A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standard deviation of 0.2 ounce. If a sample of 9 cups is selected, find the probability that the mean of the sample will be
Greater than 12.1 ounces. Find the probability if the sample is just 1 cup.
A)0.0668; 0.3085
B)0.0668; 0.8085
C)0.5668; 0.8085
D)0.5668; 0.3085

Question

Quizplus · Accepted Answer

For a sample of 9 cups, the standard error of the mean can be found using the formula: 
standard error = standard deviation / sqrt(sample size) = 0.2 / sqrt(9) = 0.0667 (rounded to 4 decimal places)

To find the probability that the mean of the sample is greater than 12.1 ounces, we need to standardize the sample mean using the formula: 
z = (sample mean - population mean) / standard error = (12.1 - 12) / 0.0667 = 1.5007

Using a standard normal distribution table or calculator, we can find the probability of a z-score being greater than 1.5007 is 0.0668 (rounded to 4 decimal places).

For a sample of just 1 cup, the standard error of the mean would be: 
standard error = standard deviation / sqrt(sample size) = 0.2 / sqrt(1) = 0.2

To find the probability that the mean of the sample is greater than 12.1 ounces, we need to standardize the sample mean using the formula: 
z = (sample mean - population mean) / standard error = (12.1 - 12) / 0.2 = 0.5

Using a standard normal distribution table or calculator, we can find the probability of a z-score being greater than 0.5 is 0.3085 (rounded to 4 decimal places). For a sample of 9 cups, the standard deviation of the sample means is:
standard deviation = standard deviation of population/sqrt(sample size) = 0.2/sqrt(9) = 0.0667

So, to find the probability that the sample mean will be greater than 12.1 ounces, we can standardize the variable:
z = (sample mean - population mean)/(standard deviation of sample means) = (12.1-12)/(0.0667) = 1.50

Using a standard normal distribution table, the probability of getting a z-score greater than 1.50 is 0.0668.

For a sample of just 1 cup, the standard deviation of the sample means is:
standard deviation = standard deviation of population/sqrt(sample size) = 0.2/sqrt(1) = 0.2

So, to find the probability that the sample mean will be greater than 12.1 ounces, we can standardize the variable:
z = (sample mean - population mean)/(standard deviation of sample means) = (12.1-12)/(0.2) = 0.50

Using a standard normal distribution table, the probability of getting a z-score greater than 0.50 is 0.3085.

A Coffee Machine Dispenses Normally Distributed Amounts of Coffee with a Mean