Answer:
(1) (c) <u>5.30 years</u>.
(2) (b) <u>0.289</u>.
(3) (b) <u>0.80</u>.
(4) (d) <u>0.50</u>.
(5) (a) <u>5.25 years</u>.
Step-by-step explanation:
Let <em>X</em> = age of the children in kindergarten on the first day of school.
The random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 4.8 years and <em>b</em> = 5.8 years.
The probability density function function of <em>X</em> is:
(1)
The expected value of a Uniform random variable is:
Compute the mean of <em>X</em> as follows:
Thus, the mean of the distribution is (c) <u>5.30 years</u>.
(2)
The standard deviation of a Uniform random variable is:
Compute the standard deviation of <em>X</em> as follows:
Thus, the standard deviation of the distribution is (b) <u>0.289</u>.
(3)
Compute the probability that a randomly selected child is older than 5 years old as follows:
![=\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E%7B5.8%7D_%7B5%7D%20%7B1%7D%5C%2C%20dx%5C%5C%3D%5Bx%5D%5E%7B5.8%7D_%7B5%7D%5C%5C%3D%285.8-5%29%5C%5C%3D0.8)
Thus, the probability that a randomly selected child is older than 5 years old is (b) <u>0.80</u>.
(4)
Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:
![=\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E%7B5.7%7D_%7B5.2%7D%20%7B1%7D%5C%2C%20dx%5C%5C%3D%5Bx%5D%5E%7B5.7%7D_%7B5.2%7D%5C%5C%3D%285.7-5.2%29%5C%5C%3D0.5)
Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) <u>0.50</u>.
(5)
It is provided that a randomly selected child is at the 45th percentile.
This implies that:
P (X < x) = 0.45
Compute the value of <em>x</em> as follows:
![[x]^{x}_{4.8}=0.45](https://tex.z-dn.net/?f=%5Bx%5D%5E%7Bx%7D_%7B4.8%7D%3D0.45)
Thus, the age of the child at the 45th percentile is (a) <u>5.25 years</u>.
