ch5_processing_of_large_data_sets.studentview

Ch5_processing_of_large_data_sets.studentview

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Unformatted text preview: iddle value is called the class mark, . mark Of course, a similar idea applies to all the other classes of the data set. Table 1 becomes…. Table Class Frequency 10-19.9 1 20-29.9 3 30-39.9 4 40-49.9 6 50-59.9 7 60-69.9 4 total 25 Table 6 Table i Class Class Class Class intervals Boundaries intervals Boundaries freq xi 9.95-19.95 Class Class mark mark fi 1 10-19.9 14.95 1 2 20-29.9 19.95-29.95 24.95 3 3 30-39.9 29.95-39.95 34.95 4 4 40-49.9 39.95-49.95 44.95 6 5 50-59.9 49.95-59.95 54.95 7 6 60-69.9 59.95-69.95 64.95 4 X XXXX 25 total XXX Note that the 3rd class mark, x3 , Note can be computed in 2 ways: can Using class limits: Using 30 + 39.9 x3 = = 34.95 2 Using class boundaries: Using 29.95 + 39.95 = 34.95 2 To compute µ and σ for x3 = the grouped data, we generalize the ideas of 2A and 3C, in a natural way, incorporating the frequency notation: notation: ∑x f ; µ= ii N σ= ∑ f (x − µ ) i i N ∑x 2 2 ii = Note that in the above summations, N i f −µ 2 runs from 1 to k ( k = number of classes, = 6, here), not from 1 to N. k ( N = ∑ fi i =1 , =25, here) =25, Example 1: Compute µ Example Compute Solution: & σ for the data of Table 6. i 1 2 σ= 2 ii N = 13.8333 f 1 14.95 223.5025 24.95 3 74.85 1867.5075 34.95 4 139.80 4886.0100 44.95 6 269.70 12123.0150 54.95 7 384.65 21136.5175 64.95 4 259.80 16874.0100 total ∑x 14.95 6 N xi2 f i 5 = xi f i 4 ii fi 3 ∑x f ∴µ = xi XXX 25 1143.75 57110.5625 1143.75 = 45.75; 25 − µ2 = 57110.5625 − 45.752 25 µ=∑ N fx Note: The formula Note motivated CASIO to call its calculators motivated (with built in statistical programs) the fx- series, such as fx-50FH. & σ for the data of Table 6, using the SD-Mode on the calculator Casio fx-50FH using Solution: ON MODE COMP MODE SD Example 2: Compute µ Example xi fi 14.95 1 24.95 3 34.95 4 44.95 6 54.95 7 64.95 4 XXX 25 14.95 24.95 34.95 34.95 64.95 DT SHIFT ; SHIFT ; SHIFT ; SHIFT SHIFT SHIFT SHIFT SHIFT SHIFT SHIFT S-SUM S-VAR S-VAR S-SUM S-SUM 3 4 4 DT DT,…………, DT n EXE x EXE xσ EXE n ∑x EXE ∑ x EXE 2 25 = N 45.75 = µ 13.8333 = σ 1143.75 = ∑ x i f i 57110.5625 = ∑ x i f i 2 ∑x f ∴µ = i i N σ= ∑x 2 ii N = 13.8333 f 1143.75 = = 45.75; 25 57110.5625 −µ = − 45.752 25 2 x Question: How much setting out MUST I show in the final Question How exam? exam? Answer: You must: s You x µ σ (1) Give a correct symbol ( or , or ) depending whether the data “population data” or “sample data” (2) Give a correct formula ( formula sheet attached to exam (2) paper ) paper (3) show a correct substitution (use SHIFT S-SUM keys) (4) give a correct answer from the calculator. E: Finding the median for grouped data E: In Ch2B, we distinguished carefully between the cases of odd N In and even N in finding the median of ungrouped data (See Note ungrouped 1 there) there) Now, when data are grouped, as in Table 2 above, we no longer Now, grouped as need to make such a distinction. The median is simply at the N − th position...
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This document was uploaded on 09/28/2013.

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