RS Aggarwal Solutions for Class 10 Maths Chapter 18 Mean ...

[Pages:37]RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

Exercise 18A

Question 1: Solution: We know that, Mean = Sum of the given observations/ Total number of observations Sum of the given observations = x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = x + x + 2 + x + 4 + x + 6 + x + 8 = 5x + 20 Total number of observations = 5 Therefore, Mean = (5x + 20)/5 Also, Mean = 11 (given) 11 = (5x + 20)/5 55 = 5x + 20 or x = 7

Question 2: Solution: Mean of 25 observations = 27 (given) Total observations = 25 Mean = Sum of the given observations/ Total number of observations 27 = (sum of 25 observations)/25 sum of 25 observations = 27 x 25 = 675 When each observation is decreased by 7, then New Sum is 675 ? 25 x 7 = 500 New Mean = 500/25= 20

Question 3: Solution:

Now, Mean = fixi / fi = 426/80 = 5.325

Question 4:

Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

Mean = fixi/ fi = 1150/40 = 28.75

Question 5: Solution:

Mean = fixi/ fi = 1980/40 = 49.5

Question 6: Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

Mean = fixi/fi = 13200/40 = 264

Question 7: Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

Mean = fixi/ fi = 8244/80 = 103.05 Used direct method as it is easy to calculate. Question 8: Solution:

Mean=fixi/fi Mean = 24 ( given) 24 = (270+25p)/(12+p) 24(12+p) = 270 + 25p or p = 18 Question 9: Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

Mean=fixi/ fi Mean = 18 (given) 18 = (752 + 20f)/(44+f) 792 + 18f = 752 + 20f or f = 20 Question 10: Solution:

Mean=fixi/ fi Mean = 54 (given) 54 = (2370 + 30p)/(39+p) 1053 + 27p = 1185 + 15p or p = 11

Exercise 18B

Question 1: Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

Where cf = cumulative frequency Median = I+{h?(N2?cf)f} Here: N = 140 N/2 = 70 cf > 70 is 140 Median class = 45-60 So, l = 45, h = 15, f = 50 and cf = cf of preceding class i.e. 65 Substitute all the value in the above formula, we get Median = 45 + {15 x (70-65)/50} = 45 + 1.5 = 46.5 Therefore, median age of diabetic patients is 46.5 years.

Question 2: Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

Where cf = cumulative frequency Median=I+{h?(N2?cf)f} Here: N = 50 N/2 = 25 Here cumulative frequency is 25 Median class = 21-28 So, l = 21, h =7, f = 11 and cf = cf of preceding class i.e. 14 Substitute all the value in the above formula, we get Median =21 + {7x (25-14)/11} = 28 Question 3: Solution:

Where cf = cumulative frequency Median=I+{h?(N2?cf)f} Here: N = 150 N/2 = 75 cf just greater than 75 is 120 Median class = 200-300 So, l = 200, h =100, f = 48 and

RS Aggarwal Solutions for Class 10 Maths Chapter 18Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph

and Ogive

cf = cf of preceding class i.e. 72 Substitute all the value in the above formula, we get Median =200+ {100 x (75-72)/48} = 200 + 6.25 = 206.25 Median of daily wages is Rs. 206.25. Question 4: Solution:

Where cf = cumulative frequency Median=I+{h?(N2?cf)f} Here: N = 49 N/2 = 24.5 cf just greater than 24.5 is 26 Median class = 15-20 So, l = 15, h = 5, f = 15 and cf = cf of preceding class i.e. 11 Substitute all the value in the above formula, we get Median = 15+ {5 x (24.5-11)/15} = 15 + 4.5

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