Question

These are the answers. Can someone help in figuring out how to get them?

15. A coffee machine is adjusted to provide a population mean of 110 ml 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, x, of those 12 cups in ml falls in the interval (1 10-2 σ1 ) (1 10 + 20% ), no adjustment is made. Otherwise, the machine is adiusted 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? How many cups should be sampled randomly so there is 99% confidence that the mean volume of the sample will lie within t2 ml of 110 ml when the machine is correctly adjusted? c) 15 (a)に-2.08, adjust, (b) 0.046, (c) 42

Answer 1

Question 15

Population mean = $\mu$ = 110 ml

Here n = 12

standard deviation = $\sigma$ = 5 ml

standard error of sample mean =  se₀ = $\sigma$/sqrt(n) = 5/sqrt(12) = 1.4434

so here sampe mean = $\bar x$ = 107.0 ml

Z = ($\bar x$ -$\mu$)/se₀ = (107 - 110)/1.4434 = -2.08

so here l Z l > 2.0 so here we would have to adjust.

(b) so here we have to find the probability that machine had not changed from its original correct setting.

Pr(110- 2$\sigma$_x < x < 110 + 2 $\sigma$_x) = Pr(Z < 2) - Pr(Z < -2)

so here looking into z table

Pr(110- 2$\sigma$_x < x < 110 + 2 $\sigma$_x) = Pr(Z < 2) - Pr(Z < -2)

= 0.97725 - 0.02275 = 0.9545

so here

Probability that it is out of the setting = 1 - 0.9545 = 0.046

(c) Here if the sampe size = n

confidence interva = 99%

critical value = NORMSINV(0.5 + 0.99/2) = 2.576

margin of error = $\sigma$/sqrt(n) = 5/sqrt(n)

so here margin of error given = 2 ml

so here

2 = 2.576 * 5/sqrt(n)

sqrt(n) = (2.576 * 5/2)

n = 41.47 or 42