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NCERT Exemplar · Q15
Q.

Find the mean and variance of the frequency distribution given below:

xx1≤x<31 \le x < 33≤x<53 \le x < 55≤x<75 \le x < 77≤x<107 \le x < 10
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Rajasthan RbseShort· 5mImportance★★★★★est
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For grouped frequency distributions, we approximate the data within each class interval by its midpoint. Using these midpoints, the mean is calculated as ∑fixi∑fi\frac{\sum f_i x_i}{\sum f_i} and the variance as ∑fixi2∑fi−(∑fixi∑fi)2\frac{\sum f_i x_i^2}{\sum f_i} - \left(\frac{\sum f_i x_i}{\sum f_i}\right)^2. For the given distribution, the mean is 4.16\boxed{4.16} and the variance is 3.99\boxed{3.99}.

When we have data presented in class intervals, as in a grouped frequency distribution, we do not know the exact value of each observation. For example, in the class 1≤x<31 \le x < 3, we know there are 6 observations, but we don't know if they are all 1, all 2, or some mix. To calculate statistical measures like the mean and variance, we need a single representative value for each class.

The most reasonable assumption we can make is that the observations within a class interval are evenly distributed, or that their average value is the midpoint of that interval. This midpoint then acts as the representative value for all observations in that class.

The mean (xˉ\bar{x}) tells us the central tendency of the data – essentially, the average value. The variance (σ2\sigma^2) measures the spread or dispersion of the data points around this mean. A higher variance indicates that the data points are more spread out from the mean, while a lower variance means they are clustered closer to the mean.

Here's how we calculate these for the given grouped frequency distribution:

  1. Identify Class Midpoints (xix_i)

    Since we don't have individual data points, we use the midpoint of each class interval as the representative value (xix_i) for that class. The midpoint is calculated as (lower limit + upper limit) / 2.

    • For 1≤x<31 \le x < 3: x1=(1+3)/2=2x_1 = (1+3)/2 = 2
    • For 3≤x<53 \le x < 5: x2=(3+5)/2=4x_2 = (3+5)/2 = 4
    • For 5≤x<75 \le x < 7: x3=(5+7)/2=6x_3 = (5+7)/2 = 6
    • For 7≤x<107 \le x < 10: x4=(7+10)/2=8.5x_4 = (7+10)/2 = 8.5
    Watch out

    Be careful with the last class interval 7≤x<107 \le x < 10. Its width is 33, not 22. Always calculate the midpoint using the actual limits of the interval.

  2. Construct an Extended Frequency Table

    To systematically calculate the mean and variance, it's helpful to create a table that includes the frequency (fif_i), midpoint (xix_i), the product fixif_i x_i, the square of the midpoint xi2x_i^2, and the product fixi2f_i x_i^2.

    Class Intervalfif_ixix_ifixif_i x_ixi2x_i^2fixi2f_i x_i^2
    1≤x<31 \le x < 36212424
    3≤x<53 \le x < 544161664
    5≤x<75 \le x < 7563036180
    7≤x<107 \le x < 1018.58.572.2572.25

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