Question #275605

Let the continuous r.v. X denote the weight (in pounds) of a package. The range of weight of packages is


between 45 and 60 pounds.


(a) Determine the probability that a package weighs more than 50 pounds.


(b) Find the mean and the variance of the weight of packages.


1
Expert's answer
2022-01-20T11:40:53-0500

(a). Probability that the pack weighs more than

 P(x>50)=1P(x50)=1P(45<x50)=14550115dx.=1115(5045)=1515=113=23.\begin{aligned} P(x>50) &=1-P(x \leq 50) \\ &=1-P(45<x \leq 50) \\ &=1-\int_{45}^{50} \frac{1}{15} \cdot d x . \\ &=1-\frac{1}{15}(50-45) \\ &=1-\frac{5}{15}=1-\frac{1}{3}=\frac{2}{3} . \end{aligned}


 (b). Mean =μ=(b+a)2=E(x). E(x)=(60+452)=52.5 Variance =E(x2)[E(x)]2E(x2)=abx2f(x)dx=b2+ab+a23=602+60×45+4523. Variance =2775(52.5)2 Variance =18.75\begin{aligned} \text { (b). Mean }&= \mu=\frac{(b+a)}{2}=E(x) . \\ \therefore &\ E(x)=\left(\frac{60+45}{2}\right)=52.5 \\ \text { Variance }=& E\left(x^{2}\right)-[E(x)]^{2} \\ &E\left(x^{2}\right)=\int_{a}^{b} x^{2} \cdot f(x) \cdot d x=\frac{b^{2}+a b+a^{2}}{3} \\ &= \frac{60^{2}+60 \times 45+45^{2}}{3 .} \\ \therefore \text { Variance }=& 2775-(52.5)^{2} \\ \text { Variance }=18.75 \end{aligned}

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