Question

a variable y of a finite population has the frequency distribution shown in the table. suppose a member is selected at random from the population and let y denote the value of the variable y for the member obtained. complete parts (a) through (d) below.
a. determine the probability distribution of the random variable y.
b. use random - variable notation to describe the events that y takes on the value 6, a value less than 6, and a value of at least 6.
the event that y takes on the value 6 can be represented as \\( \\{ y = 6 \\} \\).
the event that y takes on a value less than 6 can be represented as \\( \\{ y \lt 6 \\} \\).
the event that y takes on a value of at least 6 can be represented as \\( \\{ y \geq 6 \\} \\).
c. find \\( p ( y = 6 ) \\), \\( p ( y \lt 6 ) \\), and \\( p ( y \geq 6 ) \\).
\\( p ( y = 6 ) = .45 \\) (type an integer or a decimal. do not round.)
\\( p ( y \lt 6 ) = .3 \\) (type an integer or a decimal. do not round.)
\\( p ( y \geq 6 ) = \square \\) (type an integer or a decimal. do not round.)

Explanation:

Step1: Calculate total frequency

Total frequency \(n = 3 + 3+9 + 5=20\)

Step2: Calculate \(P(Y = 6)\)

\(P(Y = 6)=\frac{9}{20}=0.45\)

Step3: Calculate \(P(Y < 6)\)

\(P(Y < 6)=\frac{3 + 3}{20}=\frac{6}{20}=0.3\)

Step4: Calculate \(P(Y\geq6)\)

Using the formula \(P(Y\geq6)=P(Y = 6)+P(Y = 7)\)
\(P(Y = 7)=\frac{5}{20}=0.25\)
\(P(Y\geq6)=0.45+0.25 = 0.7\)

Answer:

\(P(Y = 6)=0.45\), \(P(Y < 6)=0.3\), \(P(Y\geq6)=0.7\)