Question

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Answer 1

Given:-

Mode = 67

To find:-

The missing frequency F

Solution:-

We have, mode = 67

$mode = l + ( \frac{f1 - f0}{2f1 - f0 - f2})$

We can see that mode is 67

Therefore, the model class must be 60-70

Therefore, f1 = 15, f0 = F and f2 = 12

Using the formula,

$67 = ( \frac{15 - f}{2 \times 15 - f - 12} )$

$= > 67 = ( \frac{15 - f}{30 - 12 - f} )$

$= > 67 = ( \frac{15 - f}{18 - f} )$

$= > 67(18 - f) = 15 - f$

$= > 1206 - 67f = 15 - f$

$= > 1206 - 15 = 67f - f$

$= > 1191 = 66f$

$= > f = \frac{1191}{66}$

=> f = 18.04

Hence, the value of f = 18.04

Answer 2

Answer:

Given:-

Mode = 67

To find:-

The missing frequency F

Solution:-

We have, mode = 67

mode = l + ( \frac{f1 - f0}{2f1 - f0 - f2})mode=l+(2f1−f0−f2f1−f0)

We can see that mode is 67

Therefore, the model class must be 60-70

Therefore, f1 = 15, f0 = F and f2 = 12

Using the formula,

67 = ( \frac{15 - f}{2 \times 15 - f - 12} )67=(2×15−f−1215−f)

= > 67 = ( \frac{15 - f}{30 - 12 - f} )=>67=(30−12−f15−f)

= > 67 = ( \frac{15 - f}{18 - f} )=>67=(18−f15−f)

= > 67(18 - f) = 15 - f=>67(18−f)=15−f

= > 1206 - 67f = 15 - f=>1206−67f=15−f

= > 1206 - 15 = 67f - f=>1206−15=67f−f

= > 1191 = 66f=>1191=66f

= > f = \frac{1191}{66}=>f=661191

=> f = 18.04

Hence, the value of f = 18.04