Assumed Mean Method | Statistics | Assessment | English | Grade 10

For finding the mean of grouped data using the Assumed Mean Method the formula is Mean = a + [(∑f¡d¡) / ∑f¡] , d¡ is ________

• d¡ = f¡x¡ ─ a

• d¡ = x¡ ─ f¡

• d¡ = x¡ ─ a

Answer explanation

When we calculate mean of a grouped data, by the assumed mean method , the deviation 'd' is calculated by subtracting the assumed mean from the class mark. Hence the correct answer is option 3.

If the mean of the following frequency distribution is 57.6 and the sum of the observations is 50. Find the missing frequencies x and y. Class: 0-20 20-40 40-60 60-80 80-100 100-120 Freq.(f): 7 x 12 y 8 5

• x = 8, y = 10

• x = 10, y = 8

• x = 7, y = 11

Answer explanation

With reference to the attached table, ∑f¡ = 50= 32 + x + y x + y = 50 ─ 32 = 18 x+y=18 -------- (1) ∑ f¡d¡ = 340 ─20 x +20 y ∑ f¡d¡ / ∑f¡ = (340 ─20 x +20 y) / 50 = (34 - 2x +2y) / 5 Mean = 57.6 = a + [(34 - 2x +2y) / 5] 50 + [(34 ─2 x +2 y)/5] = 57.6 34 ─2 x +2 y = (57.6 - 50 )x 5 = 38 - 2x +2y = 38 - 34 =4 - x +y = 2 -------------- (2) Adding Eq. 1 and 2, 2y = 20 y = 10 Substituting value of y in Eq. 1,we get, x +10=18 x = 8 Hence, the correct answer is Option1

The marks in English of 50 students in Class X are given below. Find the mean of the marks obtained by the students in English, using the Assumed Mean method. Marks: 0 - 20 20 - 40 40-60 60-80 80-100 Freq.(f) or No. of Students: 5 10 15 15 5

• 50

• 52

• 80

Answer explanation

a = 50 ∑ f¡ = 50 ∑ f¡d¡ = 100 ∑ f¡d¡/∑ f¡ = 100/50 = 2 Mean = a + [∑ f¡d¡/∑ f¡] = 50 + 2 = 52 Hence, the correct answer is Option 2

The daily sale of 100 vegetable vendors is given in the following table. Find the mean of the sale by assumed mean method. Daily Sales (Rs): 1000-1500 1500-2000 2000-2500 2500-3000 No. of vendors : 15 20 35 30

• 2150

• 2250

• 2350

Answer explanation

a = 2250 ∑ f¡ = 100 ∑ f¡d¡ = ─ 10000 ∑ f¡d¡/∑ f¡ = ─ 10000/100 = ─ 100 Mean = a + ∑ f¡d¡/∑ f¡ = 2250 ─ 100 = 2150 Hence, the correct answer is Option 1