A small high school holds its graduation ceremony in the gym. Because of seating

Question

A small high school holds its graduation ceremony in the gym. Because of seating constraints, students arc limited to a maximum of four tickets to graduation for family and friends. The vice principal knows that historically 30% of students want four tickets. 25% want three. 25% want two. 15% want one. and 5% want none. Let X = the number of tickets requested by a randomly selected graduating student, and assume the historical distribution applies to this rv. Find the mean and standard deviation of X. Let T = the total number of tickets requested by the 150 students graduating this year. Assuming all 150 students' requests arc independent, determine the mean and standard deviation of T. The gym can seat a maximum of 500 guests. Calculate the (approximate) probability that all students requests can be accommodated. (Express this probability in terms of T. What distribution does T have?)

Explanation / Answer

a) x p(x)

0 0.05

1 0.15

2 0.25

3 0.25

4 0.30

mean = e(x) = x1*p(x) +x2*p(x2)+........+xn*p(xn)

0*0.05 +1*0.15+2*0.25+3*0.25+4*0.30 = 0.15+0.50+0.75+1.20 = 2.60

var(x) = E(x^2) +E(X)^2

= 0*0.05+1*0.15+4*0.25+9*0.25+16*0.30 - 6.76

= 1.44

STANDARD DEV = VAR(X)^(1/2) = 1.44^(1/2) = 1.2

B) THIS PART WILL BE DONE SIMILARLY AND EVEN THE ANSWER WILL BE SAME BECAUSE EVEN IF THE TOTAL NUMBER IS GIVEN BUT THE PROBABILITY OF EACH TYPE WILL NOT CHANGE

HENCE THE MEAN AND STANDARD DEVIATION WILL REMAIN SAME.

C) 500 GUEST TOTAL

0 GUEST FOR STUDENT = 5% = 25 = PROBABILITY = 25/500

1 GUEST FOR STUDENT = 15% = 75= PROBABILITY = 75/500

2 GUES FOR STUDENT = 25% = 125 = PROBABILITY = 125/500

3 GUEST FOR STUDENT = 25= 125= PROBABILITY = 125/500

4 GUEST FOR STUDENT = 30% = 150= PROBABILITY = 150/500

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