Question

The amount X of beverage in a can labeled 12 ounces is normally distributed with mean 12.1 ounces and standard deviation 0.05 ounce. A can is selected at random.

a. Find the probability that the can contains at least 12 ounces.

b. Find the probability that the can contains between 11.9 and 12.1 ounces.

Answer 1

Given: = 12.1, = 0.05

To find the probability, we need to find the Z scores first.

Z = (X - )/

(a) P(X 12) = 1 - P(X < 12)

P(X < 12), Z = (12 - 12.1)/0.05 = -2

The probability is = 0.0228

The required probability for P(X > 21) =1 - 0.0228 = 0.9772

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(13) P(11.9 < X < 12.1) = P(X < 12.1) - P(X < 11.9)

For P(X < 12.1); Z = (12.1 – 12.1)/0.05 = 0

The probability for P(X < 12.1) = 0.5

For P(X < 11.9); Z = (11.9 – 12.1)/0.05 = -4

The probability for P(X < 11.9) = 0.00

Therefore the required probability = 0.5000 - 0.000 = 0.5000

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