Let \( X \) be the number of packages being mailed by a randomly selected customer at a shipping facility. Suppose the distribution of \( X \) is as follows:\[\begin{array}{c|cccc}x & 1 & 2 & 3 & 4 \\\hlinep(x) & 0.4 & 0.3 & 0.2 & 0.1 \\\end{array}\](a) Consider a random sample of size \( n = 2 \) (two customers), and let \( \bar{X} \) be the sample mean number of packages shipped. Obtain the probability distribution of \( \bar{X} \).(b) Refer to part (a) and calculate \( \Pr(\bar{X} \leq 2.5) \).(c) Again consider a random sample of size \( n = 2 \), but now focus on the statistic \( R \), the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of \( R \).(d) If a random sample of size \( n = 4 \) is selected, what is \( \Pr(\bar{X} \leq 1.5) \)? [Hint: you should not have to list all possible outcomes, only those for which \( \bar{x} \leq 1.5 \).]

Given Let x be the number of packages being mailed by a randomly selected customer the probability…

Answered: Let X be the number of packages being…

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 marginal distributions of two random variables X and Y, together with some values of their joint distribution, are given in the table below. Fill in the table in such a way that X and Y are independent random variables. Y= 1 Y= 3 X= 0 0.50 X= 1 0.30 X= 2 0.10 0.10 0.20 0.50 0.50 1.00
To investigate the relation between hair color and eye color, the hair color and eye color of 5383 persons were recorded. The data are given in the following table: Hair Color ... Eye Color 1 2 Light 1168 825 305 Dark 573 1312 1200 a. Determine the joint and marginal probability distributions of X and Y. b. P(X + Y = 5) C. P(1 < Xs 3, 1
The marginal distributions of two random variables X and Y, together with some values of their joint distribution, are given in the table below. Fill in the table in such a way that X and Y are independent random variables. Y= -2 Y= 0 Y= 3 X= -3 0.02 0.03 0.05 0.10 X=0 0.21 0.70 X= 3 0.06 0.20 0.20 0.30 0.50 1.00
Consider a uniform distribution from a=1to b=24. (a) Find the probability that x lies between 7 and15. (b) Find the probability that x lies between 4 and22. (c) Find the probability that x lies between 2and 11. (d) Find the probability that x lies between 6 and 16. LOADING... Click the icon to see the definition of the uniform distribution. (a) The probability that x lies between 7 and 15 is nothing. (Round to three decimal places as needed.) (b) The probability that x lies between 4 and 22 is nothing. (Round to three decimal places as needed.) (c) The probability that x lies between 2 and 11 is nothing. (Round to three decimal places as needed.) (d) The probability that x lies between 6 and 16 is nothing. (Round to three decimal places as needed.)
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Suppose that for a given computer salesperson, the probability distribution of x = the number of systems sold in one week is given by the following table. x 1 p(x) 2 3 4 5 0.06 0.10 (a) Find the mean value of x (the mean number of systems sold). 0.11 0.30 6 7 8 0.29 0.12 0.01 0.01 (b) Find the variance and standard deviation of x. (Round your standard deviation to four decimal places.) variance standard deviation How would you interpret these values? (Round your standard deviation to four decimal places.) The mean squared deviation from the mean number of systems sold in one week is A typical deviation from the mean number of systems sold in one week is (c) What is the probability that the number of systems sold is within 1 standard deviation of its mean value? (d) What is the probability that the number of systems sold is more than 2 standard deviations from the mean?

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