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Chapter 8: Q.8.4 (page 393)

Suppose that the number of units produced daily at factory A is a random variable with mean 20and standard deviation 3and the number produced at factory B is a random variable with mean 18and standard deviation of 6. Assuming independence, derive an upper bound for the probability that more units are produced today at factory B than at factory A.

Short Answer

Expert verified

An upper bound for the probability that more units are produced today at factory B than at factory A is 45/49=.9184.

Step by step solution

01

Given Information

Let XArepresents the number of units produced daily at factory Aand XBrepresents the number of units produced daily at factory B. For these random variables is given that:

μA=EXA=20,σA=3,μB=EXB=18,σA=6

Also, assume that random variables XAand XBare independent.

02

Explanation

The probability that more units are produced today at factory Bthan at factory Ais

PXB>XA=PXB-XA>0.

In that case, let's consider the random variable X=XB-XA. The random variable Xhas mean

μ=E[X]=μB-μA=-2

and, because of independence, variance

σ2=Var(X)=σB2+σA2=45

Now, using Corollary for a=2¯, we get:

P{X>-2+2⏟}≤σ2σ2+22=4545+4=.9184

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Most popular questions from this chapter

Civil engineers believe that W, the amount of weight (in units of 1000pounds) that a certain span of a bridge can withstand without structural damage resulting, is normally distributed with a mean of 400and standard deviation of40. Suppose that the weight (again, in units of 1000pounds) of a car is a random variable with a mean of 3and standard deviation.3. Approximately how many cars would have to be on the bridge span for the probability of structural damage to exceed.1?

Let f(x)be a continuous function defined for0≤x≤1. Consider the functions

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