Final answer:
The probability that a randomly selected nine-year-old child eats an average of more than 8 eggs is 0.5. The probability that a randomly selected nine-year-old child eats an average of more than 8 eggs, given that the amount is more than 5 eggs, is 0.21875. The standard deviation of the number of eggs taken by a nine-year-old child is 2.3094010767585034.
Step-by-step explanation:
To determine the probability that a randomly selected nine-year-old child eats an average of more than 8 eggs, we need to find the area under the probability density function (PDF) curve from 8 to 12. Since the distribution is uniform and the total range is 12 - 4 = 8, the probability density is 1/8 = 0.125. Therefore, the probability is (12 - 8) * 0.125 = 0.5.
For part 2, the probability that a randomly selected nine-year-old child eats an average of more than 8 eggs, given that the amount is more than 5 eggs, can be determined by finding the conditional probability. Since the distribution is uniform, the probability density is still 1/8 = 0.125. To find the conditional probability, we need to find the probability of the intersection of the two events: P(A and B) = P(A) * P(B|A). P(A) is the probability of eating more than 5 eggs, which is (12 - 5) * 0.125 = 0.875. P(B|A) is the probability of eating more than 8 eggs, given that the amount is more than 5 eggs, which is (12 - 8) * 0.125 / (12 - 5) * 0.125 = 0.25. Therefore, the conditional probability is P(B|A) = 0.875 * 0.25 = 0.21875.
To calculate the standard deviation of the number of eggs taken by a nine-year-old child, we can use the formula for the standard deviation of a uniform distribution: σ = (b - a) / sqrt(12), where a is the lower bound (4) and b is the upper bound (12). Substituting the values: σ = (12 - 4) / sqrt(12) = 2.3094010767585034.