Question

The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A first time kindergarten child is selected at random. Round all answers to two decimal places.
1. The mean of the distribution is______________

a. 5.4

b. 5.2

c. 5.3

d. 5.5

2. The standard deviation is ______________

a. 0.194

b. 0.289

c. 0.333

d. 0.261

3. The probability that the child will be older than 5 years old is_______________

a. 0.3

b. 0.8

c. 0.2

d. 0.7

4. The probability that the child will be between 5.2 and 5.7 years old is_________.

a. 0.6

b. 0.7

c. 0.8

d. 0.5

5. If such a child is at the 45th percentile, how old is that child?_______________

a. 5.25

b. 5

c. 4.75

d. 5.5

Answer 1

Answer:

(1) (c) 5.30 years.

(2) (b) 0.289.

(3) (b) 0.80.

(4) (d) 0.50.

(5) (a) 5.25 years.

Step-by-step explanation:

Let X = age of the children in kindergarten on the first day of school.

The random variable X follows a continuous Uniform distribution with parameters a = 4.8 years and b = 5.8 years.

The probability density function function of X is:

$f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a$

(1)

The expected value of a Uniform random variable is:

$E(X)=\frac{1}{2}(a+b)$

Compute the mean of X as follows:

$E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3$

Thus, the  mean of the distribution is (c) 5.30 years.

(2)

The standard deviation of a Uniform random variable is:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}$

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289$

Thus, the standard deviation of the distribution is (b) 0.289.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

$P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx$

                $=\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8$

Thus, the probability that a randomly selected child is older than 5 years old is (b) 0.80.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

$P(5.2$

                            $=\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5$

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) 0.50.

(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 x as follows:

   $P (X < x) = 0.45$

$\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45$

        $\int\limits^{x}_{4.8} {1}\, dx=0.45$

           $[x]^{x}_{4.8}=0.45$

       $x-4.8=0.45$

                $x=0.45+4.8\\x=5.25$

Thus, the age of the child at the 45th percentile is (a) 5.25 years.