Question

the numbers of trading cards owned by 10 middle - school students are given below. (note that these are already ordered from least to greatest.)
363, 396, 399, 446, 448, 492, 510, 546, 612, 628
suppose that the number 628 from this list changes to 498. answer the following.
(a) what happens to the
median?
\\(\circ\\) it decreases by \\(\square\\).
\\(\circ\\) it increases by \\(\square\\).
\\(\circ\\) it stays the same.
(b) what happens to the
mean?
\\(\circ\\) it decreases by \\(\square\\).
\\(\circ\\) it increases by \\(\square\\).
\\(\circ\\) it stays the same.

Explanation:

Part (a)

Step1: Find original median

For \( n = 10 \) (even), median is average of \( \frac{n}{2} \)-th and \( (\frac{n}{2}+1) \)-th terms.
Original data: \( 363, 396, 399, 446, 448, 492, 510, 546, 612, 628 \)
\( \frac{10}{2}=5 \)-th term: \( 448 \); \( \frac{10}{2}+1 = 6 \)-th term: \( 492 \)
Median \( = \frac{448 + 492}{2} = \frac{940}{2} = 470 \).

Step2: Find new median (628 → 498)

New data (ordered): \( 363, 396, 399, 446, 448, 492, 498, 510, 546, 612 \)
\( 5 \)-th term: \( 448 \); \( 6 \)-th term: \( 492 \)
Median \( = \frac{448 + 492}{2} = 470 \).

Step3: Compare medians

Original median = 470, new median = 470. So median stays the same.

Step1: Calculate original mean

Sum of original data:
\( 363 + 396 + 399 + 446 + 448 + 492 + 510 + 546 + 612 + 628 \)
\( = 363+396=759 \); \( 759+399=1158 \); \( 1158+446=1604 \); \( 1604+448=2052 \);
\( 2052+492=2544 \); \( 2544+510=3054 \); \( 3054+546=3600 \); \( 3600+612=4212 \); \( 4212+628=4840 \).
Mean \( = \frac{4840}{10} = 484 \).

Step2: Calculate new mean (628 → 498)

New sum: \( 4840 - 628 + 498 = 4840 - 130 = 4710 \).
New mean \( = \frac{4710}{10} = 471 \).

Step3: Find change in mean

Original mean = 484, new mean = 471.
Change: \( 471 - 484 = -13 \) (decreases by 13).

Answer:

(a):
It stays the same.

Part (b)