gusiness Statistics Page 2 of ezc@gmail.com. Printing is for personal, private u

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gusiness Statistics Page 2 of ezc@gmail.com. Printing is for personal, private use only. No part of this book may be reproduced or transmitied without n. Violators will be prosecuted CHAPTERS Discrete Probability Situations 119 8. Assume a binomial distribution with an n of 5 and a probability of success of 35 percent. Answer the following. a. What is the mean? b. What is the variance? c. What is the standard deviation? d. What is the probability of 2? What is the probability of 1 or more? e. f. What is the probability of less than 4? g. What are the 4 characteristics that must be present for a binonial situation to exist? 9. The number of people who go through the express line at a local grocery store averages 2 per minute. Assume this is a Poisson distribution and answer the following. a. What is the mean of the distribution? b. What is the probablity of 0 people going through the line in a minute? What is the probability of I person going through the line in a minute? C. d. What is the probability of 2 people going through the line in a minute? What is the probability of 5 people going through the line in a minute? e.

Explanation / Answer

Question 8

We have to use binomial distribution here. We are given

n = 5

p = 0.35

Part a

Formula for mean is given as below:

Mean = n*p = 5*0.35 = 1.75

Part b

Variance = n*p*q

Where, q = 1 – p = 1 – 0.35 = 0.65

Variance = 5*0.35*0.65 = 1.1375

Part c

Standard deviation = sqrt(n*p*q)

Standard deviation = sqrt(5*0.35*0.65)

Standard deviation = 1.06653645

Part d

P(X=x) = nCx*p^x*q^(n – x)

P(X=2) = 5C2*0.35^2*0.65^(5 – 2)

P(X=2) = 10*0.35^2*0.65^3

P(X=2) = 0.336415625

Required probability = 0.336415625

Part e

We have to find P(X1)

P(X1) = 1 – P(X=0)

P(X=0) = 5C0*0.35^0*0.65^(5 – 0)

P(X=0) = 1*1*0.65^5

P(X=0) = 0.116029063

P(X1) = 1 – P(X=0)

P(X1) = 1 – 0.116029063

P(X1) = 0.883970938

Required probability = 0.883970938

Part f

We have to find P(X<4)

P(X<4) = 1 - P(X4)

P(X4) = P(X=4) + P(X=5)

P(X=4) = 5C4*0.35^4*0.65^1 = 0.048770313

P(X=5) = 5C5*0.35^5*0.65^0 = 0.005252188

P(X4) = 0.048770313 + 0.005252188

P(X4) = 0.0540225

P(X<4) = 1 - P(X4)

P(X<4) = 1 - 0.0540225

P(X<4) = 0.9459775

Required probability = 0.9459775

Part g

Four situation needs to present are binary output, n number of trials, p probability of successes, and independence of outputs.

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