Let dataset A have observations $a_1, a_2, \dots, a_{55}$. The new dataset C is created by adding 2023 to each observation in A. So, dataset C has observations $c_1, c_2, \dots, c_{55}$, where $c_i = a_i + 2023$ for $i = 1, \dots, 55$. We need to analyze how adding a constant like 2023 affects different statistics of the dataset.

Dataset Mean Analysis

The mean of a dataset is the sum of all observations divided by the number of observations.

Mean of A ($\bar{A}$) = $\frac{1}{55} \sum_{i=1}^{55} a_i$

Mean of C ($\bar{C}$) = $\frac{1}{55} \sum_{i=1}^{55} c_i = \frac{1}{55} \sum_{i=1}^{55} (a_i + 2023)$

We can separate the summation:

$\bar{C} = \frac{1}{55} \left( \sum_{i=1}^{55} a_i + \sum_{i=1}^{55} 2023 \right) = \frac{1}{55} \sum_{i=1}^{55} a_i + \frac{1}{55} \sum_{i=1}^{55} 2023$

The first part is the mean of A, and the second part is the sum of 55 times 2023, divided by 55.

$\bar{C} = \bar{A} + \frac{55 \times 2023}{55} = \bar{A} + 2023$

So, Mean of C = 2023 + Mean of A. This statement (Option 1) is true.

Dataset Variance Analysis

Variance measures the spread or dispersion of the dataset around its mean. The variance of A ($\sigma_A^2$) is $\frac{1}{55} \sum_{i=1}^{55} (a_i - \bar{A})^2$.

The variance of C ($\sigma_C^2$) is $\frac{1}{55} \sum_{i=1}^{55} (c_i - \bar{C})^2$.

We know $c_i = a_i + 2023$ and $\bar{C} = \bar{A} + 2023$. Substitute these into the variance formula for C:

$\sigma_C^2 = \frac{1}{55} \sum_{i=1}^{55} ((a_i + 2023) - (\bar{A} + 2023))^2$

Simplify the term inside the parenthesis:

$((a_i + 2023) - (\bar{A} + 2023)) = a_i + 2023 - \bar{A} - 2023 = a_i - \bar{A}$

So, $\sigma_C^2 = \frac{1}{55} \sum_{i=1}^{55} (a_i - \bar{A})^2 = \sigma_A^2$.

Therefore, Variance of C = Variance of A. Adding a constant does not change the variance. This statement (Option 2) is true.

Dataset Median Analysis

The median is the middle value in a dataset when it is ordered. Since dataset A has 55 distinct observations, we can order them $a_{(1)} < a_{(2)} < \dots < a_{(55)}$. The median of A is the $(55+1)/2 = 28^{\text{th}}$ observation in the ordered list, i.e., $a_{(28)}$.

When we add 2023 to each observation in A to get dataset C, the order is preserved because if $a_i < a_j$, then $a_i + 2023 < a_j + 2023$. So, the ordered observations of C will be $c_{(1)} < c_{(2)} < \dots < c_{(55)}$, where $c_{(i)} = a_{(i)} + 2023$.

The median of C is the $28^{\text{th}}$ observation in the ordered list of C, i.e., $c_{(28)}$.

Median of C = $c_{(28)} = a_{(28)} + 2023$ = Median of A + 2023.

So, Median of C = 2023 + Median of A. This statement (Option 4) is true.

Dataset Maximum and Minimum (Range) Analysis

Let Max_A be the maximum value in A and Min_A be the minimum value in A. Since A has distinct observations, Max_A and Min_A are unique values.

When we add 2023 to every observation, the maximum value in C will be Max_C = Max_A + 2023, and the minimum value in C will be Min_C = Min_A + 2023.

The difference between the maximum and minimum values (also known as the range) for dataset C is:

(Maximum – Minimum) of C = Max_C – Min_C = (Max_A + 2023) – (Min_A + 2023)

= Max_A + 2023 – Min_A – 2023 = Max_A – Min_A

This is equal to (Maximum – Minimum) of A.

So, (Maximum – Minimum) of C = (Maximum – Minimum) of A. Adding a constant does not change the range.

The statement in Option 3 is (Maximum – Minimum) of C = 2023 + (Maximum – Minimum) of A. This statement is false because the range is unchanged, not increased by 2023.

Based on the analysis, the statement that is NOT true is (Maximum – Minimum) of C = 2023 + (Maximum – Minimum) of A.