Match the following:A probability distribution for a random variable X:-300.150.20XP(X=x)V✓ P(X≤0)✓ P(-3≤x≤1)✓ P(X>1)✓ P(X<-1)V-80.13-10.17A. 0.67B. 0.28C. 0.65D. 0.2010.1540.1160.09

X-8-3-10146P(X=x)0.130.150.170.200.150.110.09

Answered: Match the following: A probability…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Fill in the P(X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -4, -1, 0, 5, and 6. Value x of X -4 - 1 0 5 6 P ( X = x) 0.15 0 0.18 0.15
Fill in the P (X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -2, 3, 4, 5, and 6. Value x of X P(x x) -2 0.20 3. 4 0.10 5. 6. 0.17
Fill in the P(X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -6, 1, 2, 4, and 6. Value x of X -6 1 2 P(X = x) 0.20 4 0.16 6 0.21
For a large population of people, the probability distribution for the random variable X = the number of meals a person ate yesterday and the assigned probabilities is given below: Meals, X Probability (a) Find P(X= 5), the probability that a randomly selected person ate 5 meals yesterday. P(X= 5) = 1 3 4 5 0.17 0.35 0.24 0.17 ? (b) Find P(X ≤ 2), the probability that a randomly selected person ate 2 or fewer meals yesterday. P(X ≤ 2) = (c) Find the probability that a randomly selected person ate 1 meal or ate 5 meals yesterday. P = (d) Find P(X > 1), the probability that a randomly selected person ate more than 1 meal yesterday. P(X > 1) =
Let X be a random variable with the following probability distribution: - 4 3 5 f(x) 0.6 0.3 0.1 Find the standard deviation of X. The standard deviation is . (Round to two decimal places as needed.)
Consider the table below gives the probabilities of the distribution of random variable X. Value of X (x) P(X=x) (f(x)) 0 1 2 345 00 6 7 8 0.14 b C 0.10 0.13 0.13 d 0.02 0.01 In reference to the table, suppose that b = f(1) =0.15 and c = f(2)= 0.14. What is the probability P(2 ≤ X<7) equal to ?
Find the expected value of the random variable X, where the probability distribution of X is given by the following table: |-30 - 10 0 | 5 P(X= x)|0.15|0.25 |0.05 |0.10|0.25|0.20 10 20
Fill in the P(X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are −2, −1, 4, 5, and 6. Value x of X -2 -1 4 5 6 P ( X = x) 0.17 0.27 0.20
Which of the following tables represents a valid discrete probability distribution? X P(X=x) X P(X=X) X P(X=x) X P(X=x) 5 0.51 1 0.23 0 10 15 0.03 0.03 3 0.03 1 0.33 0.15 3 0.24 5 0.15 2 0.27 20 0.39 7 9 0.17 0.19 3 0.19 25 30 0.15 0.01 12 11 0.02 4 5 1.01 0.06 6 9 15 18 0.22 0.23 0.12 -0.03 0.43
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x 14 15 16 17 18 P(x) 0.1 0.2 0.1 0.1 0.5 Given the discrete probability distribution above, determine the following: P(14≤x<16)
Fill in the P(X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -2, 0, 4, 5, and 6. Value x of X -2 0 4 6 P(x = x ) 0.20 0.12 0.21

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Match the following: A probability distribution for a random variable X: -3 0 0.15 0.20 X P(X=x) ✓ P(X≤0) ✓ P(-3≤x≤1) ✓ P(X>1) ✓ P(X<-1) V -8 0.13 -1 0.17 A. 0.67 B. 0.28 C. 0.65 0.0.30 1 0.15 4 0.11 6 0.09