Question

please help with letters G & H and number 3 letters A-C

2. The incomplete frequency distribution table given below represents the score of an exam in a math class. (a) (5 points) Complete the frequency distribution table below Class Class Class Cumulative Percentage Limits Boundaries Midpoint Frequency Frequency Frequency 154.5, 55 — 67 61 4 167.5 4 Class 110,000 122.50% 9 13 40.00% 68 - 80 180.5 174 81 - 93 180.s, 187 16 29 23.5 93.5 94-106 40 1106.5 (b) (2 points) Find the class width for this frequency distribution table. 100 11 27.50 13 (b) Frequency (c) (2 points) How many students took this exam? 4+9+19+11=40 40 (d) (3 points) Draw the histogram. Clearly label and mark your graph. 18 16 mid Point 19 61 12 74 10 87 9 16 8 100 9 o 60 80 100 uppen FreQuericy 45 ilmit 0 4 35 70 90 (e) (3 points) Draw the ogive. Clearly label and mark your graph. cumulative 0 4 13 29 40 O 67 80 30 9 25 20 is lo 120 93 16 106 111 S OO 20 40 60 80 100 Study Guide 3 Total Points: 50 Page 3 of 4

Answer 1

Solution:

g)Pie chart

Frequency	Angle of sector=Frequency of data/Total frequency*3600
4	(4/40)*360o=36o
9	(9/40)*360o=81o
16	(16/40)*360o=144o
11	(11/40)*360o=99o
Total frequency=40	360o

The Pie chart for the above data is given below.

Pie Chart 36 99 81 144 .1.2.3.4

h)

Class limits	Frequency	Percentage
55-67	4	(4/40)*100=10%
68-80	9	(9/40)*100=22.5%
81-93	16	(16/40)*100=40%
94-106	11	(11/40)*100=27.5%
	40

(22.5+40)=62.5%=63% (round to nearest integer) of this data lie between 68 and 93.

3 a)Given range=48 and number of class=4

Class width=Range/Number of class=48/4=12

b)Given range=48 and number of class=5

Class width=Range/Number of class=48/5=9.6=10

c)Given range=48 and number of class=6

Class width=Range/Number of class=48/6=8