Respuesta :
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:
[tex]f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a<X<b,\ a<b\atop {0;\ otherwise}} \right.[/tex]
(1)
The expected value of a Uniform random variable is:
[tex]E(X)=\frac{1}{2}(a+b)[/tex]
Compute the mean of X as follows:
[tex]E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3[/tex]
Thus, the mean of the distribution is (c) 5.30 years.
(2)
The standard deviation of a Uniform random variable is:
[tex]SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}[/tex]
Compute the standard deviation of X as follows:
[tex]SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289[/tex]
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:
[tex]P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\[/tex]
[tex]=\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8[/tex]
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:
[tex]P(5.2<X<5.7)=\int\limits^{5.7}_{5.2} {\frac{1}{5.8-4.8}}\, dx\\[/tex]
[tex]=\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5[/tex]
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:
[tex]P (X < x) = 0.45[/tex]
[tex]\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45[/tex]
[tex]\int\limits^{x}_{4.8} {1}\, dx=0.45[/tex]
[tex][x]^{x}_{4.8}=0.45[/tex]
[tex]x-4.8=0.45\\[/tex]
[tex]x=0.45+4.8\\x=5.25[/tex]
Thus, the age of the child at the 45th percentile is (a) 5.25 years.
