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Extract the given data for each firm.

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Here's the solution to the problem:

a) To determine which firm has greater variability in individual wages, we compare their coefficients of variation (CV). The coefficient of variation measures relative variability and is calculated as CV=σxˉ×100%CV = \frac{\sigma}{\bar{x}} \times 100\%, where σ\sigma is the standard deviation and xˉ\bar{x} is the average wage.

  1. Extract the given data for each firm.

    For Firm A:
    Number of employees (nAn_A) = 100
    Average wage (xˉA\bar{x}_A) = Rs 3200
    Variance (σA2\sigma_A^2) = 625

    For Firm B:
    Number of employees (nBn_B) = 150
    Average wage (xˉB\bar{x}_B) = Rs 2800
    Variance (σB2\sigma_B^2) = 729

  2. Calculate the standard deviation (σ\sigma) for each firm.

    Standard deviation is the square root of the variance.
    For Firm A:

    σA=σA2=625=25\sigma_A = \sqrt{\sigma_A^2} = \sqrt{625} = 25

    For Firm B:

    σB=σB2=729=27\sigma_B = \sqrt{\sigma_B^2} = \sqrt{729} = 27
  3. Calculate the coefficient of variation (CV) for each firm.

    For Firm A:

    CVA=σAxˉA×100%=253200×100%=0.0078125×100%=0.78125%CV_A = \frac{\sigma_A}{\bar{x}_A} \times 100\% = \frac{25}{3200} \times 100\% = 0.0078125 \times 100\% = 0.78125\%

    For Firm B:

    CVB=σBxˉB×100%=272800×100%=0.009642857×100%≈0.9643%CV_B = \frac{\sigma_B}{\bar{x}_B} \times 100\% = \frac{27}{2800} \times 100\% = 0.009642857 \times 100\% \approx 0.9643\%
  4. Compare the coefficients of variation.

    Since CVB≈0.9643%CV_B \approx 0.9643\% is greater than CVA≈0.78125%CV_A \approx 0.78125\%, Firm B has greater variability in individual wages.

    b) To calculate the average monthly wage and the variance of wages of all workers from both firms combined.

  5. Calculate the combined average monthly wage (xˉcombined\bar{x}_{combined}).

    The formula for the combined mean is:

    xˉcombined=nAxˉA+nBxˉBnA+nB\bar{x}_{combined} = \frac{n_A \bar{x}_A + n_B \bar{x}_B}{n_A + n_B}

    Substitute the values:

    xˉcombined=(100×3200)+(150×2800)100+150\bar{x}_{combined} = \frac{(100 \times 3200) + (150 \times 2800)}{100 + 150} xˉcombined=320000+420000250\bar{x}_{combined} = \frac{320000 + 420000}{250} xˉcombined=740000250\bar{x}_{combined} = \frac{740000}{250} xˉcombined=2960\bar{x}_{combined} = 2960

    The combined average monthly wage is Rs 2960.

  6. Calculate the combined variance (σcombined2\sigma_{combined}^2).

    The formula for combined variance is:

    σcombined2=nA(σA2+dA2)+nB(σB2+dB2)nA+nB\sigma_{combined}^2 = \frac{n_A(\sigma_A^2 + d_A^2) + n_B(\sigma_B^2 + d_B^2)}{n_A + n_B}

    where dA=xˉA−xˉcombinedd_A = \bar{x}_A - \bar{x}_{combined} and dB=xˉB−xˉcombinedd_B = \bar{x}_B - \bar{x}_{combined}.

    First, calculate the differences (dAd_A and dBd_B):

    dA=3200−2960=240d_A = 3200 - 2960 = 240 dB=2800−2960=−160d_B = 2800 - 2960 = -160

    Next, calculate the squared differences (dA2d_A^2 and dB2d_B^2):

    dA2=(240)2=57600d_A^2 = (240)^2 = 57600 dB2=(−160)2=25600d_B^2 = (-160)^2 = 25600

    Now, substitute these values into the combined variance formula:

    σcombined2=100(625+57600)+150(729+25600)100+150\sigma_{combined}^2 = \frac{100(625 + 57600) + 150(729 + 25600)}{100 + 150} σcombined2=100(58225)+150(26329)250\sigma_{combined}^2 = \frac{100(58225) + 150(26329)}{250} σcombined2=5822500+3949350250\sigma_{combined}^2 = \frac{5822500 + 3949350}{250} σcombined2=9771850250\sigma_{combined}^2 = \frac{9771850}{250} σcombined2=39087.4\sigma_{combined}^2 = 39087.4

    The combined variance of wages is 39087.4.

    a) Firm with greater variability: Firm B\boxed{\text{Firm B}}

    b) Combined average monthly wage: Rs 2960\boxed{\text{Rs 2960}}
    Combined variance of wages: 39087.4\boxed{\text{39087.4}}

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