Tutorial on Discrete Probability Distributions.
Tutorial on Discrete Probability Distributions
Tutorial on discrete probability distributions with examples and detailed solutions.
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Bottom of Form | | Let X be a random variable that takes the numerical values X1, X2, ..., Xn with probablities p(X1), p(X2), ..., p(Xn) respectively. A discrete probability distribution consists of the values of the random variable X and their corresponding probabilities P(X).
The probabilities P(X) are such that ∑ P(X) = 1Example 1:Let the random variable X represents the number of boys in a family.
a) Construct the probability distribution for a family of two children.
b) Find the mean and standard deviation of X.Solution to Example 1: * a) We first construct a tree diagram to represent all possible distributions of boys and girls in the family. * Assuming that all the above possibilities are equally likely, the probabilities are:
P(X=2) = P(BB) = 1 / 4
P(X=1) = P(BG) + P(GB) = 1 / 4 + 1 / 4 = 1 / 2
P(X=0) = P(GG) = 1 / 4 * The discrete probability distribution of X is given by X | P(X) | 0 | 1 / 4 | 1 | 1 / 2 | 2 | 1 / 4 | * * Note that ∑ P(X) = 1 * b) The mean µ of the random variable X is defined by µ = ∑ X P(X)
= 0 * (1/4) + 1 * (1/2) + 2 * (1/4) = 1 * The standard deviation σ of the random variable X is defined by σ = Square Root [ ∑ (X- µ) 2 P(X) ]
= Square Root [ (0 - 1) 2 * (1/4) + (1 - 1) 2 * (1/2) + (2 - 1) 2 * (1/4) ]
= 1 / square root (2)Example 2:Two balanced dice are rolled. Let X be the sum of the two dice.
a) Obtain the probability distribution of X.
b) Find the mean and standard deviation of X.Solution to Example 2: * a) When the two balanced dice are rolled, there are 36 equally likely possible outcomes as shown below . * The possible values of X are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. * The possible outcomes are equally
Tutorial on discrete probability distributions with examples and detailed solutions.
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Top of Form | Web | www.analyzemath.com | |
Bottom of Form | | Let X be a random variable that takes the numerical values X1, X2, ..., Xn with probablities p(X1), p(X2), ..., p(Xn) respectively. A discrete probability distribution consists of the values of the random variable X and their corresponding probabilities P(X).
The probabilities P(X) are such that ∑ P(X) = 1Example 1:Let the random variable X represents the number of boys in a family.
a) Construct the probability distribution for a family of two children.
b) Find the mean and standard deviation of X.Solution to Example 1: * a) We first construct a tree diagram to represent all possible distributions of boys and girls in the family. * Assuming that all the above possibilities are equally likely, the probabilities are:
P(X=2) = P(BB) = 1 / 4
P(X=1) = P(BG) + P(GB) = 1 / 4 + 1 / 4 = 1 / 2
P(X=0) = P(GG) = 1 / 4 * The discrete probability distribution of X is given by X | P(X) | 0 | 1 / 4 | 1 | 1 / 2 | 2 | 1 / 4 | * * Note that ∑ P(X) = 1 * b) The mean µ of the random variable X is defined by µ = ∑ X P(X)
= 0 * (1/4) + 1 * (1/2) + 2 * (1/4) = 1 * The standard deviation σ of the random variable X is defined by σ = Square Root [ ∑ (X- µ) 2 P(X) ]
= Square Root [ (0 - 1) 2 * (1/4) + (1 - 1) 2 * (1/2) + (2 - 1) 2 * (1/4) ]
= 1 / square root (2)Example 2:Two balanced dice are rolled. Let X be the sum of the two dice.
a) Obtain the probability distribution of X.
b) Find the mean and standard deviation of X.Solution to Example 2: * a) When the two balanced dice are rolled, there are 36 equally likely possible outcomes as shown below . * The possible values of X are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. * The possible outcomes are equally