Question

Museum management. Refer to the Museum Management and Curatorship (June 2010) study of the criteria used to evaluate museum performance, Exercise 3.18 (p. 170). Recall that the managers of 30 leading museums of contemporary art were asked to provide the performance measure used most often. Of these 30 museums, 8 specified “total visitors” as the performance measure. Consider a random sample of 5 museums selected from the 30. How likely is it that none of the museums in the sample specified “total visitors” as the performance measure?

Short answer

For the given sample, there is only an 18.48% chance that none of the museums specified the total visitors as the performance measure.

Step-by-step solution

 Given information

X isthe number of times; the total visitors are selected in 5 museums.Here, the random variable x follows the hypergeometric distribution.

Calculating the chance that none of the museums specified the totalvisitors as the performance measure.

Random variable follows hypergeometric distribution whose probability can be calculated as:

$$\begin{gathered} \mathrm{P}\left(x\right)=\frac{\left(r \\ x \\ \right)\left(N-r \\ n-x \\ \right)}{\left(N \\ n \\ \right)} \end{gathered}$$

put

$N=30$

$$\begin{gathered} n=5 \\ r=8 \\ x=0 \end{gathered}$$

Hence,

$$\begin{gathered} \begin{array}{rcl}P\left(0\right)& =& \frac{\left(8 \\ 0\right)\left(30-8 \\ 5-0 \\ \right)}{\left(30 \\ 0 \\ \right)}\\ & =& \frac{\left(8 \\ 0 \\ \right)\left(22 \\ 5 \\ \right)}{\left(30 \\ 5 \\ \right)}\\ & & \end{array} \end{gathered}$$

$$\begin{gathered} P\left(0\right)=\frac{1\times 26334}{142,506} \\ =0.1848 \end{gathered}$$

Hence for the given sample, there is only an 18.48% chance that none of the museums specified the total visitors as the performance measure.Therefore, it can be concluded that it is not most likely to select that none of the museums specified “total visitors” as the performance measure.