The mean and standerd deviation deviation of some data for the time taken to complete a test are calculated with the following result s
Number of observation =25,means=18.2 s,standard deviation =3.25 s further another set of 15 obserbvation `x_(1),x_(2)...x_(15)` also in seconds is now available and we have `Sigma_(i=1)^(15) x_(i)=279 and Sigma_(i =1)^(15) x_(i)^(2)=5524 ` .Calculate the standard derivation based on all 40 observation .

Explanation

To solve the problem, we need to calculate the standard deviation based on all 40 observations. We will follow these steps:

### Step 1: Calculate the mean of the second set of observations
Given:
- Number of observations in the second set, \( n_2 = 15 \)
- Sum of the second set of observations, \( \Sigma_{i=1}^{15} x_i = 279 \)

The mean of the second set, \( \bar{x}_2 \), is calculated as follows:
\[ \bar{x}_2 = \frac{\Sigma_{i=1}^{15} x_i}{n_2} = \frac{279}{15} = 18.6 \]

### Step 2: Calculate the standard deviation of the second set of observations
Given:
- Sum of squares of the second set, \( \Sigma_{i=1}^{15} x_i^2 = 5524 \)

The formula for the standard deviation \( S_2 \) is:
\[ S_2 = \sqrt{\frac{\Sigma_{i=1}^{15} x_i^2 - \frac{(\Sigma_{i=1}^{15} x_i)^2}{n_2}}{n_2}} \]
Substituting the values:
\[ S_2 = \sqrt{\frac{5524 - \frac{(279)^2}{15}}{15}} \]
Calculating \( \frac{(279)^2}{15} \):
\[ \frac{(279)^2}{15} = \frac{77841}{15} \approx 5182.73 \]
Now substituting back:
\[ S_2 = \sqrt{\frac{5524 - 5182.73}{15}} = \sqrt{\frac{341.27}{15}} \approx \sqrt{22.75} \approx 4.77 \]

### Step 3: Calculate the combined mean of both sets
Given:
- Number of observations in the first set, \( n_1 = 25 \)
- Mean of the first set, \( \bar{x}_1 = 18.2 \)

The combined mean \( \bar{x}_c \) is calculated as:
\[ \bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2} = \frac{25 \times 18.2 + 15 \times 18.6}{40} \]
Calculating:
\[ \bar{x}_c = \frac{455 + 279}{40} = \frac{734}{40} = 18.35 \]

### Step 4: Calculate the combined standard deviation
The formula for the combined standard deviation \( S_c \) is:
\[ S_c^2 = \frac{n_1 S_1^2 + n_1 (\bar{x}_1 - \bar{x}_c)^2 + n_2 S_2^2 + n_2 (\bar{x}_2 - \bar{x}_c)^2}{n_1 + n_2} \]
Where:
- \( S_1 = 3.25 \) (standard deviation of the first set)

Calculating \( S_1^2 \) and \( S_2^2 \):
\[ S_1^2 = (3.25)^2 = 10.5625 \]
\[ S_2^2 = (4.77)^2 \approx 22.7529 \]

Now substituting into the formula:
\[ S_c^2 = \frac{25 \times 10.5625 + 25 \times (18.2 - 18.35)^2 + 15 \times 22.7529 + 15 \times (18.6 - 18.35)^2}{40} \]
Calculating each term:
- \( 25 \times 10.5625 = 264.0625 \)
- \( 25 \times (18.2 - 18.35)^2 = 25 \times (0.15)^2 = 25 \times 0.0225 = 0.5625 \)
- \( 15 \times 22.7529 = 341.294 \)
- \( 15 \times (18.6 - 18.35)^2 = 15 \times (0.25)^2 = 15 \times 0.0625 = 0.9375 \)

Now summing these:
\[ S_c^2 = \frac{264.0625 + 0.5625 + 341.294 + 0.9375}{40} = \frac{606.8565}{40} \approx 15.1714 \]

Finally, taking the square root to find \( S_c \):
\[ S_c = \sqrt{15.1714} \approx 3.89 \]

### Final Answer
The combined standard deviation based on all 40 observations is approximately \( 3.89 \).