Answer:
To find the size of the sample needed to estimate the average weight of individuals in a population with a 95% confidence level and an error of less than 1kg, we can use the following formula:
n = (Z^2 * σ^2) / (e^2)
Where:
n is the sample size
Z is the standard normal deviation corresponding to the confidence level (1.96 for 95% confidence)
σ is the standard deviation of the population (6kg in this case)
e is the desired margin of error (1kg in this case)
Plugging these values into the formula, we get:
n = (1.96^2 * 6^2) / (1^2)
Solving this equation, we get:
n = 38.4
Therefore, the sample size needed to estimate the average weight of individuals in the population with a 95% confidence level and an error of less than 1kg is approximately 38 individuals.