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Engineering Mathematics → Statistics & Regression → Mean, Median & Mode
Last updated 5 September 2026
Question The frequency distribution of the compressive strength of 20 concrete cube specimens is given in the table.
f (MPa) Number of specimens with compressive strength equal to f 23 4 28 2 22.5 5 31 5 29 4
If
μ \mu μ is the mean strength of the specimens and
σ \sigma σ is the standard deviation, the number of specimens (out of 20) with compressive strength less than
μ − 3 σ \mu - 3\sigma μ − 3 σ is
______ Solution 1. Calculate the mean (μ \mu μ μ = ∑ f i x i ∑ f i = ( 23 × 4 ) + ( 28 × 2 ) + ( 22.5 × 5 ) + ( 31 × 5 ) + ( 29 × 4 ) 20 \mu = \frac{\sum f_i x_i}{\sum f_i} = \frac{(23 \times 4) + (28 \times 2) + (22.5 \times 5) + (31 \times 5) + (29 \times 4)}{20} μ = ∑ f i ∑ f i x i = 20 ( 23 × 4 ) + ( 28 × 2 ) + ( 22.5 × 5 ) + ( 31 × 5 ) + ( 29 × 4 ) μ = 92 + 56 + 112.5 + 155 + 116 20 = 531.5 20 = 26.575 MPa \mu = \frac{92 + 56 + 112.5 + 155 + 116}{20} = \frac{531.5}{20} = 26.575 \text{ MPa} μ = 20 92 + 56 + 112.5 + 155 + 116 = 20 531.5 = 26.575 MPa 2. Calculate the variance (σ 2 \sigma^2 σ 2 σ 2 = ∑ f i x i 2 N − μ 2 \sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2 σ 2 = N ∑ f i x i 2 − μ 2 ∑ f i x i 2 = 4 ( 23 2 ) + 2 ( 28 2 ) + 5 ( 22.5 2 ) + 5 ( 31 2 ) + 4 ( 29 2 ) \sum f_i x_i^2 = 4(23^2) + 2(28^2) + 5(22.5^2) + 5(31^2) + 4(29^2) ∑ f i x i 2 = 4 ( 2 3 2 ) + 2 ( 2 8 2 ) + 5 ( 22. 5 2 ) + 5 ( 3 1 2 ) + 4 ( 2 9 2 ) ∑ f i x i 2 = 4 ( 529 ) + 2 ( 784 ) + 5 ( 506.25 ) + 5 ( 961 ) + 4 ( 841 ) = 2116 + 1568 + 2531.25 + 4805 + 3364 = 14384.25 \sum f_i x_i^2 = 4(529) + 2(784) + 5(506.25) + 5(961) + 4(841) = 2116 + 1568 + 2531.25 + 4805 + 3364 = 14384.25 ∑ f i x i 2 = 4 ( 529 ) + 2 ( 784 ) + 5 ( 506.25 ) + 5 ( 961 ) + 4 ( 841 ) = 2116 + 1568 + 2531.25 + 4805 + 3364 = 14384.25 σ 2 = 14384.25 20 − ( 26.575 ) 2 = 719.2125 − 706.230625 = 12.981875 \sigma^2 = \frac{14384.25}{20} - (26.575)^2 = 719.2125 - 706.230625 = 12.981875 σ 2 = 20 14384.25 − ( 26.575 ) 2 = 719.2125 − 706.230625 = 12.981875 3. Calculate the standard deviation (σ \sigma σ σ = 12.981875 ≈ 3.603 \sigma = \sqrt{12.981875} \approx 3.603 σ = 12.981875 ≈ 3.603 4. Calculate the threshold (μ − 3 σ \mu - 3\sigma μ − 3 σ μ − 3 σ = 26.575 − 3 ( 3.603 ) = 26.575 − 10.809 = 15.766 \mu - 3\sigma = 26.575 - 3(3.603) = 26.575 - 10.809 = 15.766 μ − 3 σ = 26.575 − 3 ( 3.603 ) = 26.575 − 10.809 = 15.766 5. Count specimens with strength < 15.766:
From the table, the minimum strength value is
22.5 22.5 22.5 MPa. Since
22.5 > 15.766 22.5 > 15.766 22.5 > 15.766 , there are
0 0 0 specimens with strength less than
μ − 3 σ \mu - 3\sigma μ − 3 σ .
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Solution 1. Calculate the mean (μ \mu μ μ = ∑ f i x i ∑ f i = ( 23 × 4 ) + ( 28 × 2 ) + ( 22.5 × 5 ) + ( 31 × 5 ) + ( 29 × 4 ) 20 \mu = \frac{\sum f_i x_i}{\sum f_i} = \frac{(23 \times 4) + (28 \times 2) + (22.5 \times 5) + (31 \times 5) + (29 \times 4)}{20} μ = ∑ f i ∑ f i x i = 20 ( 23 × 4 ) + ( 28 × 2 ) + ( 22.5 × 5 ) + ( 31 × 5 ) + ( 29 × 4 ) μ = 92 + 56 + 112.5 + 155 + 116 20 = 531.5 20 = 26.575 MPa \mu = \frac{92 + 56 + 112.5 + 155 + 116}{20} = \frac{531.5}{20} = 26.575 \text{ MPa} μ = 20 92 + 56 + 112.5 + 155 + 116 = 20 531.5 = 26.575 MPa 2. Calculate the variance (σ 2 \sigma^2 σ 2 σ 2 = ∑ f i x i 2 N − μ 2 \sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2 σ 2 = N ∑ f i x i 2 − μ 2 ∑ f i x i 2 = 4 ( 23 2 ) + 2 ( 28 2 ) + 5 ( 22.5 2 ) + 5 ( 31 2 ) + 4 ( 29 2 ) \sum f_i x_i^2 = 4(23^2) + 2(28^2) + 5(22.5^2) + 5(31^2) + 4(29^2) ∑ f i x i 2 = 4 ( 2 3 2 ) + 2 ( 2 8 2 ) + 5 ( 22. 5 2 ) + 5 ( 3 1 2 ) + 4 ( 2 9 2 ) ∑ f i x i 2 = 4 ( 529 ) + 2 ( 784 ) + 5 ( 506.25 ) + 5 ( 961 ) + 4 ( 841 ) = 2116 + 1568 + 2531.25 + 4805 + 3364 = 14384.25 \sum f_i x_i^2 = 4(529) + 2(784) + 5(506.25) + 5(961) + 4(841) = 2116 + 1568 + 2531.25 + 4805 + 3364 = 14384.25 ∑ f i x i 2 = 4 ( 529 ) + 2 ( 784 ) + 5 ( 506.25 ) + 5 ( 961 ) + 4 ( 841 ) = 2116 + 1568 + 2531.25 + 4805 + 3364 = 14384.25 σ 2 = 14384.25 20 − ( 26.575 ) 2 = 719.2125 − 706.230625 = 12.981875 \sigma^2 = \frac{14384.25}{20} - (26.575)^2 = 719.2125 - 706.230625 = 12.981875 σ 2 = 20 14384.25 − ( 26.575 ) 2 = 719.2125 − 706.230625 = 12.981875 3. Calculate the standard deviation (σ \sigma σ σ = 12.981875 ≈ 3.603 \sigma = \sqrt{12.981875} \approx 3.603 σ = 12.981875 ≈ 3.603 4. Calculate the threshold (μ − 3 σ \mu - 3\sigma μ − 3 σ μ − 3 σ = 26.575 − 3 ( 3.603 ) = 26.575 − 10.809 = 15.766 \mu - 3\sigma = 26.575 - 3(3.603) = 26.575 - 10.809 = 15.766 μ − 3 σ = 26.575 − 3 ( 3.603 ) = 26.575 − 10.809 = 15.766 5. Count specimens with strength < 15.766:
From the table, the minimum strength value is
22.5 22.5 22.5 MPa. Since
22.5 > 15.766 22.5 > 15.766 22.5 > 15.766 , there are
0 0 0 specimens with strength less than
μ − 3 σ \mu - 3\sigma μ − 3 σ .
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