เรียงข้อมูลที่ประกอบด้วยจำนวนเต็ม 7 จำนวน จากน้อยไปมาก ได้เป็น 8, a, b, c, d, e, 14 เมื่อค่าเฉลี่ยของข้อมูลชุดนี้เท่ากับ 12 และฐานนิยมคือ 14 จงหาผลบวกของจำนวนเต็มทั้งหมดที่สามารถเป... เรียงข้อมูลที่ประกอบด้วยจำนวนเต็ม 7 จำนวน จากน้อยไปมาก ได้เป็น 8, a, b, c, d, e, 14 เมื่อค่าเฉลี่ยของข้อมูลชุดนี้เท่ากับ 12 และฐานนิยมคือ 14 จงหาผลบวกของจำนวนเต็มทั้งหมดที่สามารถเป็นมัธยฐานได้ Read more

Steps to Solve

1. Calculate the sum of all numbers in the dataset.

Since the average of the 7 numbers is 12, the sum of the numbers is $7 \times 12 = 84$.

2. Express the sum in terms of the unknown variables.

We know that the sum of the numbers is also equal to $8 + a + b + c + d + e + 14 = 84$.

3. Simplify the equation.

Simplifying the equation gives us $a + b + c + d + e = 84 - 8 - 14 = 62$.

4. Consider the mode.

Since the mode is 14, and the numbers are in ascending order, at least two of the numbers $a, b, c, d, e$ must be 14. Let's denote the number of 14s as $n$. Since 14 is the mode, $n \geq 2$. Then $d = e = 14$. So, $a + b + c + 14 + 14 = 62$, which simplifies to $a + b + c = 62 - 28 = 34$.

5. Determine the constraints based on the ordering.

We know that $8 \le a \le b \le c \le 14$. Also, $c$ represents the median.

6. Find the minimum and maximum values for the median, $c$.

Since $a+b+c = 34$ and $a \le b \le c$, the minimum value for $c$ occurs when $a$ and $b$ are as large as possible while still being less than or equal to $c$. If $a = b = c$ then $3c = 34$ or $c = 34/3 \approx 11.33$. Since $a \geq 8$ and $a, b, c$ are integers, consider when $a = 8$. If $a=8$, then $b+c = 26$. Since $b \le c$, we look for a value $b \le 13$.

• if $b = 13, c=13$. This is possible: $8, 13, 13, 14, 14, 14$.
• if $b < 13$, then $c > 13$.

The median is $c$, which is the 4th value, since the list is ordered.

Let's consider the case when $a = b$ such that $2a + c = 34$. Thus, $a = (34-c)/2$. Since $8 \le a \le c$, we have $8 \le (34-c)/2 \le c$. $16 \le 34 - c \le 2c$ $c \le 34 - 16 = 18$. $34 \le 3c$ or $c \ge 34/3 \approx 11.33$. So $12 \le c \le 14$

7. List out the possible values for $a, b, c$ and check validity.

• If $c = 12$: $a + b = 34 - 12 = 22$. Since $a \le b \le c = 12$, the only solution is $a = 10$ , $b = 12$. Sequence: $8, 10, 12, 12, 14, 14, 14$. The median is 12 and the mode is 14. Since $a=10 \ge 8$, this solution is valid.

• If $c = 13$: $a + b = 34 - 13 = 21$. Since $a \le b \le c = 13$, possible solutions: $a=8, b=13$ or $a=9, b=12$, or $a=10, b=11$. Sequence: $8, 8, 13, 13, 14, 14, 14$. The median is 13 and the mode is 14. Since $a \ge 8$, all solutions are valid.

• If $c = 14$: $a + b = 34 - 14 = 20$. Since $a \le b \le c = 14$, possible solutions: $a=6, b=14$ or $a=7, b=13$ ... $a=10, b=10$. We need $a \ge 8$, $9, 11$, $10, 10$. The sequence could look like this.

• If $c = 14$: $a + b = 34 - 14 = 20$. Possible values of $a, b$: $(8, 12), (9, 11), (10, 10)$.

  • $8, 12, 14, 14, 14, 14, 14$.
  • $8, 9, 11, 14, 14, 14, 14$.The sequence could be $8, 9, 11, 14, 14, 14, 14$
  • $8, 10, 10, 14, 14, 14, 14$. The sequence could be $8, 10, 10, 14, 14, 14, 14$.
  • $8, 12, 14, 14, 14, 14, 14$

• The possible medians are now 12, 13 and 14.

8. Sum the possible values of the median. The sum of the possible values for the median is $12+13+14 = 39$.