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Statistics / Applied Statistics and Probability for Engineers 6 / Chapter 3.6 / Problem 93E

Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

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Let X be a binomial random variable with p = 0.1 and n =

Chapter 3, Problem 93E

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QUESTION:

Let X be a binomial random variable with p = 0.1 and n = 10. Calculate the following probabilities from the binomial probability mass function and from the binomial table in Appendix A and compare results.

(a) \(P(X \leq 2)\)

(b) \(P(X>8)\)

(c) \(P(X=4)\)

(d) \(P(5 \leq X \leq 7)\)

Questions & Answers

QUESTION:

Let X be a binomial random variable with p = 0.1 and n = 10. Calculate the following probabilities from the binomial probability mass function and from the binomial table in Appendix A and compare results.

(a) \(P(X \leq 2)\)

(b) \(P(X>8)\)

(c) \(P(X=4)\)

(d) \(P(5 \leq X \leq 7)\)

ANSWER:

Solution :

Step 1 of 4:

Given X be a binomial random variable with p=0.1 and n=10.

The formula for the binomial is

P(X = x) = ${nC}_{x}^{\ }$ ${p}^{x}{(1-p)}^{n-x}$

Where n=10, p=0.1 and

q = 1-p

q = 1-0.1

q =1-p

q = 0.9

Then the probability mass function is $\sum\limits_{X=0}^{x}{10C}_{x}^{\ }{(0.1)}^{x}{(0.9)}^{10-x}$

Our goal is:

We need to find the following probability,

a). P(X $\leq{}$ 2).

b). P(X>8).

c). P(X=4).

d). P(5 $\leq{}$ X $\leq{}$ 7).

a). Given P(X $\leq{}$ 2).

P(X $\leq{}$ 2) = $\sum\limits_{x=0}^{2}{10C}_{x}^{\ }{(0.1)}^{x}{(0.9)}^{10-x}$

P(X $\leq{}$ 2) = P(X=0)+P(X=1)+P(X=2)

P(X $\leq{}$ 2) = ${10C}_{0}^{\ }$ ${(0.1)}^{0}{(0.9)}^{10-0}$ + ${10C}_{1}^{\ }$ ${(0.1)}^{1}{(0.9)}^{10-1}$ + ${10C}_{2}^{\ }$ ${(0.1)}^{2}{(0.9)}^{10-2}$

P(X $\leq{}$ 2) = 0.34867+0.38742+0.193710

P(X $\leq{}$ 2) = 0.9298

Therefore, P(X $\leq{}$ 2) = 0.9298.

Now we are comparing this probability is matching with the binomial table in Appendix A table value.

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