Question

In a measurement process, groups A and B recorded 10 and 15 values, respectively. The arithmetic means and standard deviations of group A are \(\mu_A = 35\), \(\sigma_A = 0.4\) and group B are \(\mu_B = 38\), \(\sigma_B = 0.6\). The standard deviation for the combined set of group A and group B measurements is \(\underline{\hspace{2cm}}\). (rounded off to two decimal places)

Hint:

For combining groups, use the weighted variance formula based on sample sizes.

Answer 1

Correct Answer: 1.54

Standard deviation for combined groups formula: \[ \sigma_{\text{combined}} = \sqrt{\frac{(n_A - 1)\sigma_A^2 + (n_B - 1)\sigma_B^2}{n_A + n_B - 2}} \] Where: \[ n_A = 10, n_B = 15, \sigma_A = 0.4, \sigma_B = 0.6 \] Substitute the values: \[ \sigma_{\text{combined}} = \sqrt{\frac{(10 - 1) \times 0.4^2 + (15 - 1) \times 0.6^2}{10 + 15 - 2}} \] \[ = \sqrt{\frac{9 \times 0.16 + 14 \times 0.36}{23}} = \sqrt{\frac{1.44 + 5.04}{23}} = \sqrt{\frac{6.48}{23}} = \sqrt{0.2817} \] \[ = 0.53 \] Thus, \[ \boxed{0.54} \]