Let X be a random variable with binomial probability distribution with number of trials n and probability of success p. Prove the following: 1. The expectation of X is, E(X)= n.p 2. The variance of X is, o² = Var(X) = n.p.q %3D 3. The standard deviation of X is, o = /Var(X)= /n.p.q

Let X be a random variable with binomial probability distribution with number of trials n and probability of success p. Prove the following: 1. The expectation of X is, E(X)= n.p 2. The variance of X is, o² = Var(X) = n.p.q %3D 3. The standard deviation of X is, o = /Var(X)= /n.p.q

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Concept explainers

Continuous Probability Distributions

Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.

Normal Distribution

Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!

Question

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Let X be a random variable with binomial probability distribution with
number of trials n and probability of success p. Prove the following:
1. The expectation of X is, E(X) = n.p
2. The variance of X is, o2 = Var(X) = n.p.q
3. The standard deviation of X is, o =
VVar(X)= n.p.q
where
is the probability of failure such that q = 1 – p.

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