Final answer:
The probability of selecting a chicken that lays more than 270 eggs per year is 97.72%, calculated using the properties of the normal distribution and finding the complement of the probability associated with the z-score.
Step-by-step explanation:
To determine the probability of randomly selecting a chicken that lays more than 270 eggs per year, we use the normal distribution properties because the eggs laid by the chickens are normally distributed with an annual mean (μ) of 300 and a standard deviation (σ) of 15.
First, we need to calculate the z-score, which represents the number of standard deviations an element is from the mean. The formula to calculate the z-score (z) for a given value (x) is:
z = (x - μ) / σ
Here, x is 270 (the number of eggs laid by a chicken that we are interested in), so the z-score is:
z = (270 - 300) / 15
z = -30 / 15
z = -2
Now that we have the z-score, we reference the standard normal distribution table or use a software package that calculates the probability for a z-score. The table or software will give us the probability of having a z-score less than -2, which we denote as P(Z < -2). To find the probability of having more than 270 eggs, we need to subtract this value from 1 (since the total probability under the normal curve is 1).
Assuming P(Z < -2) = 0.0228 (from z-table), the probability of a chicken laying more than 270 eggs a year is:
1 - P(Z < -2) = 1 - 0.0228
1 - P(Z < -2) = 0.9772
Therefore, there is a 97.72% chance of randomly selecting a chicken that lays more than 270 eggs per year.