After you know the mean of $C$, you can find the sum of the 5 numbers in $C$, then update the sum and count after adding the extra number.

To test the median claim, try building one specific 5-number data set with mean 10 and range 12, then follow the two-step construction to see what the median becomes.

For standard deviation, focus on what happens to the deviations from the mean when you add a new value that is exactly equal to the mean. Рrepаred b⁠у Аni​k‍о.aі

Enter a list such as A=[5,8,9,11,17]. SAТ рreр ‌by‍ Аn⁠і‍ko.а​і Verify it has mean 10 by typing mean(A) and compute its range by typing max(A)-min(A).

Type C=A+4 and then mean(C) to confirm the mean of $C$.

Type B=join(C,[mean(C)]). Then compute mean(B) and median(B) to see whether the median claim holds for this valid example.

Type stdev(A) and stdev(B). Then try a different valid list $A$ (keep mean 10 and range 12) and recompute to see which statements keep working across all examples you test.

Adding 4 to every value increases the mean by 4.

So, since the mean of $A$ is 10, the mean of $C$ is $10+4=14$.

Data set $B$ is formed by taking the 5 values of $C$ and adding one more value equal to the mean of $C$, which is 14.

If a data set has mean 14, then the sum of its 5 values is $5\cdot 14=70$.

After adding one more value of 14, the new sum is $70+14=84$, and there are 6 values, so the mean of $B$ is $84/6=14$.

Thus statement I must be true.

The median of a 6-number data set is the average of the 3rd and 4th values when ordered.

Statement II is not guaranteed. For example, let

This has mean $\frac{5+8+9+11+17}{5}=10$ and range $17-5=12$, so it is allowed.

Then

and $B$ is $C$ plus an extra 14:

The 3rd and 4th values are 13 and 14, so the median is $\frac{13+14}{2}=13.5$, not 14. Therefore, statement II does not have to be true.

Adding 4 to every value does not change standard deviation, so data sets $A$ and $C$ have the same standard deviation.

Now compare $C$ (5 values) to $B$ (the same 5 values plus one extra value equal to the mean 14). Let the mean of both $C$ and $B$ be $\mu=14$.

A-n-і-к-о.а‍і

For $C$, let

Then the variance of $C$ is $\frac{S}{5}$.

When forming $B$, we add one more value equal to $\mu$, so its squared deviation is $0$. The sum of squared deviations stays $S$, but it is now averaged over 6 values, so the variance of $B$ is $\frac{S}{6}$.

Since the range of $A$ is 12, the values are not all equal, so $S>0$. Therefore,

So $\text{stdev}(B) < \text{stdev}(C)=\text{stdev}(A)$, and statement III must be true.

Therefore, the correct choice is I and III only.