SECTION 2.4 Measures of Variation 91 When a frequency distribution has classes, you can estimate the sample mean and the sample standard deviation by using the midpoint of each class. Using Midpoints of Classes The figure below shows the price ranges (in thousands of dollars) and corresponding number of homes recently listed for sale in a mid-sized U.S. city. Make a frequency distribution for the data. Then use the table to estimate the sample mean and the sample standard deviation of the data set. (Adapted from National Association of Realtors and Move, Inc.) 400–499 8 300–399 22 200–299 39 100–199 129 500 or more 19 less than 100 173 Sales Prices of Homes (in thousands of dollars) in a mid-sized city SOLUTION Begin by using a frequency distribution to organize the data. Because the class of $500 thousand or more is open-ended, you must choose a value to represent the midpoint, such as 599.5 thousand. Sales price (in thousands of dollars) x f xf 0 –99 49.5 173 8563.5 100 –199 149.5 129 19,285.5 200 –299 249.5 39 9730.5 300 –399 349.5 22 7689.0 400 – 499 449.5 8 3596.0 500 + 599.5 19 11,390.5 Σ = 390 Σ = 60,255.0 x − x (x − x)2 (x − x)2 f -105 11,025 1,907,325 -5 25 3225 95 9025 351,975 195 38,025 836,550 295 87,025 696,200 445 198,025 3,762,475 Σ = 7,557,750 x = Σxf n = 60,255.0 390 = 154.5 Sample mean Use the sum of squares to find the sample standard deviation. s = BΣ1x - x22f n - 1 = A7,557,750 389 ≈ 139.4 Sample standard deviation So, an estimate for the sample mean is $154,500 per year, and an estimate for the sample standard deviation is $139,400. TRY IT YOURSELF 9 In the frequency distribution in Example 9, 599.5 was chosen as the midpoint for the class of $500 or more. How does the sample mean and standard deviation change when the midpoint of this class is 650? Answer: Page A37 EXAMPLE 9
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