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Zorluk: OrtaMeasures of Central Tendency (Mean, Median, Mode)

A list consists of six numbers: 33, 77, 1010, 1414, 1818, and xx. If the arithmetic mean of these six numbers is equal to their median, which of the following could be the value of xx? Indicate all such values.

  1. 1-1Cevap
  2. B
    55
  3. 1111Cevap
  4. D
    1414
  5. 2020Cevap

Cevap

The valid values for xx are 1-1, 1111, and 2020.
The values 1-1, 1111, and 2020 each yield a dataset where the arithmetic mean equals the median: 1-1 gives a mean and median of 8.58.5, 1111 gives a mean and median of 10.510.5, and 2020 gives a mean and median of 1212.

Adım Adım Çözüm

1
Express the arithmetic mean in terms of xx.
Mean = 3+7+10+14+18+x6=52+x6\frac{3 + 7 + 10 + 14 + 18 + x}{6} = \frac{52 + x}{6}.
The mean of a dataset of nn numbers is the sum of all elements divided by nn.
2
Analyze the median based on the position of xx relative to the sorted known values 3,7,10,14,183, 7, 10, 14, 18.
Case 1: x7    x \le 7 \implies median = 7+102=8.5\frac{7+10}{2} = 8.5.
Case 2: 7<x<14    7 < x < 14 \implies median = x+102\frac{x+10}{2} (for 7<x107 < x \le 10) or 10+x2\frac{10+x}{2} (for 10<x<1410 < x < 14).
Case 3: x14    x \ge 14 \implies median = 10+142=12\frac{10+14}{2} = 12.
For an even number of elements (n=6n=6), the median is the average of the 3rd and 4th terms in ascending order.
3
Set the mean equal to the median for each case and solve for xx.
Case 1: 52+x6=8.5    52+x=51    x=1\frac{52+x}{6} = 8.5 \implies 52+x = 51 \implies x = -1 (valid since 17-1 \le 7).
Case 2: 52+x6=10+x2    52+x=30+3x    2x=22    x=11\frac{52+x}{6} = \frac{10+x}{2} \implies 52+x = 30+3x \implies 2x = 22 \implies x = 11 (valid since 7<11<147 < 11 < 14).
Case 3: 52+x6=12    52+x=72    x=20\frac{52+x}{6} = 12 \implies 52+x = 72 \implies x = 20 (valid since 201420 \ge 14).
This identifies all values of xx satisfying the problem constraint.

Anahtar Kavram

Evaluating mean and median of a dataset containing an unknown variable across different intervals of the variable's possible values.
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