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A fast-food restaurant uses an automated filling machine to pour its soft drinks. The machine has different settings for small, medium, and large drink cups. According to the machine鈥檚 manufacturer, when the large setting is chosen, the amount of liquid dispensed by the machine follows a Normal distribution with mean $27$ounces and standard deviation $0.8$ounces. When the medium setting is chosen, the amount of liquid dispensed follows a Normal distribution with mean $17$ounces and standard deviation $0.5$ounces. To test the manufacturer鈥檚 claim, the restaurant manager measures the amount of liquid in a random sample of $25$cups filled with the medium setting and a separate random sample of $20$ cups filled with the large setting. Let ${\overline{x}}_1-{\stackrel{炉}{x}}_2$be the difference in the sample mean amount of liquid under the two settings (large 鈥� medium). Find the probability that${\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2$ is more than $12$ ounces. Show your work.

Step by step solution

01

Given information

Sample mean$=27$

Sample mean$=17$

Standard deviation$=0.8$

Standard deviation$=0.5$

02

Explanation

The given data is written as

$\left({\stackrel{炉}{x}}_1\right)=27$

$\left({\stackrel{炉}{x}}_2\right)=17$

${蟽}_1=0.8$

${蟽}_2=0.5$

${渭}_{1-2}={渭}_1-{渭}_2$

$27-17=10$

Then, ${蟽}_1-2=\sqrt{\frac{{蟽}_1^2+{蟽}_2^2}{n_1}}$

$=\sqrt{\frac{0.8^2}{25}+\frac{0.5^2}{20}}$

$=0.205$

Now the probability greater than $12$is computed as

$P\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2>12\right)=P\left(\frac{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{{蟽}_1^2+}{{}^n}\frac{a_2^2}{n_2}}}>\frac{12-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{{蟽}_1^2+a_2^2}{n_1}\frac{a_2}{n_2}}}\right)$

$=P\left(Z>\frac{12-10}{0.205}\right)$

$=P(Z>9.76)$

$=0.$

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