1734. The examination of the quality of yarn for strength gave the following results:

| Strength of thread, g | 120-140 | 140-160 | 160-180 | 180-200 |
| :--- | :---: | :---: | :---: | :---: |
| Number of cases | 1 | 6 | 19 | 58 |


| Strength of thread, $\boldsymbol{2}$ | 200-220 | 220-240 | 240-260 | 260-280 |
| :--- | :---: | :---: | :---: | :---: |
| Number of cases | 53 | 24 | 16 | 3 |

Find the mode and median of this distribution.
Solution. Since the highest frequency $m_{k}=58$ corresponds to the interval $180-200$, then $X_{k}^{\prime}=180, m_{k-1}=19, m_{k+1}=53$ and

$$
M_{0} X=180+20 \cdot \frac{58-19}{(58-19)+(58-53)}=180+17.73=197.73
$$

To find the median, we construct the cumulative series:

| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $120-140$ | $140-160$ | $160-180$ | $180-200$ | $200-220$ | $220-240$ | $240-260$ | $260-280$ |
| 1 | 7 | 26 | 84 | 137 | 161 | 177 | 180 |

The number of the median interval $s$ is determined from the inequalities

$$
\gamma\left(X_{s}^{\prime}\right) \leqslant 0.5, \quad \gamma\left(X_{s}^{\prime \prime}\right)>0.5
$$

or, equivalently,

$$
\tilde{\gamma}\left(X_{s}^{\prime}\right) \leqslant \frac{n}{2}, \quad \tilde{\gamma}\left(X_{s}^{\prime \prime}\right)>\frac{n}{2}
$$

In our case $\frac{n}{2}=\frac{180}{2}=90, \tilde{\gamma}\left(X_{5}^{\prime}\right)=84 \leqslant 90$ and $\tilde{\gamma}\left(X_{5}^{\prime \prime}\right)=$ $=137>90$. Therefore, the median interval is $200-$ 220, and

$$
M_{\mathrm{e}} X=X_{s}^{\prime}+h \cdot \frac{\frac{n}{2}-\tilde{\gamma}\left(X_{s}^{\prime}\right)}{m_{s}}=200+20 \cdot \frac{90-84}{53}=202.26
$$