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Assumed mean (A) = 24.5 kN

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2 parts
a

24.9 kN\text{24.9 kN}

b

8.6 kN\text{8.6 kN}

Step-by-step solution

9 steps

To solve this problem, we first need to set up an extended frequency distribution table to calculate the necessary values for the mean and standard deviation using the assumed mean method.

Given:
Assumed mean (AA) = 24.5 kN24.5 \text{ kN}
Class width (cc) = 6 (e.g., 9−4+1=69-4+1=6)

The class boundaries are found by taking the midpoint between the upper limit of one class and the lower limit of the next. For example, for 9 and 10, the boundary is 9.5.
The class marks (xx) are the midpoints of the class intervals.
d=x−Ad = x - A
fdfd and fd2fd^2 are calculated for the formulas.

Here is the table:

Maximum Load (kN)Frequency (f)Class Mark (x)d=x−24.5fdfd24−936.5−18−5497210−15512.5−12−6072016−211218.5−6−7243222−271724.500028−331430.568450434−39636.5127286440−45342.51854972Total∑f=60∑fd=24∑fd2=4464\begin{array}{|c|c|c|c|c|c|} \hline \text{Maximum Load (kN)} & \text{Frequency } (f) & \text{Class Mark } (x) & d = x - 24.5 & fd & fd^2 \\ \hline 4-9 & 3 & 6.5 & -18 & -54 & 972 \\ 10-15 & 5 & 12.5 & -12 & -60 & 720 \\ 16-21 & 12 & 18.5 & -6 & -72 & 432 \\ 22-27 & 17 & 24.5 & 0 & 0 & 0 \\ 28-33 & 14 & 30.5 & 6 & 84 & 504 \\ 34-39 & 6 & 36.5 & 12 & 72 & 864 \\ 40-45 & 3 & 42.5 & 18 & 54 & 972 \\ \hline \textbf{Total} & \sum f = 60 & & & \sum fd = 24 & \sum fd^2 = 4464 \\ \hline \end{array}

Part a· 3 steps

Mean distribution

  1. Use the formula for the mean using the assumed mean method.

    The formula for the mean (xˉ\bar{x}) is:

    xˉ=A+∑fd∑f\bar{x} = A + \frac{\sum fd}{\sum f}

  2. Substitute the values from the table into the formula.

    xˉ=24.5+2460\bar{x} = 24.5 + \frac{24}{60}

  3. Calculate the mean and round to 1 decimal place.

    xˉ=24.5+0.4\bar{x} = 24.5 + 0.4 xˉ=24.9 kN\bar{x} = 24.9 \text{ kN}

    The mean distribution is 24.9 kN\boxed{\text{24.9 kN}}.

Part b

8.6 kN\text{8.6 kN}

Standard deviation of the distribution

  1. Use the formula for the standard deviation using the assumed mean method.

    The formula for the standard deviation (ss) is:

    s=∑fd2∑f−(∑fd∑f)2s = \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}

  2. Substitute the values from the table into the formula.

    s=446460−(2460)2s = \sqrt{\frac{4464}{60} - \left(\frac{24}{60}\right)^2}

  3. Calculate the standard deviation.

    s=74.4−(0.4)2s = \sqrt{74.4 - (0.4)^2} s=74.4−0.16s = \sqrt{74.4 - 0.16} s=74.24s = \sqrt{74.24}

  4. Round the standard deviation to 1 decimal place.

    s≈8.61626s \approx 8.61626 s≈8.6 kNs \approx 8.6 \text{ kN}

    The standard deviation of the distribution is 8.6 kN\boxed{\text{8.6 kN}}.

Part c· 2 steps

Estimate the mode of the distribution

  1. Identify the modal class.

    The modal class is the class with the highest frequency. From the table, the highest frequency is 17, which corresponds to the class interval 22-27.
    So, the modal class is 22-27.

  2. Identify the values for the mode formula.

    • Lower class boundary of the modal class (LL): The class before 22-27 is 16-21. The upper limit of 16-21 is 21, and the lower limit of 22-27 is 22. The boundary is 21+222=21.5\frac{21+22}{2} = 21.5. So, L=21.5L = 21.5.
    • Frequency of the modal class (fmf_m): 17.
    • Frequency of the class preceding the modal class (f1f_1):
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