Answered: Calculate the expected value of X,…

Calculate the expected value of X, E(X), for the given probability distribution. x 2 4 6 8 P(X = x) 7 20 11 20 1 20 1 20 E(X) =

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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Let X be a random variable with the following probability distribution. Value X of X P(X=x) 30 Continue 40 50 60 70 80 0.15 0.25 0.20 0.10 0.15 0.15 Complete the following. (If necessary, consult a list of formulas.) (a) Find the expectation E (X) of X. E (X) = (b) Find the variance Var(X) of X. Var(x) = 0 O S
Let X be a random variable with the following probability distribution. Value x of X P(X=x) 10 20 30 40 50 0.10 0.30 0.20 0.25 0.15 Complete the following. (If necessary, consult a list of formulas.) (a) Find the expectation E (X) of X. E (X) = 0 (b) Find the variance Var(X) of X. Var(x) =
Calculate the expected value of X, E(X), for the given probability distribution. 2 4 6. 8 6. 12 P(X = x) 20 20 20 20 E(X) :
A certain brand of cars captured 21.7% of the market. If 3 cars are selected at random, the probability distribution is given below. Find E(x). 1 2 3 P(x) 0.2 0.4 0.1 0,3 What is the expected value? E(x) = (Type an integer or a decimal.)
Let X be a random variable with the following probability distribution. Value x of X P(X=x) 3 4 5 5.0 6 0.10 0.05 0.40 0.45 Complete the following. (If necessary, consult a list of formulas.) (a) Find the expectation E (X) of X. E (X) = 0 (b) Find the variance Var(X) of X. Var(x) = 0
An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, XX, is as follows: xixi 1 2 3 4 5 6 Total P(X=xi)P(X=xi) 0.15 0.16 0.33 0.2 0.11 0.05 1 The expected value is computed, E[X]=3.11E[X]=3.11. To find the variance and standard deviation the following table was set up: xixi P(X=xi)P(X=xi) (xi−E[X])2(xi-E[X])2 (xi−E[X])2P(X=xi)(xi-E[X])2P(X=xi) 1 0.15 (1−3.11)2=(−2.11)2=4.4521(1-3.11)2=(-2.11)2=4.4521 4.4521⋅0.15=0.66784.4521⋅0.15=0.6678 2 (2−3.11)2=(−1.11)2=1.2321(2-3.11)2=(-1.11)2=1.2321 1.2321⋅0.16=0.19711.2321⋅0.16=0.1971 3 0.33 (3−3.11)2=(−0.11)2=(3-3.11)2=(-0.11)2= 0.0121⋅0.33=0.0040.0121⋅0.33=0.004 4 0.2 (4−3.11)2=(0.89)2=0.7921(4-3.11)2=(0.89)2=0.7921 0.7921⋅0.2=0.7921⋅0.2= 5 (5−3.11)2=(1.89)2=3.5721(5-3.11)2=(1.89)2=3.5721 3.5721⋅0.11=0.39293.5721⋅0.11=0.3929 6 0.05 8.3521⋅0.05=0.41768.3521⋅0.05=0.4176 E[X]=3.11E[X]=3.11 Total:…
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