Question 1: Find the mean of the following distribution:Class 0-10 10-20 20-30 30-40 40-50 Frequency 8 12 10 11 9
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Solution: Step 1: Class marks ($x_i$): 5, 15, 25, 35, 45.Step 2: $f_i x_i$: $8(5)=40$, $12(15)=180$, $10(25)=250$, $11(35)=385$, $9(45)=405$.Step 3: $\sum f_i = 50$, $\sum f_i x_i = 1260$.Step 4: Mean $\bar{x} = \frac{1260}{50} = 25.2$.Answer: 25.2.
Question 2: Find the mode of the following data:Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 10 35 52 61 38 29
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Solution: Step 1: Max frequency is 61. Modal class is 60-80.Step 2: $l=60, f_1=61, f_0=52, f_2=38, h=20$.Step 3: Mode $= 60 + (\frac{61-52}{2(61)-52-38}) \times 20 = 60 + (\frac{9}{122-90}) \times 20 = 60 + \frac{9}{32} \times 20$.Step 4: Mode $= 60 + \frac{180}{32} = 60 + 5.625 = 65.625$.Answer: 65.625.
Question 3: The mean of the following distribution is 18. Find the missing frequency $f$.Class 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Frequency 3 6 9 13 $f$ 5 4
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Solution: Step 1: $x_i$: 12, 14, 16, 18, 20, 22, 24.Step 2: $\sum f_i = 40+f$. $\sum f_i x_i = 36+84+144+234+20f+110+96 = 704+20f$.Step 3: Mean $18 = \frac{704+20f}{40+f} \Rightarrow 720+18f = 704+20f \Rightarrow 2f = 16 \Rightarrow f=8$.Answer: $f=8$.
Question 4: Find the median of the following data:Marks 0-10 10-20 20-30 30-40 40-50 No. of Students 5 8 20 15 7
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Solution: Step 1: $N=55, N/2 = 27.5$. CF: 5, 13, 33, 48, 55.Step 2: Median class is 20-30 (CF 33 > 27.5). $l=20, cf=13, f=20, h=10$.Step 3: Median $= 20 + (\frac{27.5-13}{20}) \times 10 = 20 + \frac{14.5}{2} = 20 + 7.25 = 27.25$.Answer: 27.25.
Question 5: If the median of the distribution is 28.5, find the values of $x$ and $y$. Total frequency is 60.Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 5 $x$ 20 15 $y$ 5
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Solution: Step 1: $45+x+y=60 \Rightarrow x+y=15$. Median 28.5 lies in 20-30.Step 2: $l=20, h=10, f=20, cf=5+x, N/2=30$.Step 3: $28.5 = 20 + (\frac{30-(5+x)}{20}) \times 10 \Rightarrow 8.5 = \frac{25-x}{2} \Rightarrow 17 = 25-x \Rightarrow x=8$.Step 4: $8+y=15 \Rightarrow y=7$.Answer: $x=8, y=7$.
Question 6: Calculate the mode of the following distribution:Class 10-25 25-40 40-55 55-70 70-85 85-100 Frequency 2 3 7 6 6 6
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Solution: Step 1: Max freq is 7. Modal class 40-55. $l=40, f_1=7, f_0=3, f_2=6, h=15$.Step 2: Mode $= 40 + (\frac{7-3}{14-3-6}) \times 15 = 40 + \frac{4}{5} \times 15 = 40 + 12 = 52$.Answer: 52.
Question 7: The mean of a distribution is 50 and the median is 52. Find the mode.
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Solution: Step 1: Empirical Formula: Mode $= 3(\text{Median}) - 2(\text{Mean})$.Step 2: Mode $= 3(52) - 2(50) = 156 - 100 = 56$.Answer: 56.
Question 8: Find the mean of the following data using the Assumed Mean Method:Class 0-20 20-40 40-60 60-80 80-100 Frequency 15 18 21 29 17
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Solution: Step 1: $x_i$: 10, 30, 50, 70, 90. Let $a=50$.Step 2: $d_i = x_i - 50$: -40, -20, 0, 20, 40.Step 3: $f_i d_i$: -600, -360, 0, 580, 680. $\sum f_i d_i = 300$. $\sum f_i = 100$.Step 4: Mean $= 50 + \frac{300}{100} = 53$.Answer: 53.
Question 9: Convert the following cumulative frequency distribution to a frequency distribution:Marks Below 10 Below 20 Below 30 Below 40 Below 50 No. of Students 5 12 25 38 50
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Solution: Step 1: Classes: 0-10, 10-20, 20-30, 30-40, 40-50.Step 2: Frequencies: 5, $12-5=7$, $25-12=13$, $38-25=13$, $50-38=12$.Answer: Frequencies are 5, 7, 13, 13, 12.
Question 10: Find the sum of the lower limits of the median class and modal class of the following distribution:Class 0-5 5-10 10-15 15-20 20-25 Frequency 10 15 12 20 9
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Solution: Step 1: Modal class (max freq 20) is 15-20. Lower limit = 15.Step 2: $N=66, N/2=33$. CF: 10, 25, 37... Median class is 10-15. Lower limit = 10.Step 3: Sum $= 15 + 10 = 25$.Answer: 25.