Question

The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied year round. Approximate the mean and standard deviation square footage.

Square footage	Lower Limit	Upper Limit	Frequency
0-499	0	499	9
500-999	500	999	13
1000-1499	1000	1499	36
1500-1999	1500	1999	115
2000-2499	2000	2499	125
2500-2999	2500	2999	83
3000-3499	3000	3499	45
3500-3999	3500	3999	45
4000-4499	4000	4499	22
4500-4999	4500	4999	7

(a) The mean square footage is xbar (round to the nearest integer as​ needed) = 2,418.5 (this is the answer i got)

(b) The standard deviation square footage is (round to one decimal place as​ needed.)s= 914.97 , but my homework has a correct answer of 911.4. I've attached how I got the 914.97. Please tell me what I did wrong and how can I get the right answer.  ​

Answer 1

mid-point	frequency
x	f	f*M	f*M2
249.5	9	2245.5	560252.25
749.5	13	9743.5	7302753.25
1249.5	36	44982	56205009
1749.5	115	201192.5	351986278.8
2249.5	125	281187.5	632531281.3
2749.5	83	228208.5	627459270.8
3249.5	45	146227.5	475166261.3
3749.5	45	168727.5	632643761.3
4249.5	22	93489	397281505.5
4749.5	7	33246.5	157904251.8
total	500	1209250.0	3339040625.0
mean =x̅=Σf*M/Σf=	2418.500
sample Var s2=(ΣfM2-ΣfM2/n)/(n-1)=(3339040625-(1209250)^2/500))/499=			830600.2004
Std deviation s=		√s2 =√830600.2004 =	911.3727