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The table gives the frequency distribution of marks obtained by a number of students in a test. If the mean mark is 5, calculate the: (a) the value of m; (b) median; (c) standard deviation of the distribution.

Mathematics

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Answers

3 parts
a

10\text{10}

b

5\text{5}

c

1.265\text{1.265}

Step-by-step solution

11 steps

Here's how to solve the problem:

Part a· 4 steps

The value of mm;

  1. Write down the formula for the mean.

    The mean (xˉ\bar{x}) is given by the sum of (mark ×\times frequency) divided by the sum of frequencies.

    xˉ=∑(xi×fi)∑fi\bar{x} = \frac{\sum (x_i \times f_i)}{\sum f_i}
  2. Calculate the sum of (xi×fi)(x_i \times f_i).

    ∑(xi×fi)=(3×5)+(4×m)+(5×(m+1))+(6×9)+(7×4)+(8×1)\sum (x_i \times f_i) = (3 \times 5) + (4 \times m) + (5 \times (m+1)) + (6 \times 9) + (7 \times 4) + (8 \times 1) ∑(xi×fi)=15+4m+5m+5+54+28+8\sum (x_i \times f_i) = 15 + 4m + 5m + 5 + 54 + 28 + 8 ∑(xi×fi)=110+9m\sum (x_i \times f_i) = 110 + 9m
  3. Calculate the sum of frequencies.

    ∑fi=5+m+(m+1)+9+4+1\sum f_i = 5 + m + (m+1) + 9 + 4 + 1 ∑fi=20+2m\sum f_i = 20 + 2m
  4. Substitute the sums and the given mean (xˉ=5\bar{x} = 5) into the formula and solve for mm.

    5=110+9m20+2m5 = \frac{110 + 9m}{20 + 2m}

    Multiply both sides by (20+2m)(20 + 2m):

    5(20+2m)=110+9m5(20 + 2m) = 110 + 9m 100+10m=110+9m100 + 10m = 110 + 9m

    Subtract 9m9m from both sides:

    10m−9m=110−10010m - 9m = 110 - 100 m=10m = 10

    The value of mm is 10\boxed{\text{10}}.

Part b

5\text{5}

Median;

  1. Update the frequency table with m=10m=10.

    Marks (xix_i)Frequency (fif_i)Cumulative Frequency (CF)
    355
    4105+10=155+10=15
    510+1=1110+1=1115+11=2615+11=26
    6926+9=3526+9=35
    7435+4=3935+4=39
    8139+1=4039+1=40
  2. Calculate the total number of students (NN).

    N=∑fi=40N = \sum f_i = 40
  3. Determine the position of the median.

    For discrete data, the median is at the (N+12)\left(\frac{N+1}{2}\right)-th position.

    Median position=40+12=412=20.5\text{Median position} = \frac{40+1}{2} = \frac{41}{2} = 20.5

    This means the median is the average of the 20th and 21st values.

  4. Locate the median mark using the cumulative frequency.

    From the cumulative frequency column, the 20th value falls within the mark 5 category (since CF for mark 4 is 15, and CF for mark 5 is 26). The 21st value also falls within the mark 5 category.
    Therefore, the median is 5.
    The median is 5\boxed{\text{5}}.

Part c

1.265\text{1.265}

Standard deviation of the distribution.

  1. Write down the formula for standard deviation.

    The standard deviation (σ\sigma) for grouped data is given by:

    σ=∑fixi2N−xˉ2\sigma = \sqrt{\frac{\sum f_i x_i^2}{N} - \bar{x}^2}

    We know N=40N=40 and xˉ=5\bar{x}=5.

  2. Calculate ∑fixi2\sum f_i x_i^2.

    Marks (xix_i)Frequency (fif_i)xi2x_i^2fixi2f_i x_i^2
    3595×9=455 \times 9 = 45
    4101610×16=16010 \times 16 = 160
    5112511×25=27511 \times 25 = 275
    69369×36=3249 \times 36 = 324
    74494×49=1964 \times 49 = 196
    81641×64=641 \times 64 = 64
    Total401064
    So, ∑fixi2=1064\sum f_i x_i^2 = 1064.
  3. Substitute the values into the standard deviation formula.

    σ=106440−52\sigma = \sqrt{\frac{1064}{40} - 5^2} σ=26.6−25\sigma = \sqrt{26.6 - 25} σ=1.6\sigma = \sqrt{1.6} σ≈1.2649\sigma \approx 1.2649

    Rounding to three decimal places:

    σ≈1.265\sigma \approx 1.265

    The standard deviation is 1.265\boxed{\text{1.265}}.

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