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Probability On Random Distribution

Learn continuous random variables & graph of probability density function. Practice average value or mean of any probability density function & continuous random variables.

Continuous Random Variables

  • Consider the height of an adult male chosen at random, or life of a certain randomly chosen battery.   
  • These quantities are called continuous random variable as their range over an interval of real number. ( often rounded off to nearest integer) 
  • The cholesterol level of a person is also an example of continuous random variable.
  • We are sometimes interested in say what is the probability that a randomly chosen male has height between 65 and 70 inches or blood cholesterol level of a person is between 250 and 350 etc.
  • This means that we want to find the probability that a continuous random variable takes values in a certain range. 
  • Let x denote a continuous random variable then \(P(a\leq X \leq b)\) denotes the value of probability that X lies between 'a' and 'b'.
  • Every continuous random variable X has probability density function 'f' that means

\(P(a\leq X\leq b)=\int\limits^b_af(x)dx\)

  • In the figure shown the graph of probability density function for X when

\(\to\) height in inches of adult female in U.S.

Shaded area = probability that height of woman is between 60 and 70 inches.

  • Probability density function 'f' will always satisfy
  1.  \(f(x)\geq0\,\forall x\)
  2. \(\int\limits^\infty_{-\infty}f(x)dx=1\)  because probability values always lie in [0,1]
  • A function f(x) will qualify for probability density function only if it satisfies the above two criteria.

Illustration Questions

Let \(f\) be the probability density function of continuous random variable X when X is lifetime of a light bulb in hour, then what does the following integral represent \(\int\limits^{400}_{200} f(x) dx\)  

A The probability that lifetime of bulb is more than 200 hours.

B The probability that lifetime of bulb is less than 400 hours.

C The probability that lifetime of bulb is exactly 600 hours.

D The probability that lifetime of bulb is between 200  and 400 hours.

×

If \(f\) is a probability density function then probability that X lies between 'a' and 'b' is 

\(P(a\leq X \leq b) =\int\limits_a^b f(x)dx\)

\(\therefore\) \(\int\limits_{200}^{400}f(x)dx=\) probability that X lies between 200 and 400 i.e. life time of bulb is between 200 and 400 hours.

Let \(f\) be the probability density function of continuous random variable X when X is lifetime of a light bulb in hour, then what does the following integral represent \(\int\limits^{400}_{200} f(x) dx\)  

A

The probability that lifetime of bulb is more than 200 hours.

.

B

The probability that lifetime of bulb is less than 400 hours.

C

The probability that lifetime of bulb is exactly 600 hours.

D

The probability that lifetime of bulb is between 200  and 400 hours.

Option D is Correct

Illustration Questions

Let  \(f(x)=\Big\{\begin{matrix} {\alpha(5x-x^2)} & {if} & {0\leq x\leq5} \\ 0 & if & x<0 \text{ or } x >5 \\ \end{matrix}\)  find the values of constraint \(\alpha\) if \(f\)is a probability density function.  

A \(\alpha=\)72

B \(\alpha=\).048

C \(\alpha=\) .013

D \(\alpha=-\dfrac {1}{2}\)

×

For a function \(f\)to be probability density function?

  1.  \(f(x)\geq0\,\forall x\)
  2. \(\int\limits^\infty_{-\infty}f(x)dx=1\)

 

In this case 

\(f(x)=\Big\{\begin{matrix} {\alpha(5x-x^2)} & {if} & {0\leq x\leq5} \\ 0 & if & x<0 \text{ or } x >5 \\ \end{matrix}\)

\(\therefore\)  \(f(x)\geq0\;\forall x\) as \(\alpha x^{\nearrow^+} (5-x)^{\nearrow^+}\) is positive. If \(\alpha\geq0\).

Also, \(\int\limits_{-\infty}^{\infty}f(x)=1\)

\(\Rightarrow \int\limits_{-\infty}^{\infty} 0\; dx + \int\limits_{0}^{5} \alpha(5x-x^2)\; dx + \int\limits_{5}^{\infty} 0\; dx=1\)

\(\Rightarrow \int\limits_{0}^{5} \alpha(5x-x^2)\; dx=1\)

\(\Rightarrow\left[ \alpha\left( \dfrac {5x^2}{2}-\dfrac {x^3}{3} \right) \right] _0^5=1 \)   \(\Rightarrow \alpha\left( \dfrac {125}{2}-\dfrac {125}{3} \right)=1 \)

\(\Rightarrow \alpha×\dfrac {125}{6}=1\) \(\Rightarrow \alpha=\dfrac {6}{125}=0.048\)

Let  \(f(x)=\Big\{\begin{matrix} {\alpha(5x-x^2)} & {if} & {0\leq x\leq5} \\ 0 & if & x<0 \text{ or } x >5 \\ \end{matrix}\)  find the values of constraint \(\alpha\) if \(f\)is a probability density function.  

A

\(\alpha=\)72

.

B

\(\alpha=\).048

C

\(\alpha=\) .013

D

\(\alpha=-\dfrac {1}{2}\)

Option B is Correct

Finding the value of Probability of a Random Variable being in a given Range

Let X be a continuous random variable, then \(P(a\leq X\leq b)=\int\limits_a^bf(x) dx\)

where \(f\)is the probability density function of random variable X.

Illustration Questions

Consider the probability density function of a random variable X,  \(f(x) = \begin{cases} 20x^3(1-x) & if & 0\leq x\leq 1 \\ 0 & for & {\text {all other value}} \\ \end{cases} \) Find the value of \(P\Big( 1/4\leq X\leq1/2)\)

A 0.2763

B .171875

C 1.76

D –.583

×

\(P(a\leq X \leq b)=\int\limits_a^bf(x) \,dx\) where \(f\)is the probability density function of random variable X.

\(\therefore\)  \(P\Big( 1/4\leq X\leq1/2)=\int\limits_{1/4}^{1/2}20x^3(1-x) dx\)

\(=20\int\limits_{1/4}^{1/2}\Big(x^3-x^4\Big) dx\)

\(=20 \left [ \dfrac {x^4}{4}-\dfrac {x^5}{5} \right]_{1/4}^{1/2}\)

\(=20 \left [ \left ( \dfrac {1}{16×4}-\dfrac {1}{32×5} \right)- \left(\dfrac {1}{256×4}-\dfrac {1}{1024×5} \right) \right]\)

\(=20 \left [ \left ( \dfrac {1}{64}-\dfrac {1}{160}- \dfrac {1}{1024}+\dfrac {1}{5120} \right) \right]\)

\(20[.00859375]=.171875\)

Consider the probability density function of a random variable X,  \(f(x) = \begin{cases} 20x^3(1-x) & if & 0\leq x\leq 1 \\ 0 & for & {\text {all other value}} \\ \end{cases} \) Find the value of \(P\Big( 1/4\leq X\leq1/2)\)

A

0.2763

.

B

.171875

C

1.76

D

–.583

Option B is Correct

Finding the Probabilities from the graph of Probability Density Function

  • Suppose we are given the graph of \(f\)which is probability density function of some random variable X, then \(P(a\leq X\leq b)=\int\limits_a^b\,f(x)\,dx=\) Area under \(f(x)\) between x = a and x = b.

\(P(a\leq X\leq b)=\int\limits_a^b\,f(x)\,dx=\)Shaded area

Illustration Questions

Consider the graph of probability density function \(f\)of a random variable X. \((0 \leq X\leq 20)\) From this graph find \((6 \leq X\leq 14)\)

A 1.86

B .03

C .64

D .89

×

\(P(6\leq X\leq 14)=\int\limits_6^{14}\,f(x)\,dx=\) Area under \(f(x)\) between x = 6 and x = 14

Area = Area of trapezium ABCD + DCEF

\(=\dfrac {1}{2}[AB+CD]×AD+\dfrac {1}{2}[DC+EF]×DE\)

\(=\dfrac {1}{2}[.06+.1]×4+\dfrac {1}{2}[.1+.06]×4\)

\(=.16×2+.16×2=.16×4=.64\)

\(= .64\)

Consider the graph of probability density function \(f\)of a random variable X. \((0 \leq X\leq 20)\) From this graph find \((6 \leq X\leq 14)\)

image
A

1.86

.

B

.03

C

.64

D

.89

Option C is Correct

Average value or Mean of any Probability Density Function

  • The mean of a random variable X is the long run average value of random variable X. It can be interpreted as measure of centrality of probability density function.

\(\overline x=\)mean of \(f=\dfrac {\int\limits_{-\infty}^\infty x\, f(x) dx}{\int\limits_{-\infty}^\infty \, f(x) dx}=\int\limits_{-\infty}^\infty x\, f(x) dx\)

\(\because\)  \(\int\limits_{-\infty}^\infty \, f(x) dx=1\)

\(\therefore\) \(\overline x=\mu=\int\limits_{-\infty}^{\infty}xf(x)\,dx\) = mean of \(f\)which is the Probability density function of random variable X.

Illustration Questions

Suppose the probability density function of a random variable X is given by \(f(x)= \begin {cases} 0 & if & x <0 \\ 2e^{-2x}&if&x\geq0 \end{cases}\) Find the mean of this distribution.

A \(\dfrac {3}{4}\)

B \(\dfrac {1}{2}\)

C \(\dfrac {5}{4}\)

D 6

×

Mean \(\overline x=\mu=\int\limits_{-\infty}^{\infty}xf(x)\,dx\)

In this case

\(\overline x=\mu=\int\limits_{-\infty}^{0}x×0\,dx+ \int\limits_{0}^{\infty}x×2e^{-2x}\,dx\)

\(\overline x=0+\int\limits_{0}^{\infty}2xe^{-2x}\,dx+ 2\int\limits_{0}^{\infty}x\,e^{-2x}\,dx\) ... (i)

Now consider,

\(\int x\,e^{-2x}\,dx=x×\dfrac {e^{-2x}}{-2}-\int1×\dfrac {e^{-2x}}{-2}\)(Integration by parts)

\(=-\dfrac {xe^{-2x}}{2}-\dfrac {1}{4}e^{-2x}+c\)

\(\therefore\) Value of Equation (i) becomes \(\overline x=\mu=2× \left [ \dfrac {-xe^{-2x}}{2}-\dfrac {-e^{-2x}}{4} \right ]_0^\infty\)

\(=2× \left [ 0- \left ( 0-\dfrac {1}{4} \right) \right]=2×\dfrac {1}{4}=\dfrac {1}{2}\)

\(\Rightarrow \mu=\dfrac {1}{2}\)

Suppose the probability density function of a random variable X is given by \(f(x)= \begin {cases} 0 & if & x <0 \\ 2e^{-2x}&if&x\geq0 \end{cases}\) Find the mean of this distribution.

A

\(\dfrac {3}{4}\)

.

B

\(\dfrac {1}{2}\)

C

\(\dfrac {5}{4}\)

D

6

Option B is Correct

Waiting Time Problem

  • Suppose X is a random variable which is the time you wait on hold before an agent of a company you are calling, answers your call, then it is found that.
  • \(f(t)= \begin {cases} 0& if&t<0\; (\text {as call cannot be answered before it is made})\\ ce^{-ct}& if& t \geq 0\\ \end {cases}\)
  • It is the probability density function of X where c is some constant value (t is the time at which call is answered)     
  • Also it can be verified that \(\mu=\overline x=\dfrac {1}{c}\) for this distribution.
  • So the probability that the call will be answered during first minute is  \(\int\limits_0^1 f(t) dt\) 
  • or it will be answered any time after the 4th minute is \(\int\limits_4^{\infty} f(t) dt\) 

Illustration Questions

Let X is the waiting time for a customer's call to be answered by an agent. If X has the probability density function as given below. \(f(t)= \begin {cases} 0& if&t<0\\ 0.25\,e^{-t/4}& if& t \geq 0\\ \end {cases}\) what is the probability that call will be answered during 4th minute?

A 0.7081

B 0.1045

C 1.3926

D 0.0012

×

P (Call will be answered during 4th minute)

 = \(P(3\leq x\leq 4)=\int\limits_3^4f(t) dt\)

\(\int\limits_3^4f(t) dt=\int\limits_3^40.25\,e^{-t/4} dt\)

\(=0.25 \left [ \int\limits_3^4\,e^{-t/4}\,dt \right ] =0.25× \left [ \dfrac {e^{-t/4}}{-1/4} \right ]_3^4\)

\(=-1×[e^{-1}-e{^{-3/4}}]=e{^{-3/4}}-e{^{-1}}\)

\(=0.4724-0.3679=0.1045\)

\(\therefore\) \(P(3\leq X\leq 4)=0.1045\)

Let X is the waiting time for a customer's call to be answered by an agent. If X has the probability density function as given below. \(f(t)= \begin {cases} 0& if&t<0\\ 0.25\,e^{-t/4}& if& t \geq 0\\ \end {cases}\) what is the probability that call will be answered during 4th minute?

A

0.7081

.

B

0.1045

C

1.3926

D

0.0012

Option B is Correct

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