Sure! Let's go through the steps to solve for the new mean of the remaining distribution after removing the largest value.
1. Determine the initial total sum of observations:
- Given: Mean of initial observations (\(\( \overline{x}_{initial} \)\)) = 5
- Number of initial observations (\(\( n_{initial} \)\)) = 30
- We use the formula for the mean: \(\( \overline{x}_{initial} = \frac{\text{Sum of observations}}{n_{initial}} \)\)
\(\[
\text{Sum of initial observations} = \overline{x}_{initial} \times n_{initial} = 5 \times 30 = 150
\]\)
2. Subtract the largest value from the total sum:
- The largest value to be deleted from the observations = 34
- New sum of observations after removing the largest value:
\(\[
\text{Total sum remaining} = 150 - 34 = 116
\]\)
3. Calculate the new number of observations:
- As we have deleted one value, the new number of observations:
\(\[
n_{remaining} = n_{initial} - 1 = 30 - 1 = 29
\]\)
4. Calculate the new mean:
- The new mean of the remaining distribution using the total sum of remaining observations and the new number of observations:
\(\[
\overline{x}_{remaining} = \frac{\text{Total sum remaining}}{n_{remaining}} = \frac{116}{29} = 4.0
\]\)
Therefore, the mean of the remaining distribution after deleting the largest observed value of 34 is 4.0.
