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. x 1 2 3 4 p(x) 0.1 0.2 0.3 0.4   (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.

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A: The formula of Standard deviation is,

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A: a) From the given table  P(Y=1)=0.158

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A: Given: xP(x)10.1420.17830.2240.2650.15260.02170.01980.00590.004100.001Total1

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A: Given data is

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A:

Q: Let X be the number of packages being mailed by a randomly selected customer at a certain shipping…

A: Given the distribution of X is : x1234P(x)0.20.10.40.3

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A: From the given information,       X       129 130 131 Y 15 0.12 0.42 0.06   16 0.08 0.28…

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A: Solution-: We have following table:   Y=1 Y=2 Y=3 Total X=1 - 0.08 - 0.20 X=2 - 0.32 - 0.80…

Q: Let X be the number of packages being mailed by a randomly selected customer at a certain shipping…

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

Q: Consider selecting a household at random in the rural area of a country. Define the random variable…

A: We have given that the probability distribution of X in tabulation form. The random variable is the…

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A:

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A: Solution-: We have given following table: x P(x) 0 0.04 1 0.11 2 0.18 3 0.25 4 0.12…

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A:

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A: mean=E(X)=∑x.P(X)

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Q: Let X be the number of packages being mailed by a randomly selected customer at a certain shipping…

A: Hi! Thank you for the question. As per the honor code, we are allowed to answer three sub-parts at a…

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A: From the given information, The joint mass probability function table can be shown below:   Y=1…

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A: x f(x) 0 0.001 1 0.007 2 0.034 3 0.099 4 0.188 5 X 6 0.220 7 0.136 8 0.055 9…

Q: Let X be the number of packages being mailed by a randomly selected customer at a certain shipping…

A:

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A:

Q: Use the probability distribution to complete parts​ (a) and​ (b) below.     The number of defects…

A:

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A:

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A:

Q: The following table lists the probability distribution of a discrete random variable x: x 0 1 2 3 4…

A: Solution: From the given information, the probability distribution of X is

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A:

Q: x f(x) (3) 0 0.15 1 ? 2 0.3 3 ? 4 0.2 the mean of the above distribution is known to be 2.15 (i.e.,…

A: We have given that probability distribution table.

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A: The formula of the mean is:

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A: Given:   X P(x) 0 0.266 1 0.292 2 0.238 3 0.145 4 0.04 5 0.019 total 1

Q: Height (x) (in inches) 54 P(x) 1/23 1/23 5/23 4/23 7/23 4/23 1/23 Frequency 57 1 61 4 62 63 4 67…

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A: Given   Y=0 Y=1 Y=2 Y=3 PX(X=x) X= 0 0.32 0.06 0.025 0.01   X= 1   0.02 0.05 0.07   X=2…

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A: X has a Poisson distribution with P(X>0)=0.9967

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Q: Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000…

A:

Q: The probability distribution of the number of students absent on Mondays, is as follows: X O 1 2 3…

A: (a) The probability that more than 3 students are absent:P(X > 3) = P(X = 4) + P(X = 5) + P(X =…