Find the mean score to the nearest whole number. Then, find the mean of the remaining distribution if the largest value of 34 is deleted from a set of 30 observations with a mean of 15.

The given data is:
Marks ($x$): 4, 5, 6, 7, 8, 9
Frequency ($f$): 5, 8, 5, 6, 4, 2

$$\sum f = 5 + 8 + 5 + 6 + 4 + 2 = 30$$ $$\sum fx = (4 \times 5) + (5 \times 8) + (6 \times 5) + (7 \times 6) + (8 \times 4) + (9 \times 2)$$ $$\sum fx = 20 + 40 + 30 + 42 + 32 + 18 = 182$$

$$\text{Mean} \approx 6$$

The mean score to the nearest whole number is $\boxed{\text{6}}$.

: The mean of 30 observations recorded in an experiment is 15. If the observation with the largest value of 34 is deleted, find the mean of the remaining distribution.

Given: Number of observations ($N_1$) = 30, Mean ($\bar{x}_1$) = 15.
The sum of observations ($\sum x_1$) is given by $\sum x_1 = \bar{x}_1 \times N_1$.

$$\sum x_1 = 15 \times 30 = 450$$

The observation with value 34 is deleted.

$$\text{New sum of observations} (\sum x_2) = \sum x_1 - 34 = 450 - 34 = 416$$

After deleting one observation, the new number of observations ($N_2$) is:

$$N_2 = 30 - 1 = 29$$

The new mean ($\bar{x}_2$) is given by $\bar{x}_2 = \frac{\sum x_2}{N_2}$.

$$\bar{x}_2 = \frac{416}{29} \approx 14.3448$$

The mean of the remaining distribution is $\boxed{\text{14.34}}$.