A coffee machine is adjusted to provide a population mean of 110ml of coffee per cup and a standard deviation of 5 ml. The volume of coffee per cup is assumed to have a normal distribution. The machine is checked periodically by sampling 12 cups of coffee. If the mean volume, Y, of those 12 cups in ml falls in the interval (110 - 2 cy~) < y < (110 + 2 o~), no adjustment is made. Otherwise, the machine is adjusted.

a) If a 12-cup test gives a mean volume of 107.0 ml, what should be done?

b) What fraction of the total number of 12-cup tests would lead to an adjust-

ment being made, even if the machine had not changed from its original

correct setting?

c) How many cups should be sampled randomly so there is 99% confidence

that the mean volume of the sample will lie within + 2 ml of 110ml when the machine is correctly adjusted?

Question

Nathen Powell

Mathematics

28 May, 11:17

A coffee machine is adjusted to provide a population mean of 110ml of coffee per cup and a standard deviation of 5 ml. The volume of coffee per cup is assumed to have a normal distribution. The machine is checked periodically by sampling 12 cups of coffee. If the mean volume, Y, of those 12 cups in ml falls in the interval (110 - 2 cy~) < y < (110 + 2 o~), no adjustment is made. Otherwise, the machine is adjusted.

a) If a 12-cup test gives a mean volume of 107.0 ml, what should be done?

b) What fraction of the total number of 12-cup tests would lead to an adjust-

ment being made, even if the machine had not changed from its original

correct setting?

c) How many cups should be sampled randomly so there is 99% confidence

that the mean volume of the sample will lie within + 2 ml of 110ml when the machine is correctly adjusted?

+2

Answers (1)

Know the Answer?

Elvis Joseph · Answer 1

a)

The mean volume (107) is less than the lower limit of the interval (110-2 = 108). Thus, there would be some measurements which were below the lower limit of the interval. So adjustments should be made.

b)

Standard error of sample mean = / sigma / / sqrt{n} = 5 / / sqrt{12} = 1.443376

P (adjustment is made) = P (y 110 + 2)

= P (y 112)

= P (z (112 - 110) / 1.443376)

= P (z 1.39)

= 0.0823 + 0.0823

= 0.1646

c)

z value for 99% confidence interval is 2.58

Margin of error, E = 2

Number of cups, n = (z * / sigma / E) 2

= (2.58 * 5 / 2) 2

= 41.6 / approx 42