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Chapter 7, Section 7.2, Question P9 The process for manufacturing a ball bearing results in weights that have an approximately normal distribution with mean 0.15 g and standard deviation 0.005 g. a. If you select one ball bearing at random, what is the probability that it weighs less than 0.149 g? (round to four decimal places) b. If you select four ball bearings at random, what is the probability that their mean weight is less than 0.149 g? (round to four decimal places) c. If you select ten ball bearings at random, what is the probability that their mean weight is less than 0.149 g? (round to four decimal places)

Respuesta :

LammettHash

a. Let [tex]X[/tex] be a random variable representing the weight of a ball bearing selected at random. We're told that [tex]X\sim\mathcal N(0.15,0.005^2)[/tex], so

[tex]\mathrm P(X<0.149)=\mathrm P\left(\dfrac{X-0.15}{0.005}<\dfrac{0.149-0.15}{0.005}\right)=\mathrm P(Z<-0.2)[/tex]

where [tex]Z\sim\mathcal N(0,1)[/tex]. This probability is approximately

[tex]\mathrm P(Z<-0.2)\approx0.4207[/tex]

b. Let [tex]X_i[/tex] be a random variable representing the weight of the [tex]i[/tex]-th ball that is selected, and let [tex]Y[/tex] be the mean of these 4 weights,

[tex]Y=\dfrac{X_1+X_2+X_3+X_4}4[/tex]

The sum of normally distributed random variables is a random variable that also follows a normal distribution,

[tex]X_1+X_2+X_3+X_4\sim\mathcal N(4\cdot0.15,4\cdot0.005^2)[/tex]

so that

[tex]Y\sim\mathcal N(0.15,0.005^2)[/tex]

Then

[tex]\mathrm P(Y<0.149)\approx0.4207[/tex]

c. Same as (b).

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