Probability Distributions · 2016-06-24 · Quantitaive Aptitude & Business Statistics:...

Probability Distributions

Quantitative Aptitude & Business Statistics

Quantitaive Aptitude & Business Statistics: Probability Distributions

Meaning of probability distributions Theoretical or probability distributions

refer to mathematical models of expected frequencies of a finite number of observations of a variable with associated probabilities.

Theoretical distributions are based on mathematical functions where as observed frequency distributions are based on actual observed frequencies.

Usefulness of theoretical distributions

With known parameters like mean and standard deviation of population or number of trails, chances of success and so on, probabilities of various values of a variate can be found.

Empirical models can be tested for goodness of fit with theoretical distribution available .

But such methods of testing is recommended only when proper choice form various theoretical distributions is made.

Types of theoretical distributions Theoretical distributions are classified as

follows Theoretical distributions

Discrete Continuous

Binomial, Poisson, Multinomial Normal

Meaning of Binomial Distribution Binomial Distribution is associated

with French Mathematician James Bernoulli.

Applies in situations where there are a fixed number of repeated trails of any experiment under identical conditions for which only one of the two mutually outcomes, success or failure can result in each trail.

Examples of Binomial Distribution

A fixed number of observations (trials), n e.g., 15 tosses of a coin; 20

patients; 1000 people surveyed.

Continued A binary random variable e.g., head or tail in each toss of

a coin; defective or not defective light bulb.

Generally called “success” and “failure”.

Probability of success is p, probability of failure is 1 – p

Continued Constant probability for each

observation. e.g., Probability of getting a tail

is the same each time we toss the coin.

Definition: Binomial Distribution

Binomial: Suppose that n independent experiments, or trials, are performed, where n is a fixed number, and that each experiment results in a “success” with probability p and a “failure” with probability 1-p. The total number of successes, X, is a binomial random variable with parameters n and p.

We write: X ~ Bin (n, p) {reads: “X is distributed binomially with parameters n and p}

And the probability that X=r (i.e., that there are exactly r successes) is:

rpprXP −−

== )1()(

Xpp −−

q=1-p = probability of failure

P =probability of success

n = number of trials

X = # successes out of n trials

Characteristics of Binomial Distribution

Mean=np Variance=npq SD=√npq

P(X))= XnXn

X)p1(p −−

Properties of Binomial Distribution As a p increases for a fixed n,the

binomial distribution shifts to the right.

As n increases for a fixed p,the binomial distribution shifts to the right.

As n increases for a fixed p,the mean of binomial distribution increases.

If n is large and if neither p nor q is close to zero ,the binomial distribution can be closely approximately by a normal distribution with a standard variable given by

npqnpXZ −

Shape of Binomial Distribution

Value of p Shape of Binomial Distribution

If p= 0.5 If p>0.5 If p<0.5

Symmetrical Skewed to the right Skewed to the left

Problem Q.A Coin tossed four times what

is the probability of getting A) no head B) exactly one head C) exactly two heads D) at least two heads

Solution : Number of trails (n)=4 Probability of getting ahead (P)= Probability of getting a tail (q)= Using binomial distribution ,the probability of r

successes is P (r)=

P (r)=

rnrc )q()P(n

A) Probability of getting no head P (r=0) P (r)=

B) Probability of getting exactly one head P (r=1) P (r)=

C) Probability of getting exactly two heads P (r=2) P (r)=

D) Probability of getting at least two heads P (r=2 or more) P (r)=

)4r(P)3r(P)2r(P04

=+=+==−

Q. An experiment succeeds twice as many times as it fails. Find the chance that in 6 trails ,there will be atleast 5 success.

Solution No. of trails (n)=6 Probability of Success (P)= Probability of Failures (q)=1- = P (r=5 or 6)= P (r=5)+P(r=6)

729256

1729641

24332.6

××+×=

−−

Q. Calculate the mean of the binomial distribution in each of the following alternative case

A. No. of trails=6,Probability of success=1/3

B. No. of trails=9,Probability of failure=1/3

C. probability of failure=2/3,SD=2

Solution Case (a) No. of trails=6,Probability of

success=1/3

Mean =n p=6= Case B. No. of trails=9,Probability of

failure=1/3 Mean =np=

2316 =×

6329 =×

C. probability of failure=2/3,SD=2

31nnpq

==××=

Q. Calculate the SD of the binomial distribution in each of the following alternative case

A.No.of trails=6,Probability of success=1/3

B.No.of trails=9,Probability of failure=1/3

C. probability of failure=2/3,Mean=6

A. No.of trails=6,Probability of success=1/3

B. No.of trails=9,Probability of failure=1/3

316npqSD ==××==

414.1236

319npqSD ===××==

C. probability of failure=2/3,Mean=6

Mean=np=

3118npqSD

==××==

1. Binomial distribution is ___ (a) a continuous probability

distribution (b) a continuous observed

frequency distribution (c) a discrete observed frequency

distribution (d) a discrete probability

distribution

1. Binomial distribution is ___ (a) a continuous probability

distribution (b) a continuous observed

frequency distribution (c) a discrete observed frequency

distribution (d) a discrete probability

distribution

2. Variance of binomial distribution =

(a) np (b) nq (c) npq (d) 1/npq

2. Variance of binomial distribution = ?

(a) np (b) nq (c) npq (d) 1/npq

3. If mean and standard deviation of a binomial distribution is 10 and 3 respectively; q will be ___

(a) 0.3 (b) 0.33 (c) 30 (d) 0.9

3. If mean and standard deviation of a binomial distribution is 10 and 3 respectively; q will be ___

(a) 0.3 (b) 0.33 (c) 30 (d) 0.9

4. In binomial distribution (a) mean is greater than variance (b) mean is less than variance (c) mean is equal to variance (d) none of these

5. The probability mass function of binomial distribution is given by

(a) f(x) = nCx.px.qn-x

(b) f(x) = p x. q .n-x (c) f(x) = n px .p x. q n-x (d) f(x) = nCx .pn-x .qx

5. The probability mass function of binomial distribution is given by

(a) f(x) = nCx.px.qn-x

(b) f(x) = p x. q .n-x (c) f(x) = n px .p x. q n-x (d) f(x) = nCx .pn-x .qx

Introduction to the Poisson Distribution

Poisson Distribution was originated by French mathematician Simon Denis Poisson in 1837.

The Poisson Distributions limiting form of binomial distribution

It is a discrete probability distribution . It applies in situations here the probability of

success (p) is very small and that of failure (q) Poisson distribution is for counts—if events

happen at a constant rate over time, the Poisson distribution gives the probability of X number of events occurring in time T.

Poisson Distribution, example The Poisson distribution tells you

the probability of all possible numbers of new cases, from 0 to infinity.

If X= # of new cases next month and X ~ Poisson (λ), then the probability that X=k (a particular count) is:

k λλ −

Poisson Mean and Variance

Variance and Standard Deviation

λµ =

λσ =2

λσ =

where λ = expected number of hits in a given time period

For a Poisson random variable, the variance and mean are the same!

Problems Q.1 The number of customers at the

ticket counter of a theater at a rate of 120per hour .Find the probability during a given minute.

A) no customers appear B) only one customers appears C) atleast two customers appear

Solution Average no. of customers per minute

m=120/60=2

A) no customers appear

!r)1353.0(2

!rem)rX(p

===−

1353.01

)1353.0(1!0

e2)0r(pm0

====−

B) only one customers appears

C) at least two customers appear

2706.01

)1353.0(2!1

e2)1r(pm1

====−

[ ][ ]

3235.02706.01353.01

)1r(P)0r(P1

=+−=

=+=−

Q. The mean of the poission distribution 4. Find

A) Variance B)SD C) First four moments D)β1 and β2

The mean of the poission distribution 4.

A) Variance=m=4 B) SD= D) First moment = 0 Second moment =4 Third moment =4 Fourth moment =m+3m2=52

24m ===σ1µ

23 ===

25.31652

1. An ‘n’ increases the poisson distribution

(a) shifts to the right (b) shifts to the left (c) does not shifts at all (d) none of these

1. An ‘n’ increases the poisson distribution

(a) shifts to the right (b) shifts to the left (c) does not shifts at all (d) none of these

2. In distribution mean = variance

(a) normal (b) binomial (c) poisson (d) none of these

2. In distribution mean = variance

(a) normal (b) binomial (c) poisson (d) none of these

3. In a poisson distribution (a) e=2.7138 (b) e=2.1738 (c) e=2.1783 (d) e=2.7183

4. In a poisson distribution (a) p=q (b) p>q (c) p<q (d) none of these

5.Which one is not a condition of poisson model

(a) the probability of having success in a small time interval is constant

(b) the probability of having success more than one in a small time interval is very small

(c) the probability of having success in a small interval in independent of time and also of earlier success

(d) the probability of having success in a small time interval (t, t+td) is Kt for a positive constant k.

5.Which one is not a condition of poisson model

(a) the probability of having success in a small time interval is constant

(b) the probability of having success more than one in a small time interval is very small

(c) the probability of having success in a small interval in independent of time and also of earlier success

(d) the probability of having success in a small time interval (t, t+td) is Kt for a positive constant k.

Normal distribution The normal distribution is a descriptive

model that describes real world situations.

It is defined as a continuous frequency distribution of infinite range (can take any values not just integers as in the case of binomial and Poisson distribution).

This is the most important probability distribution in statistics and important tool in analysis of epidemiological data and management science.

Characteristics of Normal Distribution

It links frequency distribution to probability distribution.

Has a Bell Shape Curve and is Symmetric.

It is Symmetric around the mean: Two halves of the curve are the

same (mirror images)

Characteristics of Normal Distribution

Hence Mean = Median The total area under the curve is 1 (or

100%) Normal Distribution has the same

shape as Standard Normal Distribution.

Q1 and Q3 are equidistant from the median and hence

Q1=Mean-0.6745SD Q3=Mean+0.6745SD and Quartile Deviation is 2/3rd or

more precisely0.6745of the SD The mean deviation is 4/5 th or more

precisely 0.7979of the SD

Distinguishing Features The mean ± 1 standard deviation

covers 68.27% of the area under the curve

The mean ± 2 standard deviation covers 95.45% of the area under the curve

The mean ± 3 standard deviation covers 99.73% of the area under the curve

Importance of Normal Distribution

It has a property of central limit theorem.

As the sample size (n) becomes large, the normal distribution serves as good approximation of many discrete distributions such as binomial and poisson.

It has many mathematical properties which makes easy to manipulate for use in social and natural science

It is useful in SQC and industrial experiments.

It is useful in finding estimates and parameters and confidence intervals and tests of significance

Characteristics of Normal Distribution Contd.,

In a Standard Normal Distribution:

The mean (μ ) = 0 and Standard deviation (σ) =1

Area under normal curve to determine the probability

Area to the right of a positive value of ‘Z’ (i.e to find above value of X)

0.5-tablevalue

Area to the left of a positive value of ‘Z’ (i.e to find below value of X)

0.5+tablevalue

Area to the right of a negative value of ‘Z’ (i.e to find above value of X)

0.5+tablevalue

Area to the left of a positive value of ‘Z’ (i.e to find below value of X)

0.5-tablevalue

Area between two positive values of ‘Z’ (i.e to find between two values of X)

Difference of their tabular values

Area between two negative values of ‘Z’ (i.e to find between two values of X)

Difference of their tabular values

Area to the right of a positive value of ‘Z’ (i.e to find between two values of X)

Sum of their tabular values

Z indicates how many standard deviations

away from the mean the point x lies. Z score is calculated to 2 decimal places.

σµ−

Problem Q.A normal distribution has

standard deviation of 200 and mean of 800.Find the area of standard normal variate in each of the following alternative cases

Case (a) for X=600 Case (b) for X below 600

Case a) Standard normal variate corresponds to 600

Area between Z=0 and Z=-1 is 0.3413

800600XXZ −=−

=σ−

Case b) Standard normal variate corresponds to 600

Area between Z=0 and Z=-1 is 0.3413

Thus, P(Z<-1)=0.50-0.3413=0.1587

800600XXZ −=−

=σ−

1. In a normal distribution is (a) mean=median=mode (b) mean>median>mode (c) mean<median<mode (d) mode=3median–2mean

2. The normal distribution is known as standard normal distribution with

(a) Mean=1 and SD=0 (b) Mean=1 and SD=1 (c) Mean = 0 and SD=1 (d) Mean=0 and SD=1

2. The normal distribution is known as standard normal distribution with

(a) Mean =1 and SD=0 (b) Mean=1 and SD=1 (c) Mean= 0 and SD=1 (d) Mean=0 and SD=-1

3. In a normal distribution skewness is ___

(a) 0 (b) >3 (c) <3 (d) <1

3. In a normal distribution skewness is ___

(a) 0 (b) >3 (c) <3 (d) <1

4. The normal curve is (a) bell-shaped (b) U shaped (c) J shaped (d) inverted J shaped

5. A normal distribution is ___

(a) unimodal (b) bimodal (c) multimodal (d) none of these

5. A normal distribution is ___

(a) unimodal (b) bimodal (c) multimodal (d) none of these

6. The probability function of normal distribution is

(a) p (x)=

(b) p (x) =

(c) p (x) = (d) none of these

σµ−

πσ2x

σ−µ

σµ−

6. The probability function of normal distribution is

(a) p (x)=

(b) p (x) =

(c) p (x) =

(d) none of these

σµ−

πσ2x

σ−µ

πσ2x

σµ−

7. The mean deviation about median of a standard normal variate is

(a) 0.675 (b) 0.675 (c) 0.80 (d) 0.80 σ

7. The mean deviation about median of a standard normal variate is

(a) 0.675 (b) 0.675 (c) 0.80 (d) 0.80 σ

8. The quartile deviation of a normal distribution with mean 10 and SD 4 is

(a) 0.675 (b) 67.50 (c) 2.70 (d) 3.20

8. The quartile deviation of a normal distribution with mean 10 and SD 4 is

(a) 0.675 (b) 67.50 (c) 2.70 (d) 3.20

9. The symbol (a) indicates the area of the standard normal curve between

(a) 0 to a (b) a to œ (c) – œ to a (d) – œ to œ

9. The symbol (a) indicates the area of the standard normal curve between

(a) 0 to a (b) a to œ (c) –œ to a (d) –œ to œ

10.The interval (m–3s ; m+3 s) curves

(a) 95% area of a normal distribution (b) 96% area of a normal distribution (c) 99% area of a normal distribution (d) all but 0.27% area of a normal

distribution

10.The interval (m–3s ; m+3 s) curves

(a) 95% area of a normal distribution (b) 96% area of a normal distribution (c) 99% area of a normal distribution (d) all but 0.27% area of a normal

distribution

11.In normal distribution the probability has a maximum value at the

(a) mode (b) mean (c) median (d) none of these

11.In normal distribution the probability has a maximum value at the

(a) mode (b) mean (c) median (d) none of these

12. In a normal distribution ___ (a) standard deviation = mean deviation (b) standard deviation = mean deviation (c) standard deviation = quartile

deviation (d) none of these

12. In a normal distribution ___ (a) standard deviation = mean deviation (b) standard deviation = mean deviation (c) standard deviation = quartile deviation (d) none of these

13. In a normal distribution, median is equal to ___

(a) (b)

2QQ 31 +

2QQ 13 −

2meanQ1 +

2meanQ3 −

13. In a normal distribution, median is equal to ___

(a) (b) (c) (d)

2QQ 31 +

2QQ 13 −

2meanQ1 +

2meanQ3 −

14. In a normal distribution is ___

(a) mean + 1’s covers 68.72% of the items

(b) mean + 2’s covers 95.54% of the items

(c) mean + 3’s covers 99.73% of the items

(d) all of above

14. In a normal distribution is ___

(a) mean + 1’s covers 68.72% of the items

(b) mean + 2’s covers 95.54% of the items

(c) mean + 3’s covers 99.73% of the items

(d) all of above

15.The total area of the normal curve is

(a) one (b) 50% (c) 0.50 (d) any value between 0 and 1

15.The total area of the normal curve is

(a) one (b) 50% (c) 0.50 (d) any value between 0 and 1

THE END

Probability Distributions

Probability Distributions · 2016-06-24 · Quantitaive Aptitude & Business Statistics:...

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