The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumulative distributionfunction:0x < 1/80.11/8 < x < 1/4F(x) =0.91/4 ≤ x < 3/813/8 ≤ xDetermine each of the following probabilities.(a) P(X≤ 1/15) = i(b) P(X≤ 1/4)= i(c) P(X ≤5/16) = i(d) P(X> 1/4)= i(e) P(X ≤ 1/2) = i

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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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The probability distribution of the number of students absent on Mondays, is as follows: X O 1 2 3 4567 f(x) 0.02 0.03 0.26 0.34 0.22 0.08 0.04 0.01 (a) What is the probability that more than 3 students are absent. (b) Compute the expected value of the random variable X. Interpret this expected value. (c) Compute the variance and standard deviation of the random variable X. (d) Compute the expected value and variance of Y : (e) Compute the covariance Cov(X, Y) = 7X+ 3. (f) What is the value of the ratio Cov(X,Y)/Var(X) ? (g) What is the value of Cov(X, Y)/Var(X) for Y = ß0 + ß₁X ? (For this exrcise, you can use excel. No need to show all calculations, just specify the formulas and present the final values.) (a) (b) (c) (d) (e) (f) (g)
The random variable X has the following discrete probability distribution: X 1 3 5 7 9 p(x) 0.1 0.2 0.4 0.2 0.1 (a) List the values X may assume. (b) What value of X is most probable? (c) Find Pr(X = 7). (d) Find Pr(X≥ 5). (e) Find Pr(X > 2).
A random variable X is distributed as Poisson distribution with λ = 3.5. Usethis Poisson distribution (Table A.3) to determine the following probabilities: (a)P(X < 5), (b) P(2 ≤ X ≤ 6), (c) P(X > 7), (d) P(X ≥ 5).
The probability distribution of the random variable X is given in the following table. Random variable, x -2 -1 0 1 2 3 P(X = x) 0.10 0.15 0.05 0.20 0.15 0.35 Find the following probabilities. (a) P(X = 0)(b) P(X ≤ 0)(c) P(-1 < X ≤ 4)(d) P(X ≥ 1)
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:…
Suppose a random variable x is best described by a uniform probability distribution with a range 0 to 5. Find the value of a that makes the following probability statements true.
You have given the following probability distribution for the discrete random variable X f(x) 1 0.1 2 0.3 3. 0.35 4 0.2 0.05 What is E(X) and V(X) ? E(X)=2.8 and V(X)=1.06 OE(X)=2.4 and V(X)=1.04 OE(X)-2.6 and V(X)-1.06 DE(X)=2.6 and V(X)-8.9 E(X)=2.8 and V(X)-8.9

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