Question

the mode of the following frequency table is 26 . find th missing frequencies if the total frequncy is 50 ( 10th grade ) ​

Answer 1

Answer:

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Answer 2

Answer:

The value of the missing frequency, F₁ = $9$.

And the value of missing frequency, F₂ = $11$.

Step-by-step explanation:

Given data,

Class               Frequency ( $f_{i}$ )

$0-10$                   $3$

$10-20$                 $F_{1}$

$20-30$                 $15$

$30-40$                 $F_{2}$

$40-50$                 $8$

$50-60$                 $4$

The total frequencies, $\sum f_{i}$ = $50$

The mode of the given frequencies = $26$

The value of the missing frequencies ( F₁ and F₂ ) =?

As we know,

• Mode = $L + h [\frac{(f_{m} -f_{1} )}{2f_{m} -f_{1}-f_{2} } ]$

Here,

• L = The lower limit ( modal class )
• h = Size of the class interval
• $f_{m}$ = Frequency ( modal class )
• $f_{1}$ = Frequency ( class prior to modal class )
• $f_{2}$ = Frequency ( class next to modal class )

As given, the mode is $26$; therefore, the modal class is $20-30$

Firstly, we have to make a table to find out the terms of the mode formula:

Class               Frequency ( $f_{i}$ )

$0-10$                   $3$

$10-20$                 $F_{1}$      -------$f_{1}$

$20-30$                 $15$      -------$f_{m}$

$30-40$                 $F_{2}$      -------$f_{2}$

$40-50$                 $8$

$50-60$                 $4$

Thus,

• L = $20$
• h = $10$
• $f_{m}$ = $15$
• $f_{1}$ = $F_{1}$
• $f_{2}$ = $F_{2}$

Now,

• Mode = $L + h [\frac{(f_{m} -f_{1} )}{2f_{m} -f_{1}-f_{2} } ]$
• $26 = 20 + 10 [\frac{15 -F_{1} )}{2(15) -F_{1}-F_{2} } ]$
• $[\frac{(150 -10F_{1} )}{30 -F_{1}-F_{2} } ] =6$
• $150 - 10F_{1} = 180 -6F_{1}- 6F_ {2}$
• $75 - 5F_{1} = 90 -3F_{1}- 3F_ {2}$
• $2F_{1} -3F_{2} = -15$   -------equation (a)

As given,

• $\sum f_{i}$ = $50$
• $3+F_{1} + 15 + F_{2} +8 +4 = 50$
• $F_{1} + F_{2} = 20$         --------equation (b)

After multiplying equation (b) by 3, we get:

• $3F_{1} + 3F_{2} = 60$  -------equation (c)

After adding equation (a) and equation (c), we get:

• $2F_{1} + 3F_{1} = 45$
• $F_{1} = 9$

After putting the value of $F_{1}$ in equation (b), we get:

• $9 + F_{2} = 20$
• $F_{2} = 11$

Hence, the value of the missing frequency, F₁ = $9$.

And the value of missing frequency, F₂ = $11$.