Question

11. Consider the following histogram of IQ scores of 7th graders. How many total students were included in the study?: * IQ's of 7th Graders 100 Frequency 60 0 20 90-99 120-129 100-109 110-119 Score Interval a. 240 Ob. 200 Oc. 180 Od. 300

Answer 1

(11) The Iq Scores of 7th graders are given to us. Score Interval Frequency (f) 90-99 60 100 - 109 80 60 110 - 119 120-129 40. If=240. Thus the total number of students = Sum of frequency Sf 240 Thus the answer is (a) 240 (12) kle have to find the sample Standard deviation, for 4,5,5, 6, 7, 7, 9, 9, 10, 12 The Size of the given Sample is, n = 10. Now we will calculate the mean of the sample. Sample mean, X= is xi hi-1 Exi 10

(4+5+5+6+7+7+9+9+10+12) To To (74) 10 7.4 Thus X = T. 4 M3 (x; - x2 Sample Standard deviation, S = n-1 x x =-7.4 (x-x)? 4 -3.4 11.56 5 -2.4 5.76 5.76 5 -2.4 1.96 6 -1.4 7 -0.4 0.16 T -0.4 0.16 1.6 2.56 9 2.56 6.76 lo 2.6 12 4.6 21.16 Zx-x)² = 58.4 S = Ž* (X;-x7² 58.4 Thus, 2.5473 = ワー)

Thus the samople standard deviation, S.2.5473 Thus the answer is (d) 2.5473 (13) Table for the number of animal tracks Collected in a park is given to us. Oppssum 15 Deer 32 Squirrel 97 Raccoon 32 41 Dog Total Animals= 217 We have to find the probability that a randomly Selected track belongs to neithey adeey not a Squirrel. That is, we have to find, P(neither a deer nor a squirrel) = 1 - PC a deey oy a squivel ) (since PCA) = 1-PCA) PC'a randomly selected is a deer) = P(deer) bumber of deer Total number of Animals 32 217 0.14 75

PC a Yaodonly selected is a squirrel) - P(Squirrel) no of Squirrels Total no. of animals 97 217 0.4470 Thus P(arandomly selected is a deey oy Squirrel) -Pa deey oy Squirrel) PCadeer) + P(Squirrel) 0.1475 + 0.4470 0.5945 0.5945 Thus, PC a randomly selected is neither a deer noy a squirrel) = 1 - P(either a deey oyas quirrel) - 1 - P(a deey oy Squirrel) = 1 - 0.5945 Thus 0.4055 the probability that avandomly selected track belongs to neither deer por a squirrel. is 0.4055 So, the answer is (b) 0.4055 -

(14) we have to find the a box & Whiskey plot. lower quartile to construct The data is given to us, ist we will arrange the given data from lower to higher values. Opossum Deey Squirrel Raccoon bog 15 32. 91 32 41 The ordered set of numbers is 15 32 , 32, 41, 97 St we will find the median (22): Median is the exactly middle value of an Ordered set of number. 15 32 32 41 97 32 is the middle → Median = 32 Lowey Quartile C92): Consider the values left of the median, 15 32 It is : Now find the median of this set of numbers

Here the median is the middle of the set of Ordered number. 15 and 32 Here the median is between Then the median mean of 15 an 32 15 +32 2 47 2 = 23.5 Median of the left of the median (Q2) = 23.5 → Lower quartile Median of 15, 32 23.5 Thus, the lower quartile = 23.5 So the answer is (b) 23.5