A machine that fills bottles with a beverage has a fill volume whose mean is 20.01 ounces, with a standard deviation of 0.02 ounces. A case consists of 24 bottles randomly sampled from the output of the machine. a. Find the mean of the total volume of the beverage in the case. b. Find the standard deviation of the total volume of the beverage in the case. c. Find the mean of the average volume per bottle of the beverage in the case. d. Find the standard deviation of the volume per bottle of the beverage in the case. e. How many bottles must be included in a case for the standard deviation of the average volume per bottle to be 0.0025 ounces?

Answer:a) The mean of the total volume of the beverage in the case of 24 bottles is 480.24 ounces.b) The standard deviation of the total volume of the beverage in the case is 0.48 ounces.c) The mean of the average volume per bottle of the beverage in the case is 20.01 ounces, the same as the expected value of a single bottle.d) The standard deviation of the volume per bottle of the beverage is 0.00417 ounces.e) The number of bottles in a case for the standard deviation of the average volume per bottle to be 0.0025 ounces must be 65 bottles. Step-by-step explanation:a) Applying the linearity of expectation, the expected value of a constant multypling the random variable is equal to the constant multiplied by the expected value of the random variable.In this case:[tex]E(24x)=24*E(x)=24*20.01=480.24[/tex]The mean of the total volume of the beverage in the case of 24 bottles is 480.24 ounces.b) The variance follows this rule: "multiplying a random variable by a constant increases the variance by the square of the constant".As the standard deviation is the square root of the variance, the linearity property applies to the standard deviationThis can be applied as[tex]\sigma=\sqrt{V(24x)}=\sqrt{24^2*V(x)}=24*\sqrt{V(x)}=24*\sigma_x=24*0.02=0.48[/tex]The standard deviation of the total volume of the beverage in the case is 0.48 ounces.c) The expected value of the average volume per bottle of the beverage in the case can be expressed as[tex]E(\bar{x})=\frac{E(24x)}{24}=\frac{24}{24} *E(x)=1*20.01=20.01[/tex]The mean of the average volume per bottle of the beverage in the case is 20.01 ounces, the same as the expected value of a single bottle.d) The standard deviation of the volume per bottle of the beverage in the case can be calculated as the standard deviation of a sample of n=24:[tex]s=\sqrt{\frac{s_x^2}{n-1} } =\frac{s_x^2}{n-1}=\frac{0.02}{\sqrt{ 24-1}}=  0.00417[/tex] The standard deviation of the volume per bottle of the beverage is 0.00417 ounces.e) To calculate how many bottles must be included in a case for the standard deviation of the average volume per bottle to be 0.0025 ounces, we can use the formula used for point d:[tex]s=\frac{s_x}{\sqrt{n-1}}=\frac{0.02}{\sqrt{n-1}} =0.0025\\\\\sqrt{n-1}=\frac{0.02}{0.0025} =8\\\\n=8^2+1=64+1=65[/tex]The number of bottles in a case for the standard deviation of the average volume per bottle to be 0.0025 ounces must be 65 bottles.