Step 1

Probability of an event is given by the formula:

\(P(A) = \frac{favorable\ outcomes}{Total\ number\ of\ outcomes}\)

\(a)P (Black) = \frac{603}{2262}\)

\(= 0.26657\)

\(= 26.657\%\)

Thus, percentage of Black is \(26.657\%\)

\(b) P (Unarmed) = \frac{165}{2262}\)

\(= 0.07294\)

\(= 7.294\%\)

Thus, the percentage of Unarmed suspects is \(7.294\%\)

Step 2

c)In order for two variables to be Independent of each other, the \(P(A and B) = P(A) \cdot P(B)\)

Let A be the event that a person's race is Black and B be the event that the suspect was Unarmed.

Then, Probability that Black people (A) are killed being Unarmed (B) is given as:

\(P(A \cap B)=P(A) \cdot P(B)\)

\(= 0.266 \cdot 0.073\)

\(= 0.01941\)

\(= 1.941\%\)

d) Percent of Black and Unarmed using the table:

Look at the intersection value of the person who belongs to Black race and Unarmed.

\(P(A \cap B) = \frac{60}{2262}\)

\(= 0.02652\)

\(= 2.652\%\)

So, the percentage that Black people are killed while being Unarmed is \(2.652\%\) Therefore, the events A and B are not independent.

Answer from (c) is significantly different from answer (d). Hence, there is a different percentage of Unarmed people being shot based on race.

e) Percentage of suspects who are white and Unarmed:

\(P (white and unarmed) = \frac{67}{2262}= 0.02962 = 2.962\%\)

Thus, the percentage suspect being White and Unarmed is \(2.962\%\)

f) Percentage of Hispanic and Unarmed:

P (Hispanic and Unarmed) \(=\frac{38}{2262} = 0.01679\)

\(= 1.679\%\)

Thus, the percentage suspect being Hispanic and Unarmed is \(1.679\%\)

Comparing the answers of d,e and f we observe that the highest percentage of unarmed people being shot is white with \(2.962\%\)

Since there are multiple subparts we will solve the first 5 parts. To get remaining subparts solved please repost the complete question and mention the subparts to be solved.