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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.

#### 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

#### 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

# 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.

#### 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 $$\int\limits_4^{\infty} f(t) dt$$

#### 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

# 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.

#### 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

#### 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 $$(6 \leq X\leq 14)$$

A

1.86

.

B

.03

C

.64

D

.89

Option C is Correct