If Z is a standard normal random variable, find the value z, for the following probabilities. (Round your answers to two decimal places.)(a) P(Z > z,) = 0.5(b) P(Z < z) = 0.8907(c) P(-z,

Ww have given that Z is standard normal random variable i.e, Z ~ N(0, 1). Note: According to…

Answered: If Z is a standard normal random…

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)
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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)
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Q1A Suppose that the probability of finding defectives within a manufactured lot of electrical fuses is 0.09. Assume the probability distribution function (PDF) is binomial. Let X be the number of defectives within the sample. In a sample of 25 fuses, find the probabilities: (a) P(X 2), (b) P(X > 0), (c) P(1 ≤ X < 4). === B Suppose the discrete random variable X has the PDF as in the table below. Find: (a) F(7), (b) E(4-3X). I 4 6 7 8 9 f(x) 0.2 0.25 0.15 0.3 0.1
Given Probabilities y = 1 y = 2 x = 1 x = 2 0.2 0.3 0.5 Find: а. Pr(X > Y) %3 b. Pr(Y = 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:…
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In a pizza takeout restaurant, the following probability distribution was obtained. The random variable z represents the number of toppings for a large pizza. The table below is partially completed. Finish it and use it to find the mean and standard deviation. P(x) | xP(x) (x – µ) | (r – µ)° (2 - 4)°P(@) 0.14 -2.66 7.0756 0.990584 0.04 0.04 -1.66 2.7556 0.110224 2 0.24 0.48 -0.66 0.4356 0.104544 3. 0.18 4 0.40 Mean = 2.66 and Standard Deviation = 1.39 Mean = 0.52 and Standard Deviation = 1.39 Mean = 2.66 and Standard Deviation = 1.94 Mean = 0.52 and Standard Deviation = 1.94 Mean = -2.66 and Standard Deviation = 1.94

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If Z is a standard normal random variable, find the value z, for the following probabilities. (Round your answers to two decimal places.) (a) P(Z > z,) = 0.5 (b) P(Z