Question #125704

The weight of competition pumpkins at the Circleville Pumpkin Show in Circleville, Ohio, can be represented by a normal distribution with a mean of 703 pounds and a standard deviation of 347 pounds.

a. Find the probability that a randomly selected pumpkin weighs at least 1,622 pounds.

b. Find the probability that a randomly selected pumpkin weighs between 465.1 and 1,622

pounds.


1
Expert's answer
2020-07-09T18:22:04-0400

Let r. v. XX be the weight of randomly choosen pumpkin. Then XN(703;3472).X\in N (703;347^2).

a. We should find P{X1622}.P\{X\geq 1622\}.

P{X1622}=1P{X<1622}.P\{X\geq 1622\}=1-P\{X<1622\}.

P{X<1622}=F(1622)P\{X<1622\}=F(1622) where F(x)=12π347xe(t703)223472dt.F(x)=\frac{1}{\sqrt{2\pi}347}\int_{-\infty}^x e^{-\frac{(t-703)^2}{2\cdot 347^2}}dt.

F(1622)=Φ(1622703347)Φ(2.648)0.9960F(1622)=\Phi(\frac{1622-703}{347})\approx\Phi(2.648)\approx 0.9960 where

Φ(x)=12πxez22dz.\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{z^2}{2}}dz.

So P{X1622}=1P{X<1622}10.9960=0.004P\{X\geq 1622\}=1-P\{X<1622\}\approx 1-0.9960=0.004.

b. We should find P{465.1<X<1622}.P\{465.1<X<1622\}.

P{465.1<X<1622}=F(1622)F(465.1).P\{465.1<X<1622\}=F(1622)-F(465.1).

F(465.1)=Φ(465.1703347)Φ(0.686)0.2464.F(465.1)=\Phi(\frac{465.1-703}{347})\approx \Phi(-0.686)\approx 0.2464.

So P{465.1<X<1622}=F(1622)F(465.1)0.99600.2464=0.7496.\text{So }P\{465.1<X<1622\}=F(1622)-F(465.1)\approx\\ \approx 0.9960-0.2464=0.7496.


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