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?

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 > 110 + 2)

= P(y < 108) + P(y > 112)

= P(z < (108 - 110)/1.443376) + P(z > (112 - 110)/1.443376)

= P(z < -1.39) + 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

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540

not me

step-by-step explanation:

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