Question

If the arithmetic mean of the frequency distribution given below is $\frac{22}{5}$ and $\sum_{i=1}^{6}f_{i}=35$, then find the value of $a^{2}+b^{3}$:}

• A. $85$
• B. $379$
• C. $559$
• D. $265$

Hint:

In frequency distribution problems with unknown frequencies, always form two equations: \[ \sum f_i=N \] and \[ \bar{x}=\frac{\sum f_i x_i}{N}. \] These two equations usually provide enough information to determine the unknown frequencies directly.

Answer 1

The Correct Option is D

Concept: For a discrete frequency distribution, the arithmetic mean is calculated using the formula \[ \bar{x}=\frac{\sum f_i x_i}{\sum f_i}, \] where:
• $x_i$ represents the observations,
• $f_i$ represents the corresponding frequencies,
• $\sum f_i$ represents the total frequency,
• $\sum f_i x_i$ represents the sum of the products of observations and their frequencies. In this problem, two unknown frequencies $a$ and $b$ are involved. Therefore, we first use the given total frequency condition to form one equation and then use the mean formula to form a second equation. Solving these simultaneous equations gives the values of $a$ and $b$, which can then be substituted into the required expression.

Step 1: Use the given total frequency condition. The frequencies are: \[ 4,\;6,\;9,\;a,\;b,\;3. \] According to the question, \[ \sum_{i=1}^{6} f_i = 35. \] Substituting all frequencies: \[ 4+6+9+a+b+3=35. \] Adding the known numerical values: \[ 22+a+b=35. \] Subtracting 22 from both sides: \[ a+b=13. \] Thus, we obtain the first linear equation: \[ a+b=13 \qquad \cdots (1) \]

Step 2: Calculate $\sum f_i x_i$. Now multiply each observation by its corresponding frequency: \[ (2)(4)=8 \] \[ (3)(6)=18 \] \[ (4)(9)=36 \] \[ (5)(a)=5a \] \[ (6)(b)=6b \] \[ (7)(3)=21 \] Adding all these products: \[ \sum f_i x_i = 8+18+36+5a+6b+21. \] Combining the constant terms: \[ \sum f_i x_i = 83+5a+6b. \]

Step 3: Apply the arithmetic mean formula. The arithmetic mean is given as \[ \frac{22}{5}. \] Using \[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i}, \] we get \[ \frac{83+5a+6b}{35} = \frac{22}{5}. \] Cross-multiplying: \[ 5(83+5a+6b) = 35\times22. \] \[ 415+25a+30b = 770. \] Subtracting 415 from both sides: \[ 25a+30b = 355. \] Dividing the entire equation by 5: \[ 5a+6b = 71. \] Thus, the second equation is \[ 5a+6b=71 \qquad \cdots (2) \]

Step 4: Solve the simultaneous equations. From equation (1), \[ a+b=13. \] Multiplying equation (1) by 5: \[ 5a+5b=65. \] Subtracting this equation from equation (2): \[ (5a+6b)-(5a+5b) = 71-65. \] \[ b=6. \] Substituting $b=6$ into equation (1): \[ a+6=13. \] \[ a=7. \] Therefore, \[ a=7,\qquad b=6. \]

Step 5: Evaluate the required expression. We need to calculate \[ a^2+b^3. \] Substituting the values obtained: \[ a^2+b^3 = 7^2+6^3. \] Calculating separately: \[ 7^2=49, \] \[ 6^3=216. \] Therefore, \[ a^2+b^3 = 49+216 = 265. \] Hence, \[ \boxed{a^2+b^3=265}. \]