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Magazine Chapter 3: Q69E (page 130)
Each of \({\rm{12}}\)refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are running. Suppose that \({\rm{7}}\) of these refrigerators have a defective compressor and the other \({\rm{5}}\) have less serious problems. If the refrigerators are examined in random order, let\({\rm{X}}\)be the number among the first \({\rm{6}}\) examined that have a defective compressor.
a. Calculate\({\rm{P(X = 4)}}\)and \(P(X拢 4)\)
b. Determine the probability that \({\rm{X}}\) exceeds its mean value by more than \({\rm{1}}\) standard deviation.
c. Consider a large shipment of \({\rm{400}}\)\({\rm{40}}\) refrigerators, of which 40 have defective compressors. If \({\rm{X}}\) is the number among \({\rm{15}}\) randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X拢5)than to use the hypergeometric \({\rm{pmf}}\).
Short Answer
a). \(\begin{array}{*{20}{c}}{{\rm{P(X = 4) = }}\frac{{{\rm{25}}}}{{{\rm{66}}}}{\rm{\gg 0}}{\rm{.3788 = 37}}{\rm{.88\% }}}\\{{\rm{P(X拢4) = }}\frac{{{\rm{29}}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.8788 = 87}}{\rm{.88\% }}}\end{array}\)
b). \({\rm{P(X > \mu + \sigma ) = }}\frac{{\rm{4}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.1212 = 12}}{\rm{.12\% }}\)
c). binominal distribution: \({\rm{n = 15}}\)and \({\rm{p = 0}}{\rm{.1}}\)
\({\rm{P(X拢5)\gg 0}}{\rm{.9978 = 99}}{\rm{.78\% }}\)
Step by step solution
Definition of probability
The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.
Calculating \({\rm{P(X = 4) and P(X£4)}}\)
Given: The number of refrigerators with faulty compressors is \({\rm{X}}\)
\(\begin{array}{*{20}{c}}{}&{{\rm{N = Population size = 12}}}\\{}&{{\rm{n = Number of draws = 6}}}\end{array}\)
\({\rm{r = }}\)Number of successes observed \({\rm{ = 7}}\)
Hypergeometric probability formula:
\({\rm{p(y) = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{r}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{\rm{N - r}}}\\{{\rm{n - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{l}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\)
Calculate at \({\rm{y = 1,2,3,4}}\) (There are \({\rm{7}}\)refrigerators with defective compressors out of \({\rm{12}}\) refrigerators, hence there can only be \({\rm{5}}\)refrigerators with no defective compressors and at least one defective compressor in the sample of \({\rm{6}}\))
\({\rm{p(0) = 0(0refrigerators with a defective compressot}}\)
\(\begin{array}{*{20}{c}}{}\\{{\rm{p(1) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{1}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 1}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{132}}}}}\end{array}\)
\(\begin{array}{*{20}{c}}{}\\{{\rm{p(2) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 2}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{5}}}{{{\rm{44}}}}}\end{array}\)
\(\begin{array}{*{20}{c}}{}\\{{\rm{p(3) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 3}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{{\rm{25}}}}{{{\rm{66}}}}}\end{array}\)
\({\rm{P(X = 4) = p(4) = }}\frac{{{\rm{(4)(0 - 4)}}}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{{\rm{2b}}}}{{{\rm{66}}}}{\rm{\gg 0}}{\rm{.3788 = 37}}{\rm{.88\% }}\)
For discontinuous or mutually exclusive occurrences, use the following addition rule:
\({\rm{P(A\;or\;B) = P(A) + P(B)}}\)
Compile the following probabilities:
\(\begin{array}{*{20}{c}}{{\rm{P(X拢4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)}}}\\{{\rm{ = 0 + }}\frac{{\rm{1}}}{{{\rm{132}}}}{\rm{ + }}\frac{{\rm{5}}}{{{\rm{44}}}}{\rm{ + }}\frac{{{\rm{25}}}}{{{\rm{66}}}}{\rm{ + }}\frac{{{\rm{25}}}}{{{\rm{66}}}}{\rm{ = }}\frac{{{\rm{116}}}}{{{\rm{132}}}}{\rm{ = }}\frac{{{\rm{29}}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.8788 = 87}}{\rm{.88\% }}}\end{array}\)
Determining the probability that \({\rm{X}}\) exceeds its mean value by more than \({\rm{1}}\) standard deviation
Given: The number of refrigerators with faulty compressors is \({\rm{X}}\)
\(\begin{array}{*{20}{c}}{}&{{\rm{N = Population size = 12}}}\\{}&{{\rm{n = Number of draws = 6}}}\end{array}\)
\({\rm{r = }}\)Number of successes observed \({\rm{ = 7}}\)
The product of the number of draws and the number of observed successes, divided by the population size, is the mean of a hypergeometric distribution:
\({\rm{\mu = }}\frac{{{\rm{n \times r}}}}{{\rm{N}}}{\rm{ = }}\frac{{{\rm{6 \times 7}}}}{{{\rm{12}}}}{\rm{ = }}\frac{{{\rm{42}}}}{{{\rm{12}}}}{\rm{ = }}\frac{{\rm{7}}}{{\rm{2}}}{\rm{ = 3}}{\rm{.5}}\)
The hypergeometric distribution's variance is:
\({{\rm{\sigma }}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}{\rm{ \times n \times }}\frac{{\rm{r}}}{{\rm{N}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{r}}}{{\rm{N}}}} \right){\rm{ = }}\frac{{{\rm{12 - 6}}}}{{{\rm{12 - 1}}}}{\rm{ \times 6 \times }}\frac{{\rm{7}}}{{{\rm{12}}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{7}}}{{{\rm{12}}}}} \right){\rm{ = }}\frac{{{\rm{35}}}}{{{\rm{44}}}}{\rm{\gg 0}}{\rm{.7955}}\)
Find the value that is one standard deviation higher than the mean. The square root of the variance is the standard deviation:
\({\rm{\mu + \sigma = 3}}{\rm{.5 + }}\sqrt {{\rm{0}}{\rm{.7955}}} {\rm{\gg 4}}{\rm{.3919}}\)
Hypergeometric probability formula:
\({\rm{p(y) = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{r}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{\rm{N - r}}}\\{{\rm{n - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{l}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\)
If \({\rm{X}}\)is equal to \({\rm{5}}\) or\({\rm{6}}\) (\({\rm{X}}\)cannot take on values bigger than \({\rm{6}}\)), it is larger than \({\rm{\mu + \sigma = 4}}{\rm{.3919}}\)
\(\begin{array}{*{20}{c}}{{\rm{p(5) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{5}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 5}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{5}}}{{{\rm{44}}}}}\\{{\rm{p(6) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{6}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 6}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{132}}}}}\end{array}\)
For discontinuous or mutually exclusive occurrences, use the following addition rule:
\({\rm{P(A\;or\;B) = P(A) + P(B)}}\)
Compile the following probabilities:
\(\begin{array}{*{20}{c}}{{\rm{P(X > \mu + \sigma ) = P(X > 4}}{\rm{.3919) = P(X = 5) + P(X = 6)}}}\\{{\rm{ = }}\frac{{\rm{5}}}{{{\rm{44}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{132}}}}{\rm{ = }}\frac{{{\rm{16}}}}{{{\rm{132}}}}{\rm{ = }}\frac{{\rm{4}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.1212 = 12}}{\rm{.12\% }}}\end{array}\)
Determining the a less tedious way to calculate (at least approximately) \({\rm{P(X\pounds5)}}\) than to use the hypergeometric \({\rm{pmf}}\).
Given: The number of refrigerators with faulty compressors is \({\rm{X}}\)
\(\begin{array}{*{20}{c}}{}&{{\rm{N = Population size = 400}}}\\{}&{{\rm{n = Number of draws = 12}}}\end{array}\)
\({\rm{r = }}\)Number of successes observed \({\rm{ = 40}}\)
When the population size is big, the binomial distribution with \({\rm{n = 15}}\)is used to approximate the hypergeometric distribution.
\({\rm{p = }}\frac{{\rm{r}}}{{\rm{N}}}{\rm{ = }}\frac{{{\rm{40}}}}{{{\rm{400}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ = 0}}{\rm{.1}}\)
Binomial probability definition:
\({\rm{P(X = k)}}{{\rm{ = }}_{\rm{n}}}{{\rm{C}}_{\rm{k}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}\)
Examine at\({\rm{k = 0,1,2,3,4,5}}\):
\({\rm{P(X = 0) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 0)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{0}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 0}}}}{\rm{\gg 0}}{\rm{.2059}}\)
\(\begin{array}{*{20}{c}}{{\rm{P(X = 1) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 1)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{1}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 1}}}}{\rm{\gg 0}}{\rm{.3432}}}\\{{\rm{P(X = 2) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 2)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 2}}}}{\rm{\gg 0}}{\rm{.2669}}}\\{{\rm{P(X = 3) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 3)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{3}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 3}}}}{\rm{\gg 0}}{\rm{.1285}}}\\{{\rm{P(X = 4) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 4)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{4}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 4}}}}{\rm{\gg 0}}{\rm{.0428}}}\end{array}\)
\({\rm{P(X = 5) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 5)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{5}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 5}}}}{\rm{\gg 0}}{\rm{.0105}}\)
For discontinuous or mutually exclusive occurrences, use the following addition rule:
\({\rm{P(A\;or\;B) = P(A) + P(B)}}\)
Compile the following probabilities:
\(\begin{array}{*{20}{c}}{{\rm{P(X拢5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)}}}\\{{\rm{ = 0}}{\rm{.2059 + 0}}{\rm{.3432 + 0}}{\rm{.2669 + 0}}{\rm{.1285 + 0}}{\rm{.0428 + 0}}{\rm{.0105\gg 0}}{\rm{.9978 = 99}}{\rm{.78\% }}}\end{array}\)
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