Step 1 :Let's denote the minimum age as \(a = 4.79\) and the maximum age as \(b = 5.73\).
Step 2 :The mean of a uniform distribution is calculated as the average of \(a\) and \(b\), so the mean is \((4.79+5.73)/2 = 5.26\).
Step 3 :The standard deviation of a uniform distribution is calculated as \(\sqrt{(b-a)^2/12}\), so the standard deviation is \(\sqrt{(5.73-4.79)^2/12} = 0.2714\).
Step 4 :The probability that the child will be older than 5 years old is equivalent to 1 minus the probability that the child will be younger than 5 years old. The probability that the child will be younger than 5 years old is \((5-4.79)/(5.73-4.79) = 0.2234\). Therefore, the probability that the child will be older than 5 years old is \(1 - 0.2234 = 0.7766\).
Step 5 :The probability that the child will be between 4.99 and 5.59 years old is calculated as \((5.59-4.99)/(5.73-4.79) = 0.6383\).
Step 6 :The age of the child at the 6th percentile is calculated as \(4.79 + 0.06*(5.73-4.79) = 4.8464\).
Step 7 :Final Answer: \(\boxed{5.26}\) is the mean of this distribution.
Step 8 :Final Answer: \(\boxed{0.2714}\) is the standard deviation.
Step 9 :Final Answer: \(\boxed{0.7766}\) is the probability that the child will be older than 5 years old.
Step 10 :Final Answer: \(\boxed{0.6383}\) is the probability that the child will be between 4.99 and 5.59 years old.
Step 11 :Final Answer: \(\boxed{4.8464}\) is the age of the child at the 6th percentile.