Question

Full-time students at a state school in New England earned between 10 and 16 credits in Spring 2012. The table below gives the distribution of the number of credits earned. # of credits 10 11 12 13 14 15 16 % of students 1 3 6 28 33 21 8 Find the approximate probability that the total number of credits earned by a random sample of 484 students from that school in that semester was more than 6750.

Answer 1

for a single student:

x	P(x)	xP(x)	x2P(x)
10	0.01	0.100	1.000
11	0.03	0.330	3.630
12	0.06	0.720	8.640
13	0.28	3.640	47.320
14	0.33	4.620	64.680
15	0.21	3.150	47.250
16	0.08	1.280	20.480
	total	13.840	193.000
	E(x) =μ=	ΣxP(x) =	13.8400
	E(x2) =	Σx2P(x) =	193.0000
	Var(x)=σ2 =	E(x2)-(E(x))2=	1.454
	std deviation=	σ= √σ2 =	1.2060

therefore for 484 students expected mean =484*13.84= 6698.56

and standard deviation =1.2060*sqrt(484)=26.532

therefore from normal approximation;approximate probability that the total number of credits earned by a random sample of 484 students from that school in that semester was more than 6750

=P(X>6750)=P(Z>(6750-6698.56)/26.532)=P(Z>1.94)=1-P(Z<1.94)=1-0.9738 =0.0262