Answer to Question #213131 in Statistics and Probability for Nazir Khan

Question

A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters,

(a) what fraction of the cups will contain more than 224 milliliters?

(b) what is the probability that a cup contains between 191 and 209 milliliters?

(c) how many cups will probably overflow if 230- milliliter cups are used for the next 1000 drinks?

(d) below what value do we get the smallest 25% of the drinks?

Answer 1

Let X represent the amount of drink distributed.

$\mu = 200 \\ \sigma=15$

(a) The fraction of the cups will contain more than 224 milliliters

$P(X>224) = 1 -P(X<224) \\ = 1 -P(Z< \frac{224-200}{15}) \\ = 1 -P(Z<1.6) \\ = 1 -0.9452 \\ = 0.0548 = 5.48 \; \%$

(b) The probability that a cup contains between 191 and 209 milliliters

$P(191<X<209) = P(X<209) -P(X<191) \\ = P(Z< \frac{209-200}{15}) -P(Z< \frac{191-200}{15}) \\ = P(Z<0.6) -P(Z< -0.6) \\ = 0.7257 -0.2743 \\ = 0.4514$

(c) The number of cups that will probably overflow if 230-milliliter cups are used for the next 1000 drinks

$P(X>230) = 1 -P(X<230) \\ = 1 -P(Z< \frac{230-200}{15}) \\ = 1 -P(Z<2) \\ = 1 -0.9772 \\ = 0.0228 \\ E(X) = n \times P \\ =1000 \times 0.0228 \\ = 22.8 ≈ 23$

(d) The value if we get the smallest 25% of the drinks

$P(X<x) = 0.25 \\ P(Z< \frac{x-200}{15}) = 0.25 \\ P(Z< -0.68) = 0.25 \\ \frac{x-200}{15} = -0.68 \\ x = -0.68 \times 15 + 200 \\ x=189.95$