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The table is a frequency distribution of ages of 200 pupils in a school. Find, correct to one decimal place, the: (a) mean; (b) standard deviation, of the distribution.

Mathematics

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Answers

2 parts
a

5.0 years\text{5.0 years}

b

2.2 years\text{2.2 years}

Step-by-step solution

4 steps

Part a· 2 steps

Mean

  1. Calculate the sum of frequencies (∑f\sum f) and the sum of (age ×\times frequency) (∑fx\sum fx).

    The total number of pupils is given as ∑f=200\sum f = 200.

    ∑fx=(1×12)+(2×14)+(3×20)+(4×45)+(5×36)+(6×20)+(7×23)+(8×16)+(9×8)+(10×6)\sum fx = (1 \times 12) + (2 \times 14) + (3 \times 20) + (4 \times 45) + (5 \times 36) + (6 \times 20) + (7 \times 23) + (8 \times 16) + (9 \times 8) + (10 \times 6) ∑fx=12+28+60+180+180+120+161+128+72+60\sum fx = 12 + 28 + 60 + 180 + 180 + 120 + 161 + 128 + 72 + 60 ∑fx=1001\sum fx = 1001

  2. Calculate the mean (xˉ\bar{x}) using the formula xˉ=∑fx∑f\bar{x} = \frac{\sum fx}{\sum f}.

    xˉ=1001200\bar{x} = \frac{1001}{200} xˉ=5.005\bar{x} = 5.005

    Rounding to one decimal place:

    xˉ=5.0\bar{x} = 5.0

    The mean age is 5.0 years\boxed{\text{5.0 years}}.

Part b

2.2 years\text{2.2 years}

Standard deviation

  1. Calculate the sum of (age2×^2 \times frequency) (∑fx2\sum fx^2).

    ∑fx2=(12×12)+(22×14)+(32×20)+(42×45)+(52×36)+(62×20)+(72×23)+(82×16)+(92×8)+(102×6)\sum fx^2 = (1^2 \times 12) + (2^2 \times 14) + (3^2 \times 20) + (4^2 \times 45) + (5^2 \times 36) + (6^2 \times 20) + (7^2 \times 23) + (8^2 \times 16) + (9^2 \times 8) + (10^2 \times 6) ∑fx2=(1×12)+(4×14)+(9×20)+(16×45)+(25×36)+(36×20)+(49×23)+(64×16)+(81×8)+(100×6)\sum fx^2 = (1 \times 12) + (4 \times 14) + (9 \times 20) + (16 \times 45) + (25 \times 36) + (36 \times 20) + (49 \times 23) + (64 \times 16) + (81 \times 8) + (100 \times 6) ∑fx2=12+56+180+720+900+720+1127+1024+648+600\sum fx^2 = 12 + 56 + 180 + 720 + 900 + 720 + 1127 + 1024 + 648 + 600 ∑fx2=5987\sum fx^2 = 5987

  2. Calculate the standard deviation (σ\sigma) using the formula σ=∑fx2∑f−xˉ2\sigma = \sqrt{\frac{\sum fx^2}{\sum f} - \bar{x}^2}.

    σ=5987200−(5.005)2\sigma = \sqrt{\frac{5987}{200} - (5.005)^2} σ=29.935−25.050025\sigma = \sqrt{29.935 - 25.050025} σ=4.884975\sigma = \sqrt{4.884975} σ≈2.2101979\sigma \approx 2.2101979

    Rounding to one decimal place:

    σ=2.2\sigma = 2.2

    The standard deviation is 2.2 years\boxed{\text{2.2 years}}.

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