Question

Here is your (discrete) data 108 83 93 118 88 58 88 65 113 103 119 101 91 92 95 86 85 73 94 61 121 71 116 78 114 119 118 131 41 70 68 98 202 155 41 93 78 96 93 110 100 61 98 39 94 86 108 75 64 50 90 123 109 69 118 85 102 86 153 109 89 134 80 104 96 129 48 81 102 31 97 130 98 105 109 77 113 62 80 99 First, sort the data. Second, build a GFDT for this table with a classwidth of 15. (Be sure your lower class limits are multiples of the classwidth.) Next, answer the following questions about the data set: 1. What is the first lower class limit in your GFDT (given the classwidth of 15)? 2. Express the fourth class as a closed interval, i.e., [ a , b ] [a,b] . 3. Give the frequency for the fourth class. 4. What is the (arithmetic) mean for this data (report accurate to one decimal place)? 5. Using the GFDT, what is the mode for this data set (Hint: Create the Frequency table first, then determine the class with the highst frequency. The mode is the middle of that class, that is (Upper class limit+Lower class Limit)/2)? Another hint ([90,104]is the class with the highest frequency) 6. What is the standard deviation for this data (report accurate to two decimal places)? 7. What is the five number summary for this data set (separate numbers with a comma)? 8. What is the IQR? 9. What usual score (i.e., non-outlier) has the largest positive z-score less than z = 2 z=2 ? (Give the data value, not the z-score.) 10. Give the fences---this would suggest something other than z-scores---for the mild outliers at the maximal end of the data set; report as a closed interval, i.e., [upper mild fence, upper extreme fence](Hint: UMF=Q3+1.5*IQR, UEF=Q3+3*IQR)

Answer 1

Sorting in ascending order

Frequency distribution

(30,45] (45,60] (60,75 (75,90] (90,105 4 2 11 4 16 22 15 6 (105.120] (120,1351 8 6 7 (135,150] (150,1651 10 (165,180] 11 (180,195] 12 (195.210] sor() Error: could not find function "sor" sort(x) [1] 88 [34] 113 114 [67] 118 118 118 119 119 121 123 129 130 131 134 153 155 202 1 31 39 41 89 91 70 71 73 75 77 78 78 80 80 81 41 48 50 58 61 62 61 64 65 85 85 86 86 86 68 69 88 98 98 98 99 100 101 102 102 103 104 105 108 108 109 109 109 110 113 93 92 93 93 G4 94 95 96 96 97 90 116 lenovo

1)The first lower vlasc limit is 30

0 (165,180] 11 (180,195] 0 12 (195,210] 1 fr=cut x,br,right=FALSE) gf dt > fr Freq (30,45] (45,60] (60,75] (75,90] (90,105] 6 4 3 11 4 16 22 (105,120] 7 15 (120,135] 8 6 (135,150] (150,165] 10 (165,1801 11 (180,1951 12 (195, 2101 5 wNH

2)The closed form of fourth class interval is [75,90]

3)The frequency for fourth class is 16

4G LTE+ 23% 21:06 X -A X - 127.5 Frequency (f Mid value (x) (2) f.d2 (5) (2) x (4) (6) = (5) x (4)| (7) Class f.d cf h 15 (3) (1) A 127.5, h = 15 (4) -24 30 45 4 37.5 6 144 4 45-60 5 75 3 52.5 -15 67.5 -4 -44 60-75 11 176 18 16 82.5 3 -48 144 34 75-90 90 105 22 97.5 -2 -44 88 56 105- 120 15 -1 -15 15 71 112.5 120 135 127.5=A 6 10 77 135-150 142.5 77 1 150 165 157.5 2 2 8 79 165- 180 172.5 3 0 79 180 195 187.5 4 0 79 195-210 202.5 5 1 25 80 n = 80 = 675 --==-

4G LTE+ 23% 21:06 Σfd h Mean X A + n 181 15 127.5 80 - 2.2625 15 127.5+ 127.5 33.9375 93.5625

4G LTE+ 23% 21:05 To find Mode Class Here, maximum frequency is 22 . The mode class is 90 - 105. . L = lower boundary point of mode class = 90 .f1 = frequency of the mode class = 22 frequency of the preceding class fo 16 - f2 frequency of the succedding class = 15 .c = class length of mode class = 15. f1-fo Z = L + C 2 1-fo-f2 22 - 16 - 15 90+ 2- 22 - 16 - 15 15 90+ 13 90 6.9231 96.9231

7)The five number summary is given as 31,76.875,94.09,109,202

4G LTE+ 22% 21:09 zomate Σf a2Σra) 2 n Population Standard deviation o h n (-181) 675 80 15 80 /675 - 409.5125 15 80 /265.4875 . 15 80 3.3186 15 = 1.8217 15 =27.3255

8)IQR = Q₃ --Q₁ =109-76.875 = 32.125

10)UMF=109 +3(32.125) = 205.375

ULF = 109 - 3(32.125) = -12.625