The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 31 liters, and standard deviation of 7.2 liters.

Question

A.) What is the probability that a daily production is less than 29.7 liters ?

B.) What is the probability that a daily production is more than 51.8 liters ?

( Round 4 decimal )

Answer 1

X ~ Normal (μ, σ²)

μ = 31 liters

σ = 7.2 liters

Find the z-score using the formula: z = (X - μ) / σ. Also, have your z-score table from the textbook or online to answer these 2 questions.

a.) z = (29.7 - 31) / 7.2 = -1.3 / 7.2 = -0.18

P(z < -0.18) = 0.4286

The probability that a daily production is less than 29.7 liters is 0.4286.

b.) z = (51.8 - 31) / 7.2 = 20.8 / 7.2 = 2.89

P(z > 2.89) = 1 - P(z ≤ 2.89) = 1 - 0.9981 = 0.0019

The probability that a daily production is more than 51.8 liters is 0.0019.