Final answer:
To find the probability of a cow producing less than 20 liters of milk, use the Z-score formula to calculate the Z-score for 20 liters. To find the probability of a cow producing between 27.3 and 38.3 liters, calculate the Z-scores for the two values and subtract the probabilities. To find the 82nd percentile of milk production, find the Z-score that corresponds to this percentile and use the Z-score formula to find the corresponding X value.
Step-by-step explanation:
To determine the probabilities in this problem, we will use the standard normal distribution. The Z-score formula is used to calculate the probability.
a.) To find the probability that a cow produces less than 20 liters, we need to calculate the Z-score for 20 liters using the formula: Z = (X - mean) / standard deviation. This gives us Z = (20 - 31) / 7.3 = -1.50. Using a Z-table or a calculator, we find that the probability is approximately 0.0668.
b.) To find the probability that a cow produces between 27.3 and 38.3 liters, we first calculate the Z-scores for both values: Z1 = (27.3 - 31) / 7.3 = -0.47 and Z2 = (38.3 - 31) / 7.3 = 1.01. Using the Z-table or a calculator, we find the corresponding probabilities for Z1 and Z2. Subtracting the probability for Z1 from Z2 gives us the probability of the range, which is approximately 0.3072.
c.) To find the 82nd percentile, we need to find the Z-score that corresponds to this percentile. Using a Z-table or a calculator, we find that the Z-score for the 82nd percentile is approximately 0.92. Then we can use the formula Z = (X - mean) / standard deviation to find the corresponding X value: 0.92 = (X - 31) / 7.3. Solving for X gives us X = 0.92 * 7.3 + 31 = 37.68 liters. Therefore, the 82nd percentile of milk production is approximately 37.68 liters.