Answer:
[tex]\begin{array}{cc}{Status} & {Probability} & {1} & {0.0845} & {2} & {0.1127} & {3} & {0.1870} & {4} & {0.1927}& {5} & {0.4231} \ \end{array}[/tex]
Step-by-step explanation:
Given
The above table
Required
The discrete probability distribution
The probability of each is calculated as:
[tex]Pr = \frac{Frequency}{Total}[/tex]
Where:
[tex]Total = 2140+ 2853 + 4734 + 4880 + 10715[/tex]
[tex]Total = 25322[/tex]
So, we have:
[tex]P(1) = \frac{2140}{25322} = 0.0845[/tex]
[tex]P(2) = \frac{2853}{25322} = 0.1127[/tex]
[tex]P(3) = \frac{4734}{25322} = 0.1870[/tex]
[tex]P(4) = \frac{4880}{25322} = 0.1927[/tex]
[tex]P(5) = \frac{10715}{25322} = 0.4231[/tex]
So, the discrete probability distribution is:
[tex]\begin{array}{cc}{Status} & {Probability} & {1} & {0.0845} & {2} & {0.1127} & {3} & {0.1870} & {4} & {0.1927}& {5} & {0.4231} \ \end{array}[/tex]