binomial, Poison and normal probability distributions

RIGOBERT NGELEJA& FRANK VENANCE

The Binomial Probability Distribution

A binomial experiment is one that possesses the following properties:

1.The experiment consists of n repeated trials;

2.Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);

3.The probability of a success, denoted by p, remains constant from trial to trial and repeated trials are independent.

The number of successes X in n trials of a binomial experiment is called a binomial random variable.

The probability distribution of the random variable X is called a binomial distribution, and is given by the formula

P(X) = Cnxp

xqn−x

combinations) in which r objects can be selected from a set of n objects,

wheren = the number of trialsx = 0, 1, 2, ... np = the probability of success in a single trialq = the probability of failure in a single trial(i.e. q = 1 − p)Cn

x is a combination

P(X) gives the probability of successes in n binomial trials.

Mean and Variance of Binomial Distribution

If p is the probability of success and q is the probability of failure in a binomial trial, then the expected number of successes in n trials (i.e. the mean value of the binomial distribution) is

E(X) = μ = np

The variance of the binomial distribution isV(X) = σ2 = npq

Note: In a binomial distribution, only 2 parameters, namely n and p, are needed to determine the probability.

A die is tossed 3 times. What is the probability of(a) No fives turning up? (b) 1 five? (c) 3 fives?

This is a binomial distribution because there are only 2 possible outcomes (we get a 5 or we don't).Now, n = 3 for each part. Let X = number of fives appearing.

(a) Here, x = 0.

EXAMPLE 1

(b) Here, x = 1.

(c) Here, x = 3.

EXAMPLE 2Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?

This is a binomial distribution because there are only 2 outcomes (the patient dies, or does not).Let X = number who recover.Here, n = 6 and x = 4. Let p = 0.25 (success - i.e. they live), q = 0.75 (failure, i.e. they die).The probability that 4 will recover:

Normal Probability Distributions

Properties of a Normal Distribution1.The normal curve is symmetrical about the mean μ;2.The mean is at the middle and divides the area into halves;3.The total area under the curve is equal to 1;It is completely determined by its mean and standard deviation σ (or variance σ2)

Note: In a normal distribution, only 2 parameters are needed, namely μ and σ2.

The Standard Normal DistributionIt makes life a lot easier for us if we standardize our normal curve, with a mean of zero and a standard deviation of 1 unit. If we have the standardized situation of μ = 0 and σ = 1, then we have:

Standard Normal Curve μ = 0, σ = 1

If we have mean μ and standard deviation σ, then Since all the values of X falling between x1 and x2 have

corresponding Z values between z1 and z2, it means:

The area under the X curve between X = x1 and X = x2 equals:

The area under the Z curve between Z = z1 and Z = z2.

Hence, we have the following equivalent probabilities: P(x1 < X < x2) = P(z1 < Z < z2)

8.13

Calculating Normal Probabilities…

P(45 < X < 60) ?

0

…mean of 50 minutes and astandard deviation of 10 minutes…

Percentages of the Area Under the Standard Normal CurveA graph of this standardized (mean 0 and variance 1) normal curve is shown.

In this graph, we have indicated the areas between the regions as follows:

-1 ≤ Z ≤ 1= 68.27%-2 ≤ Z ≤ 2= 95.45%-3 ≤ Z ≤ 3 =99.73%

EXAMPLE 1Find the area under the standard normal curve for the following, using the z-table or calculator.

(a) between z = 0 and z = 0.78(b) between z = -0.56 and z = 0(c) between z = -0.43 and z = 0.78(d) between z = 0.44 and z = 1.50(e) to the right of z = -1.33.

The Poisson Probability Distribution

The Poisson Distribution was developed by the French mathematician Simeon Denis Poisson in 1837.

The Poisson random variable satisfies the following conditions:1.The number of successes in two disjoint time intervals is independent.2.The probability of a success during a small time interval is proportional to the entire length of the time interval.

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

wherex = 0, 1, 2, 3...e = 2.71828 (but use your calculator's e button)μ = mean number of successes in the given time interval or region of space

Mean and Variance of Poisson DistributionIf μ is the average number of successes occurring in a given time interval or region in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to μ.E(X) = μand V(X) = σ2 = μ

EXAMPLE 1Vehicles pass through a junction on a busy road at an average rate of 300 per hour.a.Find the probability that none passes in a given minute.b.What is the expected number passing in two minutes?c.Find the probability that this expected number actually pass through in a given two-minute period.

The average number of cars per minute is:

(a)

(b) Expected number each 2 minutes = E(X) = 5 × 2 = 10

(c) Now, with μ = 10, we have:

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