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Probability On Normal Distribution

Evaluate Integral in normal distribution with probability density function. Practice Speed of Vehicles on the highway using standard normal distribution formula & cumulative distribution

Normal Distribution

  • Many random phenomenon such as test scores, heights of individuals, annual rainfall at a location are modeled by Normal Distribution.
  • In Normal Distribution the probability density function \(f\) for the random variable X is given by \(f(x)=\dfrac {1}{\sigma\sqrt {2\pi}}\,e\,^{-(x-\mu)^2/(2\sigma^2)}\)
  • The mean of this density function is \(\mu\)
  • \(\sigma=\) standard deviation of the X and it measures how spread out the values of X are.
  • These are the possible graph of function \(f\).

  • More the value of \(\sigma\)more the values of X are spread out.

Illustration Questions

The Normal distribution, density function of a random variable X is of the function \(f(x)=\dfrac {1}{20\sqrt {2\pi}}\,e\,^{-(x-100)^2/2×20^2}\) Find the value of \(\mu\) and \(\sigma\) for this distribution.

A \(\mu=\) 15, \(\sigma=\)20

B \(\mu=\) 5, \(\sigma=\)2

C \(\mu=\) 20, \(\sigma=\)100

D \(\mu=\) 100, \(\sigma=\)20

×

In normal distribution, the probability density function of random variable X is a member of family of function

\(f(x)=\dfrac {1}{\sigma\sqrt {2\pi}}e\;^{-(x-\mu)^2/2\sigma^2}\)

where, 

\(\mu\)= mean of distribution

\(\sigma\) = standard deviation of distribution

In this case,

\(f(x)=\dfrac {1}{20\sqrt {2\pi}}e\;^{-(x-100)^2/2×20^2}\)

Comparing with standard function we get

\(\mu=\) 100, \(\sigma=\)20

The Normal distribution, density function of a random variable X is of the function \(f(x)=\dfrac {1}{20\sqrt {2\pi}}\,e\,^{-(x-100)^2/2×20^2}\) Find the value of \(\mu\) and \(\sigma\) for this distribution.

A

\(\mu=\) 15, \(\sigma=\)20

.

B

\(\mu=\) 5, \(\sigma=\)2

C

\(\mu=\) 20, \(\sigma=\)100

D

\(\mu=\) 100, \(\sigma=\)20

Option D is Correct

Illustration Questions

It is found that height of adult males in united states are normally distributed with means 69 inches and standard deviation 2.8 inches what is the probability that an adult male chosen at random is between 70 and 75 inches tall. (Just state the interval).

A \(\int\limits_{70}^{75}\dfrac {1}{10}e\;^{-(x-100)^2/2}\; dx\)

B \(\int\limits_{5}^{10} e\;^{-x^2}\; dx\)

C \(\int\limits_{70}^{75}\dfrac {1}{7.02}e\;^{-(x-63)^2/15.68}\; dx\)

D \(\int\limits_{1}^{2}e\;^-x^2\; dx\)

×

In normal distribution, the probability density function of random variable X is a member of family of function

\(f(x)=\dfrac {1}{\sigma\sqrt {2\pi}}e\;^{-(x-\mu)^2/2\sigma^2}\)

where, 

\(\mu\)= mean of distribution

\(\sigma\) = standard deviation of distribution

In this case,

\(f(x)=\dfrac {1}{2.8\sqrt {2\pi}}e\;^{-(x-69)^2/2×2.8^2}\)

\(\Rightarrow f(x)=\dfrac {1}{7.02}e\;^{-(x-69)^2/15.68}\)

\(\therefore\)  \(P(70\leq X\leq 75)=\int\limits_{70}^{75}\dfrac {1}{7.02}e\;^{-(x-69)^2/15.68}\; dx\)

It is found that height of adult males in united states are normally distributed with means 69 inches and standard deviation 2.8 inches what is the probability that an adult male chosen at random is between 70 and 75 inches tall. (Just state the interval).

A

\(\int\limits_{70}^{75}\dfrac {1}{10}e\;^{-(x-100)^2/2}\; dx\)

.

B

\(\int\limits_{5}^{10} e\;^{-x^2}\; dx\)

C

\(\int\limits_{70}^{75}\dfrac {1}{7.02}e\;^{-(x-63)^2/15.68}\; dx\)

D

\(\int\limits_{1}^{2}e\;^-x^2\; dx\)

Option C is Correct

Speed of Vehicles on the highway (Normal distribution)

It is observed that speed of vehicles which run on a highway are normally distributed and if the maximum speed limits is known, the probability that a randomly chosen vehicle is traveling at a legal speed can be found.

Illustration Questions

The speed of vehicles on a highway with speed limit 100 km/hr are normally distributed with mean = \(\mu\)= 110 km/hr and standard deviation \(\sigma=\)10 km/hr. What is the probability that a randomly chosen vehicle is traveling at a legal speed.

A .7963

B .1584

C 1.0832

D .0014

×

The probability density function \(f\) for normal distribution is given by

\(f(x)=\dfrac {1}{\sigma\sqrt {2\pi}}e\;^{-(x-\mu)^2/2\sigma^2}\)

where,

\(\mu\)= mean of distribution

\(\sigma\) = standard deviation of distribution

In this case 

\(f(x)=\dfrac {1}{\sigma\sqrt {2\pi}}e\;^{-(x-\mu)^2/2\sigma^2}\)

\(=\dfrac {1}{10×\sqrt {2\pi}}e\;^{-(x-110)^2/2×100}\)

\(=\dfrac {1}{10×2.51}×e\;^{-(x-110)^2/200}\)

\(=\dfrac {1}{25.1}×e\;^{-(x-110)^2/200}\)

\(\therefore \) P (random vehicle is traveling at a legal speed)

= P (speed is less than 100 km/hr)

\(=P(0\leq X \leq 100)=\int\limits_0^{100}\dfrac {1}{25.1}\,e^{-(x-110)^2/200} dx\)

\(=\dfrac {1}{25.1}×\int\limits_0^{100}\,e^{-(x-110)^2/200} dx\)

\(=\dfrac {3.98}{25.1}=.1584\)

The speed of vehicles on a highway with speed limit 100 km/hr are normally distributed with mean = \(\mu\)= 110 km/hr and standard deviation \(\sigma=\)10 km/hr. What is the probability that a randomly chosen vehicle is traveling at a legal speed.

A

.7963

.

B

.1584

C

1.0832

D

.0014

Option B is Correct

Probability of the Random Variable lies within n Standard Deviation About the Mean in Normal Distribution Function

Consider the probability density function for random variable 

\(f(x)=\dfrac {1}{\sigma \sqrt{2\pi}}\,e^{-(x-\mu)^2/2\sigma^2}\)

Suppose we want to know what is the probability that random variable X lie within. one standard deviation of the mean i.e.

\(P(\mu-\sigma\leq X \leq \mu+\sigma)\)

\(=\int\limits_{\mu-\sigma}^{\mu+\sigma}\;\dfrac {1}{\sigma\sqrt {2\pi}}e^{-(x-\mu)^2/2\sigma^2}\)

\(=\int\limits_{\mu-\sigma}^{\mu+\sigma}\; \dfrac {1}{\sigma\sqrt {2\pi}} \;e^{- \left (\dfrac {x-\mu}{\sigma}\right)^2 ×\dfrac {1}{2}}\;dx\)   ...(i)

Put \(\dfrac {x-\mu}{\sigma}=t\Rightarrow dx=\sigma\; dt\)

\(\therefore\) Equation (i) becomes \(\int\limits_{-1}^{1}\dfrac {1}{\sigma \sqrt {2\pi}} \;e^{-t^2/2}\;\sigma\;dt \)

\(\dfrac {1}{ \sqrt {2\pi}}\int\limits_{-1}^{1} \;e^{-t^2/2}\;\;dt \)  which is independent of the standard deviation \(\sigma\).

  • We can similarly find the probability that random variable X lies with n standard deviation about the mean as 

\(P(\mu-n\sigma\leq X\leq \mu+n\sigma)\)\(=\dfrac {1}{ \sqrt {2\pi}}\int\limits_{-n}^{n} \;e^{-t^2/2}\;\;dt \)

Illustration Questions

For any normal distribution find the probability that random variable lies within one standard derivation about the mean

A 1.323

B .976

C .682

D .012

×

\(P(\mu-n\sigma\leq X\leq \mu+n\sigma)=\dfrac {1}{\sqrt2\,\pi}\int\limits_{-n}^{n}\,e^{-t^2/2}\,dt\)

for a normal distribution.

In this case n = 1

\(\therefore P(\mu-n\sigma\leq X\leq \mu+n\sigma)\)\(=\dfrac {1}{ \sqrt {2\pi}}\int\limits_{-1}^{1} \;e^{-t^2/2}\;\;dt \)

\(=\dfrac {1}{ 2.51}×\int\limits_{-1}^{1} \;e^{-t^2/2}\;\;dt \)

\(=\dfrac {1}{ 2.51}×1.7112=.682\)

\(\therefore\) 68% is the probability that random variable will lie within one standard deviation about the mean.

 

For any normal distribution find the probability that random variable lies within one standard derivation about the mean

A

1.323

.

B

.976

C

.682

D

.012

Option C is Correct

Cumulative Distribution Function (CDF)

Let \(X\) define a continuous random variable then we define cumulative distribution function of this random variable as

\(P(X\leq x)=\displaystyle\int \limits^x_{-\infty}f(x)\,dx=F(x)\)

It is basically the probability that the random variable \(X\) will take value less than \(x\).

Illustration Questions

A random variable \(X\) has the probability density function \(F(x)= \begin{cases} \dfrac{x^2}{9}&if \,\,0\leq x\leq3\\ \,\,0 &otherwise \end{cases}\) Find the cumulative distribution function for the random variable \(X\).

A \(F(x)=0\;\forall x\)

B \(F(x)= \begin{cases} 0&if&x < 0\\ \dfrac{x^3}{27}&if&0\leq x<3\\ 1 &if&x \geq 3 \end{cases}\)

C \(F(x)=\dfrac{x^3}{27}\,\forall x\)

D \(F(x)=1\,\forall x\)

×

Cumulative distribution function for a random variable \(X=F(x)=\displaystyle\int \limits^x_{-\infty}f(x)\,dx\)

Where \(f\) is the probability density function of \(X\).

In this case

(1)  \(F(x)=0\;\;\text{if}\;\;x<0\) (because \(f(x)=0\;\;\text{if}\;\;x<0\))

(2)  \(F(x)=\displaystyle\int \limits^0_{-\infty}f(x)\,dx+\displaystyle\int \limits^x_{0}f(x)\,dx\;\;\text{if}\;\;0\leq x<3\)

\(=0+\displaystyle\int \limits^x_{0}\dfrac{x^2}{9}\,dx=\dfrac{x^3}{27}\Bigg]^x_0\)

\(=\dfrac{x^3}{27}\)

\(\therefore\;F(x)=\dfrac{x^2}{27}\;\;\text{if}\;\;0\leq x<3\)

(3)  \(F(x)=\displaystyle\int \limits^3_{0}f(x)\,dx+\displaystyle\int \limits^4_{3}f(x)\,dx\;\;\text{if}\;\;x\geq3\)

\(=\displaystyle\int \limits^3_{0}\dfrac{x^2}{9}\,dx+0=\dfrac{x^3}{27}\Bigg]^3_0\)

\(=1\)


\(\therefore \;F(x)= \begin{cases} 0&if&x < 0\\ \dfrac{x^3}{27}&if&0\leq x<3\\ 1 &if&x \geq 3 \end{cases}\)

A random variable \(X\) has the probability density function \(F(x)= \begin{cases} \dfrac{x^2}{9}&if \,\,0\leq x\leq3\\ \,\,0 &otherwise \end{cases}\) Find the cumulative distribution function for the random variable \(X\).

A

\(F(x)=0\;\forall x\)

.

B

\(F(x)= \begin{cases} 0&if&x < 0\\ \dfrac{x^3}{27}&if&0\leq x<3\\ 1 &if&x \geq 3 \end{cases}\)

C

\(F(x)=\dfrac{x^3}{27}\,\forall x\)

D

\(F(x)=1\,\forall x\)

Option B is Correct

Illustration Questions

A random variable \(X\) has the probability density function \(f(x)=\dfrac{1}{\pi(x^2+1)}\) for all \(x\,\varepsilon R\). Find the value of \(F(1)\) where \(F(x)\) is the cumulative distribution function.

A \(\dfrac{5}{6}\)

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

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

D \(\dfrac{5}{7}\)

×

Cumulative distribution function for a random variable \(X=F(x)=\displaystyle\int \limits^x_{-\infty}f(x)\,dx\)

Where \(f\) is the probability density function of \(X\).

In this case

\(F(x)=\displaystyle\int \limits^x_{-\infty}f(x)\,dx=\displaystyle\int \limits^x_{-\infty}\dfrac{1}{\pi(x^2+1)}\;dx\)

\(=\dfrac{1}{\pi}tan^{-1}x\Bigg]^x_{-\infty}\)

\(=\dfrac{1}{\pi}\left[tan^{-1}x-tan^{-1}(-\infty)\right]\)

\(=\dfrac{1}{\pi}\left(tan^{-1}x+\dfrac{\pi}{2}\right)\)

\(=\dfrac{1}{2}+\dfrac{tan^{-1}x}{\pi}\)

\(\therefore\;F(x)=\dfrac{1}{2}+\dfrac{tan^{-1}x}{\pi}\)

 


\(\therefore \;F(1)=\dfrac{1}{2}+\dfrac{tan^{-1}1}{\pi}\)

\(=\dfrac{1}{2}+\dfrac{\pi}{4\pi}\)

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

A random variable \(X\) has the probability density function \(f(x)=\dfrac{1}{\pi(x^2+1)}\) for all \(x\,\varepsilon R\). Find the value of \(F(1)\) where \(F(x)\) is the cumulative distribution function.

A

\(\dfrac{5}{6}\)

.

B

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

C

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

D

\(\dfrac{5}{7}\)

Option B is Correct

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