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, calculate
(a) the probability that the cups contains more than 224 milliliters.
(b) the probability that a cup between 191 and 209 milliliters.
(c) number of cups that will probably overflow if 230- milliliter cups are used for the next 1000 drinks.
(d) the amount of drink in the 25th percentile of the cups.

Answer 1

Given Information:  

The average discharge per cup  = μ = 200 ml

standard deviation = σ = 15 ml

Step-by-step explanation:  

(a) the probability that the cups contains more than 224 milliliters.

z = (x - μ)/σ

z = (224 - 200)/15

z = 1.6

P(z > 1.6) = 1 - P(z < 1.6)

P(z > 1.6) = 1 - 0.9452 (from the z-table)

P(z > 1.6) = 0.0548

P(X > 224) = 0.0548

(b) the probability that a cup between 191 and 209 milliliters.

z = (x - μ)/σ

z = (191 - 200)/15

z = -0.6

z = (209 - 200)/15

z = 0.6

P( -0.6 < X < 0.6) = P( -0.6 < z < 0.6)

P( -0.6 < X < 0.6) = P( z < 0.6 ) - P( z < -0.6 )

P( -0.6 < X < 0.6) = 0.7257 - 0.2743 (from the z-table)

P( -0.6 < X < 0.6) = 0.4514

P(191 < X < 209) = 0.4514

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

z = (x - μ)/σ

z = (230 - 200)/15

z = 2

P(z > 2) = 1 - P(z < 2)

P(z > 2) = 1 - 0.9772

P(z > 2) = 0.0228

For 1000 drinks

Number of cups overflow = 0.0228*1000 = 22.8 or 23 cups

(d) the amount of drink in the 25th percentile of the cups.

P(X < x) = 0.25 (25th percentile)

P(X < (x - μ)/σ) = 0.25

P(X < -0.68) = 0.25

(x - μ)/σ = -0.68

x - μ = -0.68σ

x = -0.68σ + μ

put σ = 15 and μ = 200

x = -0.68(15) + 200

x = -10.2 + 200

x = 189.8