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Formulas: ____________________________________________________________________________________ The mean of a random variable with a discrete probability distribution is: =1 (1)+2 (2)++ ()= () =1 ____________________________________________________________________________________ The variance of a

Formulas: ____________________________________________________________________________________

The mean of a random variable with a discrete probability distribution is:

=1 (1)+2 (2)++ ()= ()

=1

____________________________________________________________________________________

The variance of a discrete probability distribution:

2 =( )2 ()

=1

____________________________________________________________________________________

A shortcut formula for variance:

2 =2 ()

=1

2

____________________________________________________________________________________

The standard deviation of a probability distribution:

=2

____________________________________________________________________________________

Binomial Distribution:

( = )= , 0

____________________________________________________________________________________

Mean of the Binomial Distribution:

https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php

https://inst-fs-iad-prod.inscloudgate.net/files/fc254946-5cbe-4ed1-b31d-e17d2e0ed04b/z_table.pdf?download=1&token=eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzUxMiJ9.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.9j_EVZYDVn_rSSXZbgINsCGLuc_LhK1c4SH1AnD-CD1P3EXJ44O6OHw14cV7r0uBxBBt3tKUDTolXYfgMCwxDw

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Problem 1. Determine whether the distribution represents a probability distribution. If it does not, state why. (a) X 5 7 9 P(X) 1.1 0.2 0.3 (b) X - 20 - 30 - 40 - 50 P(X) 0.35 0.35 -0.1 0.4 (c) X 2 3 P(X) 0.5 0.3 0.3 (d) X - 2 - 1 0 2 P(X) 0.15 0.3 0.3 0.05 0.2 Problem 2. The number of cups of coffee a customer drinks with an evening meal at a restaurant is shown in the table, with the corresponding probabilities. Find the mean, variance, and standard deviation of the distribution. Xi P(X;) X; . P(X;) X7 . P(X1) 0 0.31 0.42 2 0.21 3 0.04 4 0.02 Total:Problem 3. A survey found that 10% of Americans believe that they have seen an UFO. For sample of 15 people find probability that: (a) exactly 4 believe they have seen an UFO. (b) at least 13 believe they have seen an UFO. Problem 4. In a recent year, 13% of businesses have eliminated jobs. In a random sample of 300 businesses, find the mean, variance, and standard deviation for the number of those who eliminated jobs. 2Problem 7. A survey found that the American family generates an average of 17.2 pounds of glass garbage each year. Assume variable is normally distributed with standard deviation of 2.5 pounds. (a) If one family is selected at random find the probability that it will generate less than 20 pounds of glass garbage in a year. (b) If a sample of 10 families is randomly selected, find the probability that the mean of the sample will be less than 20 pounds. Problem 8. The average teacher's salary in North Dakota is $37,764. Assume a normal distribution with standard deviation of $5, 100. For a sample of 40 teachers, find the probability that the sample mean is (a) greater than $35,000 (b) between $37,000 and $40,000Problem 5. Using the standard normal distribution find: (a) P(Z > 2.07) = (b) P(-0.16

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