At a large university students have either a final exam or a final paper at the end of a course. The table below lists the distribution of the number of final exams that students at the university will take, and their associated probabilities.

a. What are the mean and standard deviation of this distribution?

X 0 1 2 3

P (X) 0.05 0.25 0.40 0.30

a.

Mean of given distribution=1.95

Standard deviation of given distribution=0.865

Step-by-step explanation:

a.

Mean of given distribution = E (x) = ∑[x*P (x) ]

Mean of given distribution=0*0.05+1*0.25+2*0.4+3*0.3

Mean of given distribution=0+0.25+0.8+0.9

Mean of given distribution=1.95

Standard deviation of given distribution=σx=√[ E (x²) - (E (x)) ²]

E (x²) = ∑x²p (x) = 0²*0.05+1²*0.25+2²*0.4+3²0.3

E (x²) = 0+0.25+1.6+2.7

E (x²) = 4.55

Standard deviation of given distribution=√[4.55 - (1.95) ²]

Standard deviation of given distribution=√[4.55-3.803]

Standard deviation of given distribution=√0.748

Standard deviation of given distribution=0.865