Question

The following tegenstagram represents the core of a random sa of severth-grade dents Os are measured to the nearest whole number. The frequency of each class is wove each rectangle Use the higram to wwersporta) Pool ON osobo (a) How many students were sampled idents Delamine the class with The class width is (c) identify the classes and the frequencies. Choose the correct answer below OA 60-70, 1:70-80, 380-90, 12, 90-100, 45: 100-110, 12, 110-120, 41; 120-130, 33; 130-140, 7140-150, 4150-160,2 OB. 60-69, 1; 70-79, 30-0, 12, 30-0, 45100-109, 52, 110-110, 41: 120-129, 33, 130-139. 7: 140-149, 4150-159.2 OC 65, 1:75, 3, 5, 12, 55, 45: 105, 52: 115, 41: 125,33; 135, 7, 145,4 155.2 (d) Which class has the highest frequency OA 105 OB. 100-109 OC 90.99

Answer 1

1 3 12 45 41 33 7 2 From the given histogram, for better understanding the problem we forom this table containing the classes, and commesponding frequencies first - Class 60-69 70-79 80-89 90-99 100-109 110-11911204129 130-1391140-149 150 - 159 Frequency 52 4 Table 1. To obtaim total number of students sampled, we have to sum the observed frequencies with respect to each class, so here No. of Students = 1+3+12+45 +52+41+33+4+4+2 = 200 Answer: 200 studenta, (6) By class width, we mean the difference between upper and lower boundaries of a class interval. In this case, for class 60-69 the comesponding class boundary is, 59.5 - 69.5, for 70-79 the class boundary is, 69.5-79.5, and so on. .: Class-cwidth = 69.5-59.5 +10 --- Answer

B: 60-69:1, 40-49:2,80-89:12,90-99:45 2100-109:52, 110-119:41 100-109. So, corect answer (e) We see that from the table 1, lowest frequen y = 1, and 120729:33, 130-139:7 () From table 1,we can easily identify the correct answer... > 140-149:49 150–159:2 (d) We see that, from the table 1, highest frequency - 52, and the coresponding class is is: B.100-109, the corresponding class is 60-69. So, the corect answer is, A. 60-69. @) we want to find, pencentage of students with IQ at least 130. .. we have to considera from the class 130–139, uplo the class 150 159. calculation is shown below - Class with Frequency IQ score atleast 130 130-139 - 140-149 4 150-159 2 Total 7+4+2= 13 .: Required percentage =( 136 X100 (100) 7 6.59% Answer: 6.5% @) There is no student, who have an IQ of 164, because- D. No, because there are no bars on frequencies greater than an IQ of 160. -Answer