1743 Chapter 4 Probability Distribution PDF

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BAMS1743 QUANTITATIVE METHODSCHAPTER 5: PROBABILITY DISTRIBUTIONPROBABILITY DISTRIBUTION A frequency distribution gives an analysis of the number of times each particular value occurs in a set of items. A probability distribution simply replaces the actual numbers (frequencies) with proportions of ...

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BAMS1743 QUANTITATIVE METHODS CHAPTER 5: PROBABILITY DISTRIBUTION PROBABILITY DISTRIBUTION  A frequency distribution gives an analysis of the number of times each particular value occurs in a set of items. A probability distribution simply replaces the actual numbers (frequencies) with proportions of the total.  For example, in a statistics test, the marks out of ten awarded to 50 students might be as follows: Marks out of 10 0 1 2 3 4 5 6 7 8 9 10 Total

Number of students (Freq. distribution) 0 0 1 2 4 10 15 10 6 2 0 50

Proportion or prob. (prob. distribution) 0.00 0.00 0.02 0.04 0.08 0.20 0.30 0.20 0.12 0.04 0.00 1.00

 A probability distribution of a discrete random variable is called a discrete probability distribution. The discrete probability distributions:  the Binomial distribution;  the Poisson distribution.  A probability distribution of a continuous random variable is called a continuous probability distribution. A very important continuous probability distribution is the normal distribution.

BINOMIAL DISTRIBUTION A Binomial experiment must satisfy the following 4 conditions: 1. There are n identical trials  an experiment is repeated n times under identical conditions 2. Each trial has only two possible outcomes  success or failure  Success or failure can be interpreted in any way that is convenient. For example (i) situation good or bad (ii) item defective or non-defective (iii) company gain profit or loss 3. The probabilities of the two outcomes remain constant for each independent trial.  The probability of success is denoted by p  The probability of failure is denoted by q  p+q=1 4. The trials are independent  Outcome of one trial does not affect the outcome of another trial Let random variable X ≡ number of successes in n independent trials  X is having Binomial distribution with parameters n and p  denoted by X ~ B(n, p) The probability of obtaining exactly x successes in n trials is given by

P( X  x )  nCx px (1 p)n x ; x  0, 1, , n

Where

n ≡ total number of trials p ≡ probability of success x ≡ number of successes in n trials

 For a Binomial distribution,  Mean,  = np  Variance, 2 = np(1 – p) = npq  Standard deviation,  =

npq

Example: Suppose we are interested in the number of heads showing face up on the three tosses of a coin. The possible results are (0 head), (1 head), (2 heads) and (3 heads). (a)

Find the probability for each number of heads.

(b)

If X denotes the number of heads, find the mean and standard deviation of X.

Solution: Let X ≡ number of heads showing face up on the three tosses of a coin. X ~ B (n = 3, p = 0.5) (a) P( X  x)  nC p x(1 p) nx; x

x  0, 1, , n

 3Cx(0.5)x (0.5)3x;

X=x 0 1 2 3

x  0, 1, 2, 3

P( X = x )

 3C (0.5)0 (0.5)3  0.125 0  3C (0.5)1(0.5)2  0.375 1  3C (0.5)2 (0.5)1  0.375 2 3  C (0.5)3 (0.5)0  0.125 3 1.000

(b)

Mean,  = np = 3 × 0.5 = 1.5 Standard deviation,

  npq 

3  0.5  0.5  0.75  0.8660

Example: The probability that a college student will graduate is 0.4. Determine the probability that out of 5 students a)

none,

b)

1,

c)

at least 1,

d)

all will graduate

Solution: Let X ≡ number of students that will graduate out of 5 students X ~ B( n = 5, p = 0.4 )

P ( X  x )  5C x 0.4x (0.6)5 x ; x  0,1, ,5

a)

P( X  0)  5C 0.40 (0.6)5  0.07776  0.0778 0

b)

P( X  1)  5C 0.41(0.6) 4  0.2592 1

c)

P( X  1)  P( X  1)  P( X  2)  P( X  3)  P ( X  4)  P ( X  5) Or

P( X  1)  1 P( X  1)  1  P( X  0)  1  0.0778  0.9222 d)

P( X  5) 

5

C5 0.45 (0.6)0  0.01024  0.0102

Example: A new-car salesperson knows from past experience that she will make a sale to about 20% of her customers. Find the probability that in five randomly selected attempts, she makes sale to (a) exactly three customers; (b) at most one customer; (c) at least one customer; (d) fewer than three customers; (e) more than three customers. If X denotes the number of sales made out of the five attempts, find the mean and standard deviation of X. (Answer: (a) 0.0512, (b) 0.7373, (c) 0.6723 (d) 0.9421, (e) 0.0067, 1.0, 0.8944) Solution: Let X ≡ number of sales made out of 5 attempts X ~ B( n = 5, p = 0.2 )

P( X  x )  5Cx (0.2)x (0.8)5x;

x  0, 1, 2, 3, 4, 5

Remarks: Students may encounter inequalities as listed: Inequality sign At least x≥ At most x≤ Less than / fewer than x< Exceed / more than x> Not more than x≤ Not less than x≥ Notes:  The mean of binomial distribution is also equal to its expected value.  Standard deviation of a probability distribution is a measure of dispersion. When  is small, the probability that we will get a value close to the mean is high and when  is large, we are more likely to get a value far away from the mean. POISSON DISTRIBUTION

A discrete random variable having probability function of the form

e   x P( X  x)  ; x  0,1,2,  x! where  can take any positive value, is said to follow the Poisson distribution. If an event is randomly scatted in time (or space) and has mean number of occurrence  in a given interval of time (or space) and X is the random variable “the number of occurrences in the given interval”.  X is a Poisson distribution with parameter   denoted by X ~ Po () and x = 0, 1, 2, ... Note: If  is not given, then  = np For a Poisson distribution with parameter ,  mean =  and variance =  Example: The mean number of bacteria per ml of a liquid is known to be 4. Assuming that the number of bacteria follows a Poisson distribution, find the probability that, in 1 ml of liquid, there will be a) no b) 4 c) less than 3 bacteria Solution: Let X ≡ the number of bacteria in 1 ml of liquid. e 4 4 x ; x  0,1,2,   X ~ Po (4) where P( X  x)  x! a) b)

c)

Example: Based on information obtained in previous example, find the probability that a) in 3 ml of liquid, there will be less than 2 bacteria. b) in 0.5 ml of liquid, there will be more than 2 bacteria. Solution: a)

In 1 ml of liquid we expect 4 bacteria  In 3 ml of liquid we expect 4 × 3 = 12 bacteria Let Y ≡ the number of bacteria in 3 ml of liquid. e1212 y ; y  0,1,2,   Y ~ Po (12) where P (Y  y )  y!

P(Y  2)  P(Y  0)  P(Y  1) e 12120 e 12121   0! 1! 5  7.987 10 b)

Poisson approximation to the Binomial distribution  A Binomial distribution with parameters n and p can be approximated by a Poisson distribution with parameter  = np if n is large and p is small (such that np < 5).  The approximation gets better as n   and p  0  X ~ B(n, p)  X ~ Po (  np) when n   and p  0

Example: Eggs are packed in boxes of 500. On average, 0.8% of the eggs are found to be broken when the eggs are unpacked. Find the probability that in a box of 500 eggs (i) exactly 3 (ii) less than 2 (iii) more than 2 will be broken. Solution:

NORMAL DISTRIBUTION The normal distribution is an important probability distribution that is often applied to continuous variables. The Characteristics of a Normal Distribution  The normal curve representing the normal probability distribution is often described as being bell-shaped and is symmetrical about its central mean.

f ( x)

X Mean  The two tails of the normal curve extend indefinitely in both directions, i.e. they never actually touch the X-axis.  Total area under the normal curve equals to total probability, i.e. one.  The normal curve has a single peak at the exact center of the distribution.  The mean, median and mode of the distribution are equal and located at the peak.  A normal distribution is completely specified by two parameters the mean () and the standard deviation () ( x ) 2 2  2  1  f ( x)  e  2 2 where    x   , e = 2.7183 and  = 3.142  If the random variable X (for e.g. height, weight etc.) has a normal distribution with mean () and standard deviation () then X is denoted by X ~ N (, 2).  Every different combination of  and  would generate a different normal probability distribution. The shape of the normal curve depends on the value of standard deviation; a larger standard deviation results a flatter, wider spread distribution. The value of the mean determines the position of the normal curve.  The relative dispersion of frequencies around the mean can be measured exactly in terms of standard deviations. Examples: (a) About 68.26% of frequencies have a value within   . (b) About 95.45% of frequencies have a value within   2. (c) About 99.73% of frequencies have a value within   3. STANDARD NORMAL DISTRIBUTION  Normal distribution with mean,  = 0 and standard deviation,  = 1 is known as standard normal distribution.  Any normally distributed set of values x can be converted into the standard normal values z by using the formula x μ z σ which indicates the number of standard deviations away from the mean.

X  If the random variable X ~ N ( , 2 ) then Z  μ and Z ~ N (0, 12 ) is σ known as a standard normal variable or standardized normal variable.  By determining the Z value using the above formula, we can find the area or the probability under the normal curve by referring to the standard normal distribution tables.

Note: (a) Area under the whole curve = 100% = 1 z

0

(b) P( Z > 0 ) = P( Z < 0 ) = 50% = 0.5

z 0

(c) P( Z  z ) = P( Z > z )

z

0

(d) P( Z < –z ) = P( Z > z ) z 0

(e) P( Z < z ) = 1 – P( Z > z ) (f) P( Z > –z ) = 1 – P( Z > z )

z

0

z

0

E.g. Take z = 1, P ( Z > 1 ) = 0.1587

0

z

E.g. Take z = 2.77, P ( Z > 2.77 ) = 0.00280

z

0

Example: The lifetime of an electrical component is known to follow a normal distribution with a mean 2000 hours and a standard deviation 200 hours. If a component is randomly selected find the probability that the component will last (a) (b) (c) (d) (e) (f ) (g) (h)

more than 2200 hours; less than 1800 hours; at least 1900 hours; not more than 2100 hours; between 1800 and 2200 hours; between 2100 and 2300 hours; between 1700 and 1900 hours; between 2000 and 2400 hours.

Solution: Let X ≡ lifetime of electrical component (hours) X ~ N ( = 2000,  2 = 200 2 ) (a)

(b)

(c)

(d)

P ( X > 2200 ) = P ( Z > 2200  2000 ) 200 =P(Z>1) = 0.1587 P ( X < 1800 ) = P( Z < 1800  2000 ) 200 = P( Z < – 1 ) = P( Z > 1 ) = 0.1587 P ( X ≥ 1900 ) = P ( Z ≥ 1900  2000) 200 = P ( Z ≥ – 0.5 ) = P ( Z < 0.5 ) = 1 – P ( Z > 0.5 ) = 1 – 0.3085 = 0.6915

0

0

0

z

z

z

P ( X ≤ 2100 ) = P( Z ≤ 2100  2000 ) 200 z 0

= P ( Z ≤ 0.5 ) = 1 – P ( Z > 0.5 ) = 1 – 0.3085 = 0.6915 (e)

(f)

(g)

(h)

P ( 1800 < X < 2200 ) = P( 1800  2000 < Z < 2200  2000 ) 200 200 =P(–11) = 1 – 2(0.1587) = 0.6826 P ( 2100 < X < 2300 ) = P ( 2100  2000 < Z < 2300  2000 ) 200 200 = P ( 0.5 < Z < 1.5 ) = P ( Z > 0.5 ) – P ( Z > 1.5 ) = 0.3085 – 0.0668 = 0.2417 P ( 1700 < X < 1900 ) = P ( 1700  2000 < Z < 1900  2000 ) 200 200 = P (– 1.5 < Z < – 0.5 ) = P ( 0.5 < Z < 1.5 ) = 0.2417 P ( 2000 < X < 2400 ) = P ( 2000  2000 < Z < 2400  2000 ) 200 200 =P(0 2 ) = 0.5 – 0.02275 = 0.47725

0

0

0

0

z

z

z

z

Example: Refer to the previous Example: (a) Determine the lifetime of the component such that only 10% of the components will fail before that time. (b) If the mean life time of the component is to be increased such that 20% of the components will last for more than 2200 hours, find the new mean life. Solution: (a)

(b)

Let k ≡ the required lifetime (hours) P( X < k ) = 0.1 P( Z < k  2000 ) = 0.1 200 From table, k  2000  1.2816 200 k   1.2816  200  2000  1743.68 hours  1744 hours

Let 1 ≡ the mean life required P( X > 2200 ) = 0.2 2200   1 ) = 0.2 P( Z > 200 2200   1  0.8416 From table, 200

z 0

0

1  0.8416  200  2200 1  2031.68 hours  2032 hours

Example: A local bank has 4 teller windows. By monitoring the customer waiting times, the mean is found to be 3.8 min, the standard deviation is 1.5 min, and the distribution is normal. Past experience has shown that customers become irritated when the waiting time exceeds 5 min. (a) What proportion of customers will be irritated by the present system? (b) By adding a teller to the system, the percentage of customers irritated would be reduced to only 10%. What is the mean waiting time with an extra teller? Assuming that the standard deviation remains unchanged.

z

Solution:

Example: The grade-point averages of students in a large college are normally distributed, with a mean 2.4 and a standard deviation 0.5. (a) What proportion of the students possesses grade-point average in excess of 3.0? (b) If students who possess grade-point average below 1.9 are excluded from the college, what percentage of the students will be excluded? (c) Suppose that three students are selected at random from the student population. What is the probability that all three possess grade-point average in excess of 3.0? Solution:

Normal approximation to the Binomial distribution  In a Binomial situation when np > 5 and nq > 5, the Normal distribution with mean=  =np and standard deviation=  = np(1  p) can be used to approximate the Binomial distribution. 2  X ~ B (n , p )  X ~ N ( ,  ) when np > 5 and nq > 5 Continuity correction factor  The addition and/or subtraction of 0.5 from the value(s) of x when the Normal distribution is used as an approximation to the Binomial distribution, where x is the number of successes in n trials Remarks: Binomial P( X = 2 ) P( 3  X  5 ) P( 3  X  5 ) P( 3  X  5 ) P( 3  X  5 ) P( X < 4 ) P( X  4 ) P( X  4 ) P( X > 4 ) P( X  0 ) P( X > 0 ) P( X = 0 )

Normal P( 1.5 < X < 2.5 ) P( 2.5 < X < 5.5 ) P( 3.5 < X < 5.5 ) P( 2.5 < X < 4.5 ) P( 3.5 < X < 4.5 ) P( X < 3.5 ) P( X < 4.5 ) P( X > 3.5 ) P( X > 4.5 ) P( X > – 0.5 ) P( X > 0.5 ) P( – 0.5 < X < 0.5 )

Example: Find the probability that 200 tosses of a bias coin will result in less than 51 heads in which the probability of getting a head is 0.2 for each toss. Solution:

Example: Find the probability of getting between 3 and 6 heads inclusive in 10 tosses of a fair coin by using (a) the Binomial distribution, (b) the Normal approximation to the Binomial distribution. Solution:

Example: Find the probability that 200 tosses of a fair coin will result in (a) between 80 and 120 heads inclusive, (b) less than 90 heads, (c) less than 85 or more than 115 heads. Solution:

16

The Normal Approximation to the Poisson distribution    

2 If X ~ Po ( ) then mean    and variance    . For large , X ~ N ( ,  ) approximately. Generally, we require  > 20 for a good approximation. X ~ Po ( )  X ~ N ( , ) when  is large.

Example: The number of calls received by an office switchboard per hour follows a Poisson distribution with parameter 30. Using the normal approximation to the Poisson distribution, find the probability that in one hour, there are (a) more than 23 calls, (b) between 25 and 28 calls inclusive, (c) 34 calls. Solution:

17

Taken from Statistical Tables by J Murdoch and JA Barnes 18

Taken from Statistical Tables by J Murdoch and JA Barnes

19

BAMS1743 QUANTITATIVE METHODS Tutorial 5 (Probability Distribution) 1.

85% of households in an area have a notebook. If six households are randomly selected from the area, what is the probability that the number having a notebook is (a) (b) (c) (d) (e) (f)

2.

On a rainy day, 10% of the production employees at Super Steel are absent from work. Ten production employees are to be selected at random for a study on absenteeism. (a) (b) (c) (d)

3

What is the random variable in this problem? Is the random variable discrete or continuous? Why? What is the probability that none of the 10 production employees selected at random on a rainy day is absent? Compute the mean, variance and standard deviation of the distribution.

A factory finds that, on the average, 20% of the bolts produced by a given machine will be defective for certain specified requirements. If 10 bolts are selected at random from the day’s production of this machine, find the probability that (a) (b) (c)

4.

exactly four, at least four, at most five, between two and five inclusive, exceed four, fewer than four.

exactly 2 will be defective; 2 or more will be defective; more than 5 but at most 8 will be defective.

Customers arrive randomly at a department store at an average rate of 3.4 per minute. Assuming the customers arrivals form a Poisson distribution, calculate the probability that (a) (b) (c) (d)

no customers arrive in any particular minute, exactly one customer arrives in any particular minute, two or more customers arrive in any particular minute, One or more customers arrive in any 30-second period. 20

5.

An insurance company receives on average 2 claims per week from a certain factory. Assuming that the number of claims follows a Poisson distribution, find the probability that (a) (b) (c)

6.

Cars arrive at a petrol station at an average rate of 30 per hour. Assuming that the number of cars arriving at the petrol station follows a Poisson distribution, find the probability that (a) (b) (c) (d) (e)

7.

it receives more than 3 claims in a given week, it receives more than 2 claims in a given fortnight, It receives no claims on a given day, assuming that the factory operates on a 5-day week.

No cars arrive during a particular 5 minutes interval, More than 3 cars arrive during a 5 minutes interval, More than 2 cars arrive during a 15 minutes interval, In a period of half an hour, 10 cars arrive, Less than 3 cars arrive during a 10 minutes interval.

The probability that a particular make of light bulb is faulty is 0.01. The light bulbs are packed in boxes of 100. Find the probability that in a certain box there are (a) (b) (c)

no faulty light bulbs, 2 faulty light bulbs, More than 3 faulty light bulbs.

8.

A sample of 200 items is taken from a large batch, which is known to produce defective with probability 0.005. Find the probabilities that there are 0, 1, 2, 3, 4, and 5 defectives in the sample using a Poisson approximation to the binomial.

9.

Orders received by a particular firm have amounts that follow a normal distribution with a mean of RM103.60 and a standard deviation of RM8.75. (a) (b) (c) (d)

What percentage of orders will have amount below RM120.05? What percentage of orders will have amount over RM92.75? What will be the amount such that approximately 25% of orders are for greater amounts? Above what amount will 90% of orders lie? 21

10. A manufacturer makes chocolate bars with a mean weight of 110 grams and a standard deviation of 2 grams. The weights are normally distributed. (a) (b)

What proportion of the bars is likely to be less in weig...