The Hypergeometric Distribution. In this exercise, we discuss the hypergeometric distribution in more detail. When sampling is done without replacement from a finite population, the hypergeometric distribution is the exact probability distribution for the number of members sampled that have a specified attribute. The hypergeometric probability formula is

$P\left(X=x\right)=\frac{\left(\begin{array}{c}Np\\ x\end{array}\right)\left(\begin{array}{c}N\left(1-p\right)\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$,

where Xdenotes the number of members sampled that have the specified attribute, Nis the population size, nis the sample size, and pis the population proportion.

To illustrate, suppose that a customer purchases 4 fuses from a shipment of 250, of which 94 % are not defective. Let a success correspond to a fuse that is not defective.

(a) Determine N, n, and p.

(b) Apply the hypergeometric probability formula to determine the probability distribution of the number of nondefective fuses that the customer gets.

Key Fact 5.6 shows that a hypergeometric distribution can be approximated by a binomial distribution, provided the sample size does not exceed 5% of the population size. In particular, you can use the binomial probability formula

$P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}$

with $n=4\mathrm{and}p=0.94$, to approximate the probability distribution of the number of nondefective fuses that the customer gets.

(c) Obtain the binomial distribution with parameters $n=4\mathrm{and}p=0.94$.

(d) Compare the hypergeometric distribution that you obtained in part (b) with the binomial distribution that you obtained in part (c).

Step by step solution

01

Part (a) Step 1. Given information.

The given hypergeometric probability formula is:

$P\left(X=x\right)=\frac{\left(\begin{array}{c}Np\\ x\end{array}\right)\left(\begin{array}{c}N\left(1-p\right)\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

A consumer orders four fuses from a batch of 250, 94 percent of which are not defective.

02

Part (a) Step 2. Calculate N, n, and p.

From the above information, we can determine the values of N, n, and p:

$$\begin{gathered} N=250 \\ n=4 \\ p=0.94 \end{gathered}$$

03

Part (b) Step 1. Calculate the probability distribution.

$$\begin{gathered} P\left(X=x\right)=\frac{\left(\begin{array}{c}Np\\ x\end{array}\right)\left(\begin{array}{c}N\left(1-p\right)\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)} \\ =\frac{\left(\begin{array}{c}250\times 0.94\\ x\end{array}\right)\left(\begin{array}{c}250\left(1-0.94\right)\\ n-x\end{array}\right)}{\left(\begin{array}{c}250\\ 4\end{array}\right)} \\ =\frac{\left(\begin{array}{c}235\\ x\end{array}\right)\times \left(\begin{array}{c}15\\ n-x\end{array}\right)}{158882750} \end{gathered}$$

Now put the values of xas 1, 2, 3, and 4.

The probability distribution of the number of non-defective fuses the customer will receive is:

X	$P\left(X=x\right)$
0	$$\begin{gathered} \frac{\left(\begin{array}{c}235\\ 0\end{array}\right)\left(\begin{array}{c}15\\ 4-0\end{array}\right)}{158882750}=\frac{1365}{158882750} \\ =0.000009 \end{gathered}$$
1	$$\begin{gathered} \frac{\left(\begin{array}{c}235\\ 1\end{array}\right)\left(\begin{array}{c}15\\ 4-1\end{array}\right)}{158882750}=\frac{455}{158882750} \\ =0.000673 \end{gathered}$$
2	role="math" localid="1652970602695" $\frac{\left(\begin{array}{c}235\\ 2\end{array}\right)\left(\begin{array}{c}15\\ 4-2\end{array}\right)}{158882750}=0.018170$
3	$\frac{\left(\begin{array}{c}235\\ 3\end{array}\right)\left(\begin{array}{c}15\\ 4-3\end{array}\right)}{158882750}=0.201606$
4	$\frac{\left(\begin{array}{c}235\\ 4\end{array}\right)\left(\begin{array}{c}15\\ 4-4\end{array}\right)}{158882750}=0.779542$

04

Part (c) Step 1. With given parameters, obtain the binomial distribution.

The formula for binomial distribution is:

$$\begin{gathered} P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x} \\ n=4\mathrm{and}p=0.94 \\ P\left(X=0\right)=\left(\begin{array}{c}4\\ 0\end{array}\right){\left(0.94\right)}^0{\left(0.06\right)}^{4-0} \\ =0.000013 \\ P\left(X=1\right)=\left(\begin{array}{c}4\\ 1\end{array}\right){\left(0.94\right)}^1{\left(0.06\right)}^{4-1} \\ =0.000812 \\ P\left(X=2\right)=\left(\begin{array}{c}4\\ 2\end{array}\right){\left(0.94\right)}^2{\left(0.06\right)}^{4-2} \\ =0.019086 \\ P\left(X=3\right)=\left(\begin{array}{c}4\\ 3\end{array}\right){\left(0.94\right)}^3{\left(0.06\right)}^{4-3} \\ =0.1999340 \\ P\left(X=4\right)=\left(\begin{array}{c}4\\ 4\end{array}\right){\left(0.94\right)}^4{\left(0.06\right)}^{4-4} \\ =0.780749 \end{gathered}$$

X	0	1	2	3	4
P(X)	0.000013	0.000812	0.019086	0.199340	0.780749

05

Part (d) Step 1. Compare parts (c) and (d).

Parts (c) and (d) show that there is less difference between the results obtained by using hypergeometric distribution and binomial distribution.

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