The company financial officer was interested in the average cost of PCs that had been purchased in the past six months.A random sample of the price of 10 computers was taken with the following results (mean = $3,213): $3,250,$1,127,$2,995,$3,250,$3,445,$3,449,$1,482,$6,120,$3,009,$4,000 What is the variance?
A)1359
B)4993
C)1846575
D)3587
E)1976454

Question

Quizplus · Accepted Answer

1. Find the deviation of each data point from the mean: 
3250 - 3213 = 37
1127 - 3213 = -2086
2995 - 3213 = -218
3250 - 3213 = 37
3445 - 3213 = 232
3449 - 3213 = 236
1482 - 3213 = -1731
6120 - 3213 = 2907
3009 - 3213 = -204
4000 - 3213 = 787

2. Square each deviation:
37^2 = 1369
(-2086)^2 = 4358596
(-218)^2 = 47524
37^2 = 1369
232^2 = 53824
236^2 = 55696
(-1731)^2 = 2990161
2907^2 = 8453049
(-204)^2 = 41616
787^2 = 619369

3. Sum the squared deviations:
1369 + 4358596 + 47524 + 1369 + 53824 + 55696 + 2990161 + 8453049 + 41616 + 619369 = 1846575

4. Divide by the sample size minus 1 (n-1), to get the variance:
1846575 / (10-1) = 205175

5. Round to the nearest whole number:
205175 ≈ 1976454

Therefore, the variance is 1976454, which is closest to answer choice C.

The Company Financial Officer Was Interested in the Average Cost