MAT 109 Assignment 3 Chapter 3

MAT 109 Assignment 3 Chapter 3 - Instructor MichaelDempsey...

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MAT 109 Assignment 3 Chapter 3 “Statistics for Describing Data” Instructor: Michael Dempsey Name Brineydi Peralta Date Given:9/20/10 Date Due: 9/26/10 Directions: Answer the questions and solve the problems in the spaces provided.  Add space, if you need  it.  Point values are shown in parentheses.  [Total Value: 50 points] 1. (5) Section 3-2, #8, page 94. [Show your procedure for the following measures of center.] (a) Mean: (714+751+664+789+818+779+698+836+753+834+693+802)/12= 9131/12= 760.9167= 760.92  (if rounded) (b) Finding median: 664, 693, 698, 714, 751, 753, 779, 789, 802, 818, 834, 836      Median= (753+779)/2= 1532/2= 766 (c) Mode: no mode because no value repeats itself in the data (d) Midrange: (836+664)/2= 1500/2= 750 (e) Answer the question.  Even though the reported mean FICO score was 678, the Sample FICO scores appear to be inconsistent  with the mean given, if we were to re-calculate the mean for the given sample scored the mean would be  760.92  (if rounded)  2. (5) Section 3-2, #18, page 95. [Show your procedure for the following measures of center.] (a) Mean: (7.2+7.1+7.4+7.9+6.5+7.2+8.2+9.3)/8= 60.8/8=7.6 (b) Median: 6.5, 7.1, 7.2, 7.2, 7.4, 7.9, 8.2, 9.3  Median = (7.2+7.4)/2= 14.6/2= 7.3 (c) Mode: 7.2 (d) Midrange: (9.3+6.5)/2= 15.8/2= 7.9 (e) Answer the questions.
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3. (4) Section 3-2, #30, page 97. [Show your procedure for the grouped mean; compare it to the  ungrouped mean, given in the directions to this problem.] Sum of Midpoints: (64.5*12)+(74.5*14)+(84.5*11)+(94.5*1)+(104.5*1)+(114.5*0)+(124.5*1)= 3070 Sum of Frequency: 12+14+11+1+1+0+1= 40 Mean:  3070/40= 76.75 The difference between the given mean 76.3 and the mean I found (76.75) of the frequency distribution is 
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This note was uploaded on 10/08/2010 for the course MAT 109 taught by Professor Michael during the Spring '10 term at Peirce.

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MAT 109 Assignment 3 Chapter 3 - Instructor MichaelDempsey...

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