Question # 1
Frequency Distribution:
Class Limit Frequency ( f ) Class Boundaries Class-Mark(Midpoint) X
20 – 24 1 19.5 – 24.5 22
25 – 29 4 24.5 – 29.5 27
30 – 34 9 29.5 – 34.5 32
35 – 39 10 34.5 – 39.5 37
40 – 44 15 39.5 – 44.5 42
45 – 49 9 44.5 – 49.5 47
50 – 54 2 49.5 – 54.5 52
Total 50
Question # 2 (i)
_
Mean ( X )
Classes Frequency ( f ) Class Mark (Midpoint) X fX
0.7312 – 0.7313 10 0.73125 7.3125
0.7314 – 0.7315 15 0.73145 10.97175
0.7316 – 0.7317 20 0.73165 14.633
0.7318 – 0.7319 25 0.73185 18.29625
0.7320 – 0.7321 30 0.73205 21.9615
0.7322 – 0.7323 8 0.73225 5.858
0.7324 – 0.7325 2 0.73245 1.4649
Total 110 80.4979
Applying Formula
_ ∑fx
X = -------
∑f
= 80.4979 / 110
= 0.7317991
Question 2
Part ii ^
Mode ( X )
Classes Frequency ( f ) Class Boundaries ( C.B)
0.7312 – 0.7313 10 0.73115 – 0.73135
0.7314 – 0.7315 15 0.73135 – 0.73155
0.7316 – 0.7317 20 0.73155 – 0.73175
0.7318 – 0.7319 25 0.73175 – 0.73195
0.7320 – 0.7321 30 0.73195 – 0.73215
0.7322 – 0.7323 8 0.73215 – 0.73235
0.7324 – 0.7325 2 0.73235 – 0.73255
Total 110
Applying Formula
^ ( fm – fi )
Mode = X = L + -------------------------- * h
( fm – fi ) + ( fm – f2)
(30 - 25)
= 0.73195 +--------------------------- * 0.0002
(30 – 25) + (30 – 8)
5
= 0.73195 + ----------------- *0.0002
(5) + (22)
5
= 0.73195 + ----------- * 0.0002
27
= 0.73195 + 0.1851852 * 0.0002
= 0.73195 + 0.000037
= 0.731987
Working:
Class Mark is obtained by adding the sum of two limits of the and dividing by 2.
e.g. 0.7312 + 0.7313 /2 = 0.73125
fX is obtained X multiplying with (Frequency) f.
e.g. 10*0.73125 =7.3125
The Mode lies somewhere between 0.73195 and 0.73215.
By applying formula we noted that
fm = 30, fi = 25, f2 = 8
Frequency Distribution:
Class Limit Frequency ( f ) Class Boundaries Class-Mark(Midpoint) X
20 – 24 1 19.5 – 24.5 22
25 – 29 4 24.5 – 29.5 27
30 – 34 9 29.5 – 34.5 32
35 – 39 10 34.5 – 39.5 37
40 – 44 15 39.5 – 44.5 42
45 – 49 9 44.5 – 49.5 47
50 – 54 2 49.5 – 54.5 52
Total 50
Question # 2 (i)
_
Mean ( X )
Classes Frequency ( f ) Class Mark (Midpoint) X fX
0.7312 – 0.7313 10 0.73125 7.3125
0.7314 – 0.7315 15 0.73145 10.97175
0.7316 – 0.7317 20 0.73165 14.633
0.7318 – 0.7319 25 0.73185 18.29625
0.7320 – 0.7321 30 0.73205 21.9615
0.7322 – 0.7323 8 0.73225 5.858
0.7324 – 0.7325 2 0.73245 1.4649
Total 110 80.4979
Applying Formula
_ ∑fx
X = -------
∑f
= 80.4979 / 110
= 0.7317991
Question 2
Part ii ^
Mode ( X )
Classes Frequency ( f ) Class Boundaries ( C.B)
0.7312 – 0.7313 10 0.73115 – 0.73135
0.7314 – 0.7315 15 0.73135 – 0.73155
0.7316 – 0.7317 20 0.73155 – 0.73175
0.7318 – 0.7319 25 0.73175 – 0.73195
0.7320 – 0.7321 30 0.73195 – 0.73215
0.7322 – 0.7323 8 0.73215 – 0.73235
0.7324 – 0.7325 2 0.73235 – 0.73255
Total 110
Applying Formula
^ ( fm – fi )
Mode = X = L + -------------------------- * h
( fm – fi ) + ( fm – f2)
(30 - 25)
= 0.73195 +--------------------------- * 0.0002
(30 – 25) + (30 – 8)
5
= 0.73195 + ----------------- *0.0002
(5) + (22)
5
= 0.73195 + ----------- * 0.0002
27
= 0.73195 + 0.1851852 * 0.0002
= 0.73195 + 0.000037
= 0.731987
Working:
Class Mark is obtained by adding the sum of two limits of the and dividing by 2.
e.g. 0.7312 + 0.7313 /2 = 0.73125
fX is obtained X multiplying with (Frequency) f.
e.g. 10*0.73125 =7.3125
The Mode lies somewhere between 0.73195 and 0.73215.
By applying formula we noted that
fm = 30, fi = 25, f2 = 8
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