Question 4
The age of children in kindergarten on the first day of school is uniformly distributed between 4.79 and 5.73 years old. A first time kindergarten child is selected at random. Round answers to 4 decimal places if possible.
a. The mean of this distribution is
b. The standard deviation is
c. What is the probability that the child will be older than 5 years old?
d. The probability that the child will be between 4.99 and 5.59 years old is
e. If such a child is at the 6th percentile, how old is that child? years old.
Hint:
Written Hint

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.