tristan records the number of customers who visit the store each hour on a saturday. his data representing…

tristan records the number of customers who visit the store each hour on a saturday. his data representing the first seven hours are 15, 23, 12, 28, 20, 18, and 23. how many customers visited the store during the eighth hour if the median number of customers per hour did not change?\n○ 16\n○ 19\n○ 20\n○ 23

tristan records the number of customers who visit the store each hour on a saturday. his data representing the first seven hours are 15, 23, 12, 28, 20, 18, and 23. how many customers visited the store during the eighth hour if the median number of customers per hour did not change?\n○ 16\n○ 19\n○ 20\n○ 23

Answer

Explanation:

Step1: Find median of 7 values

First, sort the data: (12, 15, 18, 20, 23, 23, 28).
For 7 (odd) values, median is the 4th term: (20).

Step2: Analyze median for 8 values

With 8 (even) values, median is average of 4th and 5th terms.
We need median to remain (20), so (\frac{\text{4th term} + \text{5th term}}{2} = 20).
After adding an 8th value, sort the new data. Let the 8th value be (x).

Sorting the original data + (x), the 4th term should be (20) (or close) and 5th term such that their average is (20).

If (x \leq 20), sorted data: (12, 15, 18, x, 20, 23, 23, 28) (wait, no—8 terms: positions 1 - 8. Wait, original 7 terms: positions 1 - 7. Adding 8th, positions 1 - 8.

Wait, original sorted: (12, 15, 18, 20, 23, 23, 28) (7 terms). Adding (x), sort all 8.

To have median (average of 4th and 5th) = 20, we need (\text{4th term} + \text{5th term} = 40).

Original 4th term (in 7 terms) is 20. When adding (x), if (x \leq 20), the 4th term could be (x) or 20? Wait, let's list possible sorted 8-term data:

Case 1: (x \leq 20): sorted as (12, 15, 18, x, 20, 23, 23, 28) → 4th term (x), 5th term (20). Then (\frac{x + 20}{2} = 20) → (x = 20).

Case 2: (20 < x \leq 23): sorted as (12, 15, 18, 20, x, 23, 23, 28) → 4th term (20), 5th term (x). Then (\frac{20 + x}{2} = 20) → (x = 20) (but (x > 20) here, contradiction).

Case 3: (x > 23): sorted as (12, 15, 18, 20, 23, 23, x, 28) → 4th term (20), 5th term (23). Average: (\frac{20 + 23}{2} = 21.5 \neq 20). Not good.

So only when (x \leq 20), and 4th term is (x) or 20? Wait, no—let's take the option 20. If (x = 20), sorted data: (12, 15, 18, 20, 20, 23, 23, 28). Now 4th term: 20, 5th term: 20. Average: (\frac{20 + 20}{2} = 20). Perfect.

Check other options:

  • 16: sorted: (12, 15, 16, 18, 20, 23, 23, 28). 4th term 18, 5th term 20. Average: (\frac{18 + 20}{2} = 19 \neq 20).

  • 19: sorted: (12, 15, 18, 19, 20, 23, 23, 28). 4th term 19, 5th term 20. Average: (\frac{19 + 20}{2} = 19.5 \neq 20).

  • 23: sorted: (12, 15, 18, 20, 23, 23, 23, 28). 4th term 20, 5th term 23. Average: (\frac{20 + 23}{2} = 21.5 \neq 20).

Only when (x = 20), median remains 20.

Answer: 20 (Option: 20)