(\frac{\sqrt{(a^2 + 1)^2} - 2. \sqrt{a^2 + 1}. \sqrt{a^2 -1} + \sqrt{(a^2 -1)^2}}{(a^2 + 1) - ( a^2 - 1)}\)</p>
<p>=\(\frac{a^2+1-2\sqrt{{(a^2)}^2-{(1)^2}}+a^2 -1}{2}\)</p>
<p>=\(\frac{2a^2-2\sqrt{{a^4}-{1}}}{2}\)</p>
<p>= \(\frac{2( a^2 - \sqrt{a^4 -1)}}{2}\)</p>
<p>= \(a^2 - \sqrt{a^4 - 1}\)</p>
Q8:
Simplify:
\(\frac{3\sqrt{2}}{\sqrt{6}-{\sqrt{3}}}\) - \(\frac{4\sqrt{3}}{\sqrt{6}-{\sqrt{2}}}\) + \(\frac{2\sqrt{3}}{\sqrt{6}+{\sqrt{2}}}\)
Type: Long Difficulty: Easy
<p>\(\frac{3\sqrt{2}}{\sqrt{6}-{\sqrt{3}}}\) - \(\frac{4\sqrt{3}}{\sqrt{6}-{\sqrt{2}}}\) + \(\frac{2\sqrt{3}}{\sqrt{6}+{\sqrt{2}}}\)</p>
<p>= \(\frac{3\sqrt{2}}{\sqrt{6}-{\sqrt{3}}}\)\(\times\) \(\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}\) - \(\frac{4\sqrt{3}}{\sqrt{6}-{\sqrt{2}}}\)\(\times\) \(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}\) + \(\frac{2\sqrt{3}}{\sqrt{6}+{\sqrt{2}}}\) \(\times\) \(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}\)</p>
<p>=(\(\frac{3\sqrt{12}+ 3\sqrt{6}}{\sqrt{6^2}-\sqrt{3^2}}\) ) - (\(\frac{4\sqrt{18}+ 4\sqrt{6}}{\sqrt{6^2}-\sqrt{2^2}}\) ) + (\(\frac{2\sqrt{18}- 2\sqrt{6}}{\sqrt{6^2}-\sqrt{2^2}}\) )</p>
<p>= (\(\frac{3\sqrt{12}+3\sqrt{6}}{{6}-{3}}\) )- (\(\frac{4\sqrt{18}+ 4\sqrt{6}}{{6}-{2}}\) )+ (\(\frac{2\sqrt{18}- 2\sqrt{6}}{{6}-{2}}\) )</p>
<p>= (\(\frac{3(\sqrt{12}+{\sqrt6}}{{3}}\) ) - (\(\frac{4(\sqrt{18}+{\sqrt6}}{{4}}\) ) + (\(\frac{2(\sqrt{18}-{\sqrt6}}{{4}}\) )</p>
<p>= \(\sqrt{12}\) + \(\sqrt{6}\) - \(\sqrt{18}\) - \(\sqrt{6}\) + (\(\frac{\sqrt{18}-{\sqrt6}}{{2}}\) )</p>
<p>= \(\frac{2\times2\sqrt{3} - 2 \times 3 \sqrt{2} + 3\sqrt{2} - \sqrt{6}}{{2}}\)</p>
<p>= (\(\frac{4\sqrt{3}-{3\sqrt2}-{\sqrt{6}}} {{4}}\) )</p>
Q9:
Rationalize the denominator.
\(\frac{1}{\sqrt{7} -1}\)
Type: Short Difficulty: Easy
<p>\(\frac{1}{\sqrt{7} -1}\)</p>
<p>= \(\frac{1}{\sqrt{7} -1}\) \(\times\) \(\frac{\sqrt{7} +1}{\sqrt{7}+1}\)</p>
<p>= \(\frac{\sqrt{7} +1}{(\sqrt{7})-(1)^2}\)</p>
<p>= \(\frac{\sqrt{7} +1}{7-1}\)</p>
<p>= \(\frac{\sqrt{7} +1}{6}\)</p>
<p> </p>
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Distribution Function,Continuous Distribution,Probability Density Function
Distribution Function
Definition:
let X be a discrete random variable with pmf \(p(x_i)\) ; i=1,2,....,n then the function F for all integers x defined by
$$F(x)=P(X \leq x)=\sum_{{x_i \leq x}} p(x_i) \; \; \; \; \; \; \; \; \; ;- \infty < x < \infty$$
is called discrete distribution function of X,the summation is taken over all values of the r.v. X.
The distribution function of a r.v. X can be easily determined from the probability distribution of the r.v. X. Assuming the r.v. X takes real values \( x_1,x_2,....,x_n\) with respective probabilities \(p_1,p_2,......,p_n\) and \(x_1 < x_2 <...... < x_n\) then
$$P(X <x_1)=0, \; \; \; \; \; \; \; P(X \leq x_1)=P(X < x_1)+P(X=x_1)=p_1$$
$$P(X < x_2)=p_1, \; \; \; \; \; \; P(X \leq x_2)=P(X < x_2)+P(X=x_2)=p_1 +p_2$$
$$P(X < x_3)=p_1+p_2, \; \; \; P(X \leq x_3)=P(X < x_3)+P(X=x_3)=p_1 +p_2+p_3$$
$$and \; so \; on \; .Thus,$$
$$P(X < x_n)=P(X < x_n)+P(X=x_n)=p_1 +p_2+........+p_{n-1}+p_n=1$$
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Example
Determine the distribution function from the probability distributiongiven below:
| x | 0 | 1 | 2 | 3 |
| p(x) | \(\frac {1}{8}\) | \(\frac {3}{8}\) | \(\frac {3}{8}\) | \(\frac {1}{8}\) |
Also draw a graph of the distribution function.
Solution
The distribution function of the given probability distribution of r.v. X is determined as follows:
$$ For \;x=0, \; F(0)=p(0)=\frac{1}{8}$$
$$ For \;x=1, \; F(1)=p(0)+p(1)=\frac{1}{8}+\frac {3}{8}=\frac{4}{8}$$
$$ For \;x=2, \; F(2)=p(0)+p(1)+p(2)=\frac{1}{8}+\frac {3}{8}+\frac{3}{8}=\frac{7}{8}$$
$$ For \;x=3, \; F(3)=p(0)+p(1)+p(2)+p(3)=\frac{1}{8}+\frac {3}{8}+\frac{3}{8}+\frac{1}{8}=1$$
$$Hence, \; the \; required \; distribution \; function \; of \; X \; is \; given \; by$$
References
Sukubhattu,Narendra Prasad. Probability and Inference-I. Asmita Books Publishers & Distributors (P)Ltd. 2013
Lesson
Random Variables
Subject
Statistics
Grade
Bachelor of Science
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