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问题及解答

[Exer8-2] Exercise 77 of Book {Devore2017B} P.137

Posted by haifeng on 2020-04-07 08:02:40 last update 2020-04-07 08:02:40 | Edit | Answers (1)

An article in the Los Angeles Times (Dec. 3, 1993) reports that $1$ in $200$ people carry the defective gene that causes inherited colon cancer. In a sample of $1000$ individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that

  • (a) Between $5$ and $8$ (inclusive) carry the gene.
  • (b) At least $8$ carry the gene.
     

 

1

Posted by haifeng on 2020-04-08 11:41:34

Let $X$=number of people who carry the defective gene that causes inherited colon cancer.

By the report of the article in the Los Angeles Times (Dec. 3, 1993), the expected value or mean number of the people who carry the defective gene in the sample of 1000 individuals would be equal to 5.

Hence, the approximate distribution of the number who carry this gene corresponding to this sample is Poisson distribution: Poisson(5). That is, the pmf of the rv $X$ is

\[
p(x;\lambda)=p(x;5)=\frac{e^{-5}5^x}{x!},\quad x=0,1,2,\ldots
\]

 


(a) The approximate probability that $X$ between $5$ and $8$ (inclusive) carry the defective gene is

\[
P(5\leqslant X\leqslant 8)=\sum_{x=5}^{8}p(x;5)=\sum_{x=5}^{8}\frac{e^{-5}5^x}{x!}\approx 0.491413
\]

 

Do calculation by Calculator.exe

>> sum(5^x/(x!),x,5,8)
in> sum(5^x/(x!),x,5,8)

out> 72.932168

------------------------

>> exp(5)
in> exp(5)
out> 148.413158

------------------------

>> 72.932168/148.413158
in> 72.932168/148.413158

out> 0.491413

 


(b)

The approximate probability that $X$ will be at least $8$ (inclusive) who carry the defective gene is

\[
\begin{split}
P(X\geqslant 8)&=1-P(X < 8)=1-P(X\leqslant 7)\\
&=1-\sum_{x=0}^{7}p(x;5)=1-\sum_{x=0}^{7}\frac{e^{-5}5^x}{x!}\\
&\approx 1-0.866628\\
&=0.133372
\end{split}
\]