Statement-1 : The variance of the variates 112, 116, 120,125,132 about their A.M. is 48.8.
Statement-2 : `sigma=sqrt((Sigma_(i=1)^(n) f_i(x_i-barx)^2)/(Sigma_(i=1)^(n) f_i)`

Answer

To solve the problem, we need to verify the statements regarding the variance of the given variates and the formula for variance. Let's go through the steps systematically.

### Step 1: Calculate the Arithmetic Mean (A.M.)
The arithmetic mean (A.M.) is calculated using the formula:

\[ \text{A.M.} = \frac{\text{Sum of observations}}{\text{Number of observations}} \]

The observations given are: 112, 116, 120, 125, 132.

Calculating the sum:

\[ \text{Sum} = 112 + 116 + 120 + 125 + 132 = 605 \]

Number of observations (n) = 5.

Now, substituting into the formula:

\[ \text{A.M.} = \frac{605}{5} = 121 \]

### Step 2: Calculate the Variance
Variance is calculated using the formula:

\[ \text{Variance} = \frac{\sum (x_i - \bar{x})^2}{n} \]

Where \(x_i\) are the observations and \(\bar{x}\) is the mean we calculated.

Now, we calculate each term \((x_i - \bar{x})^2\):

1. For \(x_1 = 112\):
\[ (112 - 121)^2 = (-9)^2 = 81 \]

2. For \(x_2 = 116\):
\[ (116 - 121)^2 = (-5)^2 = 25 \]

3. For \(x_3 = 120\):
\[ (120 - 121)^2 = (-1)^2 = 1 \]

4. For \(x_4 = 125\):
\[ (125 - 121)^2 = (4)^2 = 16 \]

5. For \(x_5 = 132\):
\[ (132 - 121)^2 = (11)^2 = 121 \]

Now, summing these squared differences:

\[ \text{Sum of squared differences} = 81 + 25 + 1 + 16 + 121 = 244 \]

Now, substituting into the variance formula:

\[ \text{Variance} = \frac{244}{5} = 48.8 \]

### Conclusion
From our calculations, we find that the variance of the variates 112, 116, 120, 125, and 132 about their A.M. is indeed 48.8. Therefore, Statement-1 is true.

### Statement-2 Verification
The formula given in Statement-2 is:

\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n} f_i (x_i - \bar{x})^2}{\sum_{i=1}^{n} f_i}} \]

This is indeed the correct formula for calculating the standard deviation when considering frequency distributions. In our case, since all observations have a frequency of 1, the formula simplifies to the one we used for variance.

### Final Answer
Both Statement-1 and Statement-2 are true.

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