PROBABILITY DISTRIBUTIONS A listing of all the values the random variable can assume with their corresponding probabilities.

PROBABILITY DISTRIBUTIONS A random variable does not mean that the values can be anything ( a random number) Random variables have a well defined set of outcomes and well defined probabilities for the occurrence of each outcome. The random refers to the fact that the outcomes happen by chance -- that is, you don’t know which outcome will occur next.

PROBABILITY DISTRIBUTIONS Example : Rolling of a die X 1 2 3 4 5 6 sum p(x) 1/6 1/6 1/6 1/6 1/6 1/6 6/6=1

PROBABILITY DISTRIBUTIONS There are two types of random variables: that is , discrete and continuous. DRV can assume a countable number of values. Example: The number of sales made by a a salesperson in a given week. The number of customers waiting to be served in a restaurant at a particular time.

PROBABILITY DISTRIBUTIONS Two types of probability distributions Discrete Continuous

PROBABILITY DISTRIBUTIONS Discrete variable can take only a limited number of values, which can be listed. Continuous variable can take on any value within a given range, so we cannot list all the possible values.

PROBABILITY DISTRIBUTIONS The probability distribution of a discrete random variable is a graph, table or formula that specifies the probability associated with each possible value the random variable can assume.

PROBABILITY DISTRIBUTIONS Example : Tossing two coins and let x be the number of heads observed. Solution: X 0 1 2 p(x) 1/4 1/2 1/4 p(x) p(x) 1/2 1/2 1/4 1/4 x 0 1 2 x 0 1 2

PROBABILITY DISTRIBUTIONS The mean or expected value, of a discrete random variable x is = E(x) = All x The expected value is the mean of the probability distribution, a measure of its central tendency.

PROBABILITY DISTRIBUTIONS The variance of a discrete random variable x is All x

PROBABILITY DISTRIBUTIONS Expected values of discrete random variables Example: Examine the probability distribution for x ( the number of heads observed in the tossing of two fair coins) In a large number of experiments, 1/4 should result in x=0, 1/2 in x=1 and 1/4 in x=2 heads. Therefore, the average number of heads is =0(1/4) + 1(1/2) + 2(1/4)

Example 1: Consider a random variable X with the following probability distribution: p(x) = 0.05x, x = 2, 3, 4, 5, or 6 Express the probability distribution in tabular form. b) Find the following probabilities: P(X≥4) P(X > 4) P(3≤X≤5) P(2 < X < 4) P(X = 4.5)

Solution - Example 1 a) X 2 3 4 5 6 p(x) 0.10 0.15 0.20 0.25 0.30 b) 0.75 0.55 0.60 0.15 0.00

Example 2 Let X and Y be two independent random variables with the following probability distributions: X 0 1 2 p(x) 0.50 0.30 0.20 Y 0 1 2 p(Y) 0.40 0.50 0.10 Calculate E(X) and E(Y) Calculate V(X) and V(Y)

PROBABILITY DISTRIBUTIONS Binomial Probabilities A binomial experiment is an experiment which satisfies these four conditions: 1. A fixed number of trials 2. Each trial is independent of the others. 3. There are only two outcomes 4. The probability of each outcome remains constant from trial to trial.

PROBABILITY DISTRIBUTIONS Binomial Probabilities A binomial experiment has a fixed number of independent trials, each with only two outcomes. Examples: 1. Asking 300 people if they watch TV news. 2. Rolling a die to see if a 5 appears

PROBABILITY DISTRIBUTIONS Binomial Probabilities The Binomial Formula( Probability Distribution) Where: p = probability of a success q = 1-p = probability of failure r = number of success desired n = number of trials

PROBABILITY DISTRIBUTIONS Binomial Probabilities Mean: Standard Deviation:

Example 3: An insurance representative has appointments with 4 prospective clients tomorrow. From the past experience she knows that the probability of making a sale on any appointment is 1 in 5 or 0.20. What is the probability that she will sell a policy to 3 out of 4 prospective clients?

Outline of Answer P(sale)=0.2 P(No sale)=1-0.2=0.8 n=4 P(3 sales)= =0.0256

Example 4: A company is planning to sell a new product in four areas North, South, East and West. The probability that the product will be successful in an area is 0.3. Success in one area will be independent of success or failure in the other areas. What is the probability of success in no areas, one area, two areas, three areas, and four areas?

Outline of Answer P (success)=p=0.3; q=1-p=0.7; n=4 Use Binomial Tables. P(0 successes)=P(0)=0.2401 P(1)=0.4116 P(2)=0.2646 P(3)=0.0756 P(4)=0.0081

Example 5: Of the total output produced in a factory, 8% is defective Example 5: Of the total output produced in a factory, 8% is defective. In a batch of 500 units, what is the mean expected number of defective items, and what is the standard deviation?

Outline of Answer N=500; p=0.8 Mean = np = 500*0.8 = 40 units Standard Deviation =

PROBABILITY DISTRIBUTIONS Poisson Distribution A probability distribution used when a density of items is distributed over a period of time. The sample size needs to be large and the probability of success to be small.

PROBABILITY DISTRIBUTIONS Poisson Distribution Characteristics: 1. The experiment consists of counting the number of times a particular event occurs during a given unit of time or in a given area or volume. 2. The probability that an event occurs in a given unit of time, area or volume is the same for all the units.

PROBABILITY DISTRIBUTIONS 3. The number of events that occur in one unit of time, area or volume is independent of the number that occur in other units. 4. The mean (or expected) number of events in each unit will be denoted by

PROBABILITY DISTRIBUTIONS Poisson Formula(Probability Distribution) Where: = probability of exactly x occurrences = 2.71828, constant.

Example 6: Alden and Associates write weekend trip insurance at a very nominal charge. Records show that the probability that a motorist will have an accident during weekend and file a claim is 0.0005. Suppose Alden wrote 400 policies for the forthcoming weekend. What is the probability that exactly two claims will be filed?

Outline of Answer p=0.0005 n=400 Mean=λ=np=0.2 Using Poisson Tables,

Example 7: It is thought that, on an average, 0 Example 7: It is thought that, on an average, 0.01% of the workforce in a particular industry will suffer from an acute form of industrial disease. Hot Flush Inc., employs a workforce of 2,000 men, and the senior medical adviser of the company has stated that if three people in the workforce suffer from the disease, the cases should be treated as a sign that there are unacceptable health hazards. Calculate the probability that three or more men in the workforce will catch the disease.

Outline of Answer p=0.0001 n=2000 Mean=λ=np=0.2 Using Poisson Tables, P(3 or more)=1-P(2 or less) =1-[P(0)+P(1)+P(2)] =1-[0.8187+0.1637+0.0164]=0.0012

PROBABILITY DISTRIBUTIONS The Poisson distribution can be a reasonable approximation of the binomial under the following conditions: 1. n is large, and 2. p is small. The rule most often used is when n is greater than or equal to 20 and p is less or equal to 0.05.

Normal Distribution This is the most important continuous distribution. Many distributions can be approximated by a normal distribution. The normal distribution is the cornerstone distribution of statistical inference.

Normal Distribution A random variable X with mean m and variance s2 is normally distributed if its probability density function is given by

Finding Normal Probabilities Two facts help calculate normal probabilities: The normal distribution is symmetrical. Any normal distribution can be transformed into a specific normal distribution called… “STANDARD NORMAL DISTRIBUTION” Example The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?

Finding Normal Probabilities Solution If X denotes the assembly time of a computer, we seek the probability P(45<X<60). This probability can be calculated by creating a new normal variable the standard normal variable. Every normal variable with some m and s, can be transformed into this Z. Therefore, once probabilities for Z are calculated, probabilities of any normal variable can be found. E(Z) = m = 0 V(Z) = s2 = 1

Finding Normal Probabilities Example - continued 45 - 50 X - m 60 - 50 P(45<X<60) = P( < < ) 10 s 10 = P(-0.5 < Z < 1) To complete the calculation we need to compute the probability under the standard normal distribution

Finding Normal Probabilities Example - continued P(45<X<60) = P( < < ) 45 X 60 - m - 50 s 10 = P(-.5 < Z < 1) We need to find the shaded area z0 = 1 z0 = -.5

Finding Normal Probabilities Example - continued P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328 .1915 .1915 .1915 .1915 .3413 -.5 .5 1.0

Example 8 The amount of soda pop in each bottle is normally distributed with a mean of 32.2 ounces and a standard deviation of .3 ounces. Find the probability that a bottle bought by a customer will contain more than 32 ounces. Solution The random variable X is the amount of soda in a bottle. 0.7486 x = 32 m = 32.2

Finding Values of Z Sometimes we need to find the value of Z for a given probability We use the notation zA to express a Z value for which P(Z > zA) = A A zA

Finding Values of Z Example 9 Solution Determine z exceeded by 5% of the population Determine z such that 5% of the population is below Solution z.05 is defined as the z value for which the area on its right under the standard normal curve is .05. 0.45 0.05 0.05 -Z0.05 Z0.05 1.645

Characteristics of Normal Probability Distribution The curve is bell shaped, that is, it has the same shape on either side of the vertical line from mean. It has a single peak. As such it is unimodal. The mean is located at the centre of the distribution. The distribution is symmetrical. The two tails of the distribution extend indefinitely but never touch the horizontal axis (asymptotic).

The mean, median and mode have the same value, that is, mean = median = mode. The percentage distribution of area under standard normal curve is broadly as follows: ± 1s =68.27%; ± 2s =95.44% and ± 3s =99.73%. The units for the standard normal distribution curve are denoted by Z and are called the Z values or Z scores. They are also called standard units or standard scores. The Z score is known as a ‘standardised’ variable because it has a zero mean and a standard deviation of one.

Bell-shaped curve

Area under the normal curve, between  ± k  k Area in % 1 68.26 1.96 95.00 2 95.46 2.58 99.00 3 99.73

Normal approximation to the Binomial Normal approximation to the binomial works best when the number of experiments (sample size) is large, and the probability of success, p, is close to 0.5. For the approximation to provide good results two conditions should be met: np 5; n(1 - p) 5

Example 10 If X is a normal random variable with a mean of 100 and a standard deviation of 10, find the following probabilities: P(X ≥ 128) P(X ≤ 113) P(87 ≤ X ≤ 98)

If Z is a standard normal random variable, find the value z for which: Example 11 If Z is a standard normal random variable, find the value z for which: the area between 0 and z is 0.3729 the area to the right of z is 0.7123 the area to the left of z is 0.1736 the area to the left of z is 0.7673 the area to the right of z is 0.1841 the area between –z and z is 0.6630

Examples 12. Suppose the owner of a bakery knows that the daily demand for his whole meal bread is a random variable having the mean of 400 loaves and the standard deviation is 20. What is the probability that the demand for its bread will exceed 450 loaves? 13. The average monthly sales of 5000 firms are normally distributed with mean and standard deviation of Rs.36,000 and Rs.10,000 respectively. Find a) the number of firms having sales over Rs.40,000 b) the percentage of firms having sales between Rs.38,500 and Rs.41,000 c) the number of firms having sales between Rs.30,000 and Rs.40,000