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.
| 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 | |
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