Which of the following is a synonym for 'جگنو'?

English
Which of the following is a synonym for 'جگنو'?

ScanSolve AI Answer

Answer

51.31

Mean (اوسط)

Step 1: Identify midpoints xix_i for each class interval.

Classxi01051020152030253040354050455060556070657080758090859010095\begin{array}{c|c} Class & x_i \\ \hline 0{-}10 & 5 \\ 10{-}20 & 15 \\ 20{-}30 & 25 \\ 30{-}40 & 35 \\ 40{-}50 & 45 \\ 50{-}60 & 55 \\ 60{-}70 & 65 \\ 70{-}80 & 75 \\ 80{-}90 & 85 \\ 90{-}100 & 95 \\ \end{array}

Step 2: Compute fixif_i x_i for each class.

5×5=25,7×15=105,8×25=200,12×35=420,15×45=675,18×55=990,14×65=910,11×75=825,10×85=850,3×95=285.\begin{align*} 5 \times 5 &= 25, \\ 7 \times 15 &= 105, \\ 8 \times 25 &= 200, \\ 12 \times 35 &= 420, \\ 15 \times 45 &= 675, \\ 18 \times 55 &= 990, \\ 14 \times 65 &= 910, \\ 11 \times 75 &= 825, \\ 10 \times 85 &= 850, \\ 3 \times 95 &= 285. \end{align*}

Step 3: Sum the fixif_i x_i values.

25+105=130,130+200=330,330+420=750,750+675=1425,1425+990=2415,2415+910=3325,3325+825=4150,4150+850=5000,5000+285=5285.\begin{align*} 25 + 105 &= 130, \\ 130 + 200 &= 330, \\ 330 + 420 &= 750, \\ 750 + 675 &= 1425, \\ 1425 + 990 &= 2415, \\ 2415 + 910 &= 3325, \\ 3325 + 825 &= 4150, \\ 4150 + 850 &= 5000, \\ 5000 + 285 &= 5285. \end{align*}

fixi=5285\sum f_i x_i = 5285, N=103N = 103.

Step 4: Compute mean.

xˉ=5285103.\bar{x} = \frac{5285}{103}.

103×51=5253103 \times 51 = 5253, 52855253=325285 - 5253 = 32, so xˉ=51+3210351.31\bar{x} = 51 + \frac{32}{103} \approx 51.31.

51.31

Median (وسطی)

Step 1: N=103N = 103, median position =N2=51.5=\frac{N}{2} = 51.5.

Step 2: Locate median class from cumulative frequencies. Cumulative frequency before 506050{-}60 is 4747, reaches 6565 in 506050{-}60. Median class is 506050{-}60.

Step 3: Apply median formula

Median=L+(N2Cf)h,Median = L + \left( \frac{\frac{N}{2} - C}{f} \right) h,

where L=50L=50, C=47C=47, f=18f=18, h=10h=10.

51.54718=4.518=0.25,0.25×10=2.5.\frac{51.5 - 47}{18} = \frac{4.5}{18} = 0.25, \quad 0.25 \times 10 = 2.5.

Median =50+2.5=52.5= 50 + 2.5 = 52.5.

52.5

Mode (وضع عام)

Step 1: Modal class is 506050{-}60 (highest f=18f=18).

Step 2: Apply mode formula

Mode=L+(f1f02f1f0f2)h,Mode = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h,

where L=50L=50, f1=18f_1=18, f0=15f_0=15, f2=14f_2=14, h=10h=10.

18152×181514=33629=370.4286,\frac{18 - 15}{2 \times 18 - 15 - 14} = \frac{3}{36 - 29} = \frac{3}{7} \approx 0.4286, 0.4286×104.29.0.4286 \times 10 \approx 4.29.

Mode =50+4.29=54.29= 50 + 4.29 = 54.29.

54.29

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Quick Answer

Mean (اوسط) Step 1: Identify midpoints x_i for each class interval.

Which of the following is a synonym for 'جگنو'?
English

This English question involves literary analysis, grammar, or writing skills. The detailed response below provides a well-structured answer with supporting evidence and clear explanations.

ScanSolve AI Answer
Mean (اوسط) Step 1: Identify midpoints x_i for each class interval. c|c Class & x_i \\ 0-10 & 5 \\ 10-20 & 15 \\ 20-30 & 25 \\ 30-40 & 35 \\ 40-50 & 45 \\ 50-60 & 55 \\ 60-70 & 65 \\ 70-80 & 75 \\ 80-90 & 85 \\ 90-100 & 95 \\ Step 2: Compute f_i x_i for each class. 5 × 5 &= 25, \\ 7 × 15 &= 105, \\ 8 × 25 &= 200, \\ 12 × 35 &= 420, \\ 15 × 45 &= 675, \\ 18 × 55 &= 990, \\ 14 × 65 &= 910, \\ 11 × 75 &= 825, \\ 10 × 85 &= 850, \\ 3 × 95 &= 285. Step 3: Sum the f_i x_i values. 25 + 105 &= 130, \\ 130 + 200 &= 330, \\ 330 + 420 &= 750, \\ 750 + 675 &= 1425, \\ 1425 + 990 &= 2415, \\ 2415 + 910 &= 3325, \\ 3325 + 825 &= 4150, \\ 4150 + 850 &= 5000, \\ 5000 + 285 &= 5285. f_i x_i = 5285, N = 103. Step 4: Compute mean. x = (5285)/(103). 103 × 51 = 5253, 5285 - 5253 = 32, so x = 51 + (32)/(103) ≈ 51.31. 51.31 Median (وسطی) Step 1: N = 103, median position =(N)/(2) = 51.5. Step 2: Locate median class from cumulative frequencies. Cumulative frequency before 50-60 is 47, reaches 65 in 50-60. Median class is 50-60. Step 3: Apply median formula Median = L + ( (N)/(2) - Cf ) h, where L=50, C=47, f=18, h=10. (51.5 - 47)/(18) = (4.5)/(18) = 0.25, 0.25 × 10 = 2.5. Median = 50 + 2.5 = 52.5. 52.5 Mode (وضع عام) Step 1: Modal class is 50-60 (highest f=18). Step 2: Apply mode formula Mode = L + ( (f_1 - f_0)/(2f_1 - f_0 - f_2) ) h, where L=50, f_1=18, f_0=15, f_2=14, h=10. (18 - 15)/(2 × 18 - 15 - 14) = (3)/(36 - 29) = (3)/(7) ≈ 0.4286, 0.4286 × 10 ≈ 4.29. Mode = 50 + 4.29 = 54.29. 54.29