Statistics question

Question

A drinks machine dispenses coffee into cups. A sign on the machine

indicates that each cup contains 50 ml of coffee. The machine actually

dispenses a mean amount of 55 ml per cup and 10% of the cup contain

less than the amount stated on the sign. Assuming that the amount of

coffee dispensed into each cup is normally distributed, find

(i) the standard deviation of the amount of coffee dispensed per cup

in ml.

(ii) the percentage of cups that contain more than 51 ml.

Following complaints, the owners of the machine made adjustments.

Only 2.5 % of the cups now contain less than 50 ml. The standard

deviation of the amount dispensed is reduced to 3 ml. Assuming that the

amount of coffee dispensed is still normally distributed,

(iii) find the new mean amount of coffee per cup.

Answer 1

(i) mean = 55, and 10% contain less than 50 ml. The Z value which has 10% of area to the left of it is -1.28

(on a TI-84 : invnorm, area = 0.10, mean = 0, standard deviation = 1)

Z = (50 - 55) / σ = -1.28 giving σ = 3.91

(ii) p (X>51). normalcdf. lower = 51, upper = 1E99, mean = 55, standard deviation = 3.91

0.8469, so 84.7 %

(iii) Z value that has area 2.5% to the left is : -1.96

So Z value corresponding to 50 ml = -1.96

If mean = μ then (50 - μ) / 3 = -1.96 giving 50 - μ = -5.88 so μ = 55.88