91影视

A discrete random variable is one that can only take on a finite number of different values.

\({\rm{n = 10}}\)

The probability is calculated by dividing the number of positive outcomes by the total number of potential outcomes (\({\rm{15}}\) of the \({\rm{50}}\) businesses violate at least one regulation):

\(\begin{array}{c}{\rm{p = }}\frac{{{\rm{\# of favourable outcomes}}}}{{{\rm{\# of possible outcomes}}}}\\{\rm{ = }}\frac{{{\rm{15}}}}{{{\rm{50}}}}\\{\rm{ = }}\frac{{\rm{3}}}{{{\rm{10}}}}\\{\rm{ = 0}}{\rm{.3}}\end{array}\)

Because the sample of \({\rm{10}}\) represents more than \({\rm{10\% }}\) of the population, we cannot use the binomial distribution; instead, we may use the hypergeometric distribution in this case:

The value of N is the population size which is \({\rm{50}}\).

The value of n is the number of draws which is \({\rm{10}}\).

The value of r is the number of observed successes which is \({\rm{15}}\).

Hypergeometric probability formula:

\({\rm{p(y) = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{r}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{N - r}}}\\{{\rm{n - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\)

with

\(\left( {\begin{array}{*{20}{l}}{\rm{x}}\\{\rm{y}}\end{array}} \right){\rm{ = }}\frac{{{\rm{x!}}}}{{{\rm{y!(x - y)!}}}}\)

Fill in the blanks with the following values:

\(\begin{array}{c}{\rm{p(y) = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{15}}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{50 - 15}}}\\{{\rm{10 - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{10}}}\end{array}} \right)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{15}}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{35}}}\\{{\rm{10 - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{10}}}\end{array}} \right)}}\end{array}\)

Therefore, the value is: \(\begin{array}{c}{\rm{p(y) = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{15}}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{50 - 15}}}\\{{\rm{10 - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{10}}}\end{array}} \right)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{15}}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{35}}}\\{{\rm{10 - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{50}}}\\{{\rm{10}}}\end{array}} \right)}}\end{array}\).

\(\begin{array}{c}{\rm{p = }}\frac{{{\rm{\# of favourable outcomes}}}}{{{\rm{\# of possible outcomes}}}}\\{\rm{ = }}\frac{{{\rm{150}}}}{{{\rm{500}}}}\\{\rm{ = }}\frac{{\rm{3}}}{{{\rm{10}}}}\\{\rm{ = 0}}{\rm{.3}}\end{array}\)

Because the sample of \({\rm{10}}\) represents more than \({\rm{10\% }}\) of the population, we cannot use the binomial distribution; instead, we may use the hypergeometric distribution in this case:

The definition of binomial probability is:

\(\begin{aligned}{\rm{P(X = k)}}\\&{{\rm{ = }}_{\rm{n}}}{{\rm{C}}_{\rm{k}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}\\&{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}\end{aligned}\)

Replace n with \({\rm{10}}\) (since we'll pick \({\rm{10}}\) firms at random) and p with \({\rm{0}}{\rm{.30}}\):

\(\begin{aligned}{\rm{P(X = k) = }}\frac{{{\rm{10!}}}}{{{\rm{k!(10 - k)!}}}}{\rm{ \times 0}}{\rm{.3}}{{\rm{0}}^{\rm{k}}}{\rm{ \times (1 - 0}}{\rm{.30}}{{\rm{)}}^{{\rm{10 - k}}}}\\{\rm{ = }}\frac{{{\rm{10!}}}}{{{\rm{k!(10 - k)!}}}}{\rm{ \times 0}}{\rm{.3}}{{\rm{0}}^{\rm{k}}}{\rm{ \times 0}}{\rm{.7}}{{\rm{0}}^{{\rm{10 - k}}}}\end{aligned}\)

Therefore, the value is: \({\rm{P(X = k) = }}\frac{{{\rm{10!}}}}{{{\rm{k!(10 - k)!}}}}{\rm{ \times 0}}{\rm{.3}}{{\rm{0}}^{\rm{k}}}{\rm{ \times 0}}{\rm{.7}}{{\rm{0}}^{{\rm{10 - k}}}}\).

(C) EXACT

The number of draws multiplied by the number of successes, divided by the population size, is the mean of a hypergeometric distribution:

\(\begin{aligned}{\rm{E(X) = \mu }}\\&{\rm{ = }}\frac{{{\rm{n \times r}}}}{{\rm{N}}}\\&{\rm{ = }}\frac{{{\rm{10 \times 15}}}}{{{\rm{50}}}}\\&{\rm{ = }}\frac{{{\rm{150}}}}{{{\rm{50}}}}\\&{\rm{ = 3}}\end{aligned}\)

The hypergeometric distribution's variance is then:

\(\begin{aligned}{\rm{V(X) = }}{{\rm{\sigma }}^{\rm{2}}}\\&{\rm{ = n \times }}\frac{{\rm{r}}}{{\rm{N}}}{\rm{ \times }}\frac{{{\rm{N - r}}}}{{\rm{N}}}{\rm{ \times }}\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}\\&{\rm{ = 10 \times }}\frac{{{\rm{15}}}}{{{\rm{50}}}}{\rm{ \times }}\frac{{{\rm{50 - 15}}}}{{{\rm{50}}}}{\rm{ \times }}\frac{{{\rm{50 - 10}}}}{{{\rm{50 - 1}}}}\\&{\rm{ = }}\frac{{{\rm{12}}}}{{\rm{7}}}\\ &\approx {\rm{1}}{\rm{.7143}}\end{aligned}\)

APPROXIMATED

The product of the sample size n and the probability p is the mean of a binomial distribution:

\(\begin{aligned}{\rm{E(X) = \mu }}\\&{\rm{ = np}}\\&{\rm{ = 10(0}}{\rm{.30)}}\\&{\rm{ = 3}}\end{aligned}\)

The square root of the product of the sample size, probability of success, and probability of failure is the standard deviation of the binomial distribution. The variance equals the standard deviation squared.

\(\begin{aligned}{\rm{V(X) = }}{{\rm{\sigma }}^{\rm{2}}}\\&{\rm{ = npq}}\\&{\rm{ = np(1 - p)}}\\&{\rm{ = 10(0}}{\rm{.30)(1 - 0}}{\rm{.30)}}\\ &\approx {\rm{2}}{\rm{.1}}\end{aligned}\)

Therefore, the values are:

\(\begin{aligned}{\rm{E(X) = 3 and}}\\{\rm{V(x) }} \approx {\rm{ 1}}{\rm{.7143}}\\{\rm{E(X) = 3 and}}\\{\rm{V(x) }} \approx {\rm{ 2}}{\rm{.1}}\end{aligned}\)