Let X be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of X is as follows.X142 3p(x) 0.2 0.1 0.4 0.3(a) Consider a random sample of size n = 2 (two customers), and let X be the sample mean number of packages shipped. Obtain the probability distribution of X.2X11.5P(x) 0.040.040.12(b) Refer to part (a) and calculate P(X ≤ 2.5).2.533.54× 0.04×0.16× .15× 0.09the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of(c) Again consider a random sample of size n = 2, but now focus on the statistic R =R. [Hint: Calculate the value of R for each outcome and use the probabilities from part (a).]RP(R)23(d) If a random sample of size n = 4 is selected, what is P(X ≤ 1.5)? [Hint: You should not have to list all possible outcomes, only those for which x ≤ 1.5.]

Answer a).X Bar11.522.533.54P(X Bar)0.040.040.170.200.140.240.09b). 0.45…

Answered: Let X be the number of packages being…

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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The joint probability distribution function of X and Y is given by: 1 2 3 1 0.05 0.06 0.10 X 2 0.13 0.21 0.30 a. Find the conditional distribution of Y given X = 1 b. Find the conditional distribution of X given Y = 2 5 0.02 0.03 0.10
The probability distribution of random variable X is defined as follows: X 0 1 2 3 4 Probability 0 0.3 0.1 0.3 0.3 Find the expected value and the standard deviation of X.
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Let X represent the full height of a certain species of tree. Assume that X has a normal probability distribution with a mean of 58.2 ft and a standard deviation of 4.8 ft. A tree of this type grows in my backyard, and it stands 56.8 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter. P(X<56.8)= My neighbor also has a tree of this type growing in her backyard, but hers stands 61.1 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers. P(X>61.1)= Enter your answers as decimals accurate to 4 decimal places.
Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 1 2 4 Probability 0.264 0.289 0.238 0.156 0.037 0.016 ... (a) Find the mean, variance, and standard deviation of the probability distribution. The mean is 1.5. (Round to one decimal place as needed.) The variance is (Round to one decimal place as needed.)
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a newspaper article, the mean of the x distribution is about $46 and the estimated standard deviation is about $9. (a) MULTIPLE CHOICE Consider a random sample of n = 40 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution? A-The sampling distribution of x is approximately normal with mean ?x = 46 and standard error ?x = $0.23. B-The sampling distribution of x is not normal. C-The sampling distribution of x is approximately normal with mean ?x = 46 and standard error ?x = $1.42. D-The sampling distribution of x is approximately normal with mean ?x = 46 and standard error ?x = $9. (MULTIPLE CHOICE)Is it…
Let the random variable T follow a Student's t distribution with 9 degrees of freedom. State the distribution of the random variable T. Group of answer choices T∼Student′st(9) T∼Student′st(8) T∼Normal(0,1) T∼Student′st(0,1) Compute the probability that T is between −0.1 and 0.6. True or False: The larger the degrees of freedom in a Student's t distribution, the closer to a standard normal distribution the Student's t distribution becomes.
A snack company makes packages of grapes and honeydew slices. The weight of an individual package of grapes, G, is approximately Normally distributed with a mean of 3.25 ounces and a standard deviation of 0.91 ounces. The weight of an individual package of honeydew slices, H, is approximately Normally distributed with a mean of 4.51 ounces and a standard deviation of 2.02 ounces. Assume G and H are independent random variables. Let D = G – H. What is the probability that a randomly selected package of grapes weighs more than a randomly selected package of honeydew slices? 0.127 0.230 0.284 0.334

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