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NCERT Exemplar · Q9
Q.

The frequency distribution:

xxA2A3A4A5A6A
ff211111

where A is a positive integer, has a variance of 160. Determine the value of A.

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We calculate the mean and variance of the given frequency distribution in terms of A, equate the variance to 160, and solve for A, finding that A = 7.

When we analyse a set of data, especially a frequency distribution, two fundamental measures help us understand its characteristics: the mean and the variance. The mean (μ\mu or xˉ\bar{x}) tells us the central tendency or the average value of the data. The variance (σ2\sigma^2) quantifies the spread or dispersion of the data points around this mean. A higher variance means the data points are more spread out, while a lower variance indicates they are clustered closer to the mean.

For a frequency distribution where xix_i are the data values and fif_i are their corresponding frequencies, the formulas are adapted to account for how often each value appears:

  • Mean: μ=∑fixi∑fi\mu = \frac{\sum f_i x_i}{\sum f_i}
  • Variance: σ2=∑fi(xi−μ)2∑fi\sigma^2 = \frac{\sum f_i (x_i - \mu)^2}{\sum f_i}
    Tip

    A more computationally convenient formula for variance, often called the shortcut formula, is:

    σ2=∑fixi2∑fi−(∑fixi∑fi)2\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \left(\frac{\sum f_i x_i}{\sum f_i}\right)^2

    This formula avoids calculating deviations (xi−μ)(x_i - \mu) for each data point, which can involve fractions and make calculations cumbersome. We will use this shortcut formula.

Let's apply these concepts to the given problem.

  1. Identify the data and total frequency:

    The given frequency distribution is:

    xxA2A3A4A5A6A
    ff211111

    The data values are xi∈{A,2A,3A,4A,5A,6A}x_i \in \{A, 2A, 3A, 4A, 5A, 6A\} and their corresponding frequencies are fi∈{2,1,1,1,1,1}f_i \in \{2, 1, 1, 1, 1, 1\}.

    First, we calculate the total frequency, N=∑fiN = \sum f_i:

    N=2+1+1+1+1+1=7N = 2 + 1 + 1 + 1 + 1 + 1 = 7

  2. Calculate the mean (μ\mu):

    We need ∑fixi\sum f_i x_i:

    ∑fixi=(2×A)+(1×2A)+(1×3A)+(1×4A)+(1×5A)+(1×6A)\sum f_i x_i = (2 \times A) + (1 \times 2A) + (1 \times 3A) + (1 \times 4A) + (1 \times 5A) + (1 \times 6A)

    ∑fixi=2A+2A+3A+4A+5A+6A=22A\sum f_i x_i = 2A + 2A + 3A + 4A + 5A + 6A = 22A

    Now, calculate the mean:

    μ=∑fixiN=22A7\mu = \frac{\sum f_i x_i}{N} = \frac{22A}{7}

  3. Calculate the sum of squared values multiplied by frequencies (∑fixi2\sum f_i x_i^2):

    This is a crucial step for the shortcut variance formula.

    ∑fixi2=(2×A2)+(1×(2A)2)+(1×(3A)2)+(1×(4A)2)+(1×(5A)2)+(1×(6A)2)\sum f_i x_i^2 = (2 \times A^2) + (1 \times (2A)^2) + (1 \times (3A)^2) + (1 \times (4A)^2) + (1 \times (5A)^2) + (1 \times (6A)^2) …

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