Q.If \(\sum \limits_{i=1}^n (x_i - 7) = -8\) and \(\sum \limits_{i=1}^n (x_i + 3) = 72\), then what are the values of \(\bar{x}\) (the mean of \(x_i\)) and \(n\) (the number of terms)? (a) \(\bar{x}=5, n=8\) (b) \(\bar{x}=6, n=8\) (c) \(\bar{x}=4, n=7\) (d) \(\bar{x}=8, n=6\)
Answer: B
\(\sum \limits_{i=1}^n (x_i - 7) = -8\) Or, \(\sum \limits_{i=1}^n x_i - \sum \limits_{i=1}^n 7 = -8\) Or, \(\sum \limits_{i=1}^n x_i - 7n = -8\) — (i) Again, \(\sum \limits_{i=1}^n (x_i + 3) = 72\) Or, \(\sum \limits_{i=1}^n x_i + \sum \limits_{i=1}^n 3 = 72\) Or, \(\sum \limits_{i=1}^n x_i + 3n = 72\) — (ii) Subtracting (i) from (ii), we get: \(10n = 80\) Or, \(n = 8\) Substituting \(n = 8\) into equation (i), we get: \(\sum \limits_{i=1}^n x_i - 7 \times 8 = -8\) Or, \(\sum \limits_{i=1}^n x_i = -8 + 56\) Or, \(\sum \limits_{i=1}^n x_i = 48\) \(\therefore \bar{x} = \cfrac{\sum \limits_{i=1}^n x_i}{n} = \cfrac{48}{8} = 6\) \(\therefore \bar{x} = 6, n = 8\)
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