Statistics
Mean and Standard Deviation
Grade 11
Question:
<p>Let \(x_1, x_2, \ldots, x_n\) be \(n\) observations. Let \(w_i = lx_i + k\) for \(i = 1, 2, \ldots, n\), where \(l\) and \(k\) are constants. The mean of \(x_i\)'s is 48 and their standard deviation is 12. Also, the mean of \(w_i\)'s is 55 and standard deviation of \(w_i\)'s is 15. The values of \(l\) and \(k\) should be</p>
<p>\(l=2.5, k=5\)</p>
<p>\(l=-1.25, k=5\)</p>
<p>\(l=2.5, k=-5\)</p>
<p>\(l=1.25, k=-5\)</p>
Step-by-Step Solution
Key Concept: When data is transformed linearly as w_i = lx_i + k, the mean transforms as Mean(w) = l·Mean(x) + k, while standard deviation transforms as SD(w) = |l|·SD(x) (the additive constant k doesn't affect spread). Use both equations simultaneously to solve for l and k.
<p><strong>Step 1:</strong> Apply the transformation property for mean.</p><p>Mean of w_i = l·Mean(x_i) + k</p><p>55 = l(48) + k ... (equation 1)</p><p><strong>Step 2:</strong> Apply the transformation property for standard deviation.</p><p>SD of w_i = |l|·SD(x_i)</p><p>15 = |l|(12)</p><p>|l| = 15/12 = 5/4</p><p>Therefore, l = 5/4 or l = -5/4</p><p><strong>Step 3:</strong> Substitute l = 5/4 into equation 1.</p><p>55 = (5/4)(48) + k</p><p>55 = 60 + k</p><p>k = -5</p><p><strong>Step 4:</strong> Verify with l = -5/4.</p><p>55 = (-5/4)(48) + k</p><p>55 = -60 + k</p><p>k = 115</p><p><strong>Step 5:</strong> Since the question asks for 'the values' (typically implying the primary/positive solution in standard form), and l = 5/4, k = -5 is the conventional answer.</p><p>∴ Answer: <strong>l = 5/4 (or 1.25), k = -5</strong> [Option D]</p>
Correct Answer: D