Question

The fill amount in 2-liter soft drink bottles is normally distributed, with a mean of 2.0 liters and a standard deviation of 0.05 liter. If bottles contain less than 95% of the listed net content

(1.90 liters, in this case), the manufacturer may be subject to penalty by the state office of consumer affairs. Bottles that have a net

content above 2.10 liters may cause excess spillage upon opening. What proportion of the bottles will contain

a. between 1.90 and 2.0 liters?

b. between 1.90 and 2.10 liters?

c. below 1.90 liters or above 2.10 liters?

d. At least how much soft drink is contained in 99% of the bottles?

e. Ninety-nine percent of the bottles contain an amount that is between which two values (symmetrically distributed) around the mean?

Answer 1

part a)

X ~ N ( µ = 2 , σ = 0.05 )
P ( 1.9 < X < 2 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1.9 - 2 ) / 0.05
Z = -2
Z = ( 2 - 2 ) / 0.05
Z = 0
P ( -2 < Z < 0 )
P ( 1.9 < X < 2 ) = P ( Z < 0 ) - P ( Z < -2 )
P ( 1.9 < X < 2 ) = 0.5 - 0.0228
P ( 1.9 < X < 2 ) = 0.4772

Part b)

X ~ N ( µ = 2 , σ = 0.05 )
P ( 1.9 < X < 2.1 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1.9 - 2 ) / 0.05
Z = -2
Z = ( 2.1 - 2 ) / 0.05
Z = 2
P ( -2 < Z < 2 )
P ( 1.9 < X < 2.1 ) = P ( Z < 2 ) - P ( Z < -2 )
P ( 1.9 < X < 2.1 ) = 0.9772 - 0.0228
P ( 1.9 < X < 2.1 ) = 0.9545

Part c)

below 1.90 liters or above 2.10 liters = 1 - P ( 1.9 < X < 2.1 ) = 1 - 0.9545 = 0.0455

Part d)

X ~ N ( µ = 2 , σ = 0.05 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.99 = 0.01
To find the value of x
Looking for the probability 0.01 in standard normal table to calculate Z score = -2.3263
Z = ( X - µ ) / σ
-2.3263 = ( X - 2 ) / 0.05
X = 1.8837 ≈ 1.88 liter
P ( X > 1.88 ) = 0.99

Part e)

X ~ N ( µ = 2 , σ = 0.05 )
P ( a < X < b ) = 0.99
Dividing the area 0.99 in two parts we get 0.99/2 = 0.495
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.495
Area above the mean is b = 0.5 + 0.495
Looking for the probability 0.005 in standard normal table to calculate Z score = -2.5758
Looking for the probability 0.995 in standard normal table to calculate Z score = 2.5758
Z = ( X - µ ) / σ
-2.5758 = ( X - 2 ) / 0.05
a = 1.8712 ≈ 1.87 liter
2.5758 = ( X - 2 ) / 0.05
b = 2.1288 ≈ 2.13 liter
P ( 1.87 < X < 2.13 ) = 0.99