1737. Let the dairy farm of a collective farm have data on the milk yield of cows for the lactation period:

| Milk yield, kg | Number of cows | Milk yield, kg | Number of cows |
| :---: | :---: | :---: | :---: |
| $400-600$ | 1 | $1600-1800$ | 14 |
| $600-800$ | 3 | $1800-2000$ | 12 |
| $800-1000$ | 6 | $2000-2200$ | 10 |
| $1000-1200$ | 11 | $2200-2400$ | 6 |
| $1200-1400$ | 15 | 2400 and above | 2 |
| $1400-1600$ | 20 |  |  |

Find the arithmetic mean of the milk yield of cows on this farm.
Solution. The arithmetic mean will be found using the simplified formula

$$
\bar{X}=k \bar{U}+x_{0} \quad\left(U=\frac{X-x_{0}}{k}\right)
$$

by setting $x_{0}=1500$ and $k=200$. All necessary calculations are presented in the following table:

| Interval $X_{i}^{\prime}-X_{i}^{\prime \prime}$ | Frequency  $m_{i}$ | Midpoint of interval $x_{i}$ | $u_{i}=\frac{x_{i}-x_{0}}{k}$ | $m_{i} u_{i}$ |
| :---: | :---: | :---: | :---: | :---: |
| $400-600$ | 1 | 500 | -5 | -5 |
| $600-800$ | 3 | 700 | -4 | -12 |
| $800-1000$ | 6 | 900 | -3 | -18 |
| $1000-1200$ | 11 | 1100 | -2 | -22 |
| $1200-100$ | 15 | 1300 | 0 | -15 |
| $1400-1600$ | 20 | 1500 | 0 | 14 |
| $1600-1800$ | 14 | 1700 | 2 | 24 |
| $1800-2000$ | 12 | 1900 | 4 | 30 |
| $2000-200$ | 10 | 2100 | 5 | 24 |
| $2200-2400$ | 6 | 2300 | 5 | 10 |
| 2400 and above | 2 | 2500 | 5 | 30 |
| Sum | 100 |  |  |  |

$$
\bar{U}=\frac{30}{100}=0.3, \quad \bar{X}=k \bar{U}+x_{0}=200 \cdot 0.3+1500=1560
$$