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Zorluk: OrtaMeasures of Central Tendency for Ungrouped Data

The mean of four positive integers is 99. If the mode of the set of numbers is 1212 and their median is 1010, what is the value of the smallest integer?

  1. 44Cevap
  2. B
    22
  3. C
    66
  4. D
    55

Cevap

The smallest integer is 4.
Let the four positive integers in ascending order be abcda \le b \le c \le d. The sum of the four numbers is 4×9=364 \times 9 = 36. For 1212 to be the mode, it must appear at least twice, so c=12c = 12 and d=12d = 12. The median of four numbers is b+c2=10\frac{b + c}{2} = 10, which gives b+122=10\frac{b + 12}{2} = 10, so b=8b = 8. Substituting these into the sum equation a+8+12+12=36a + 8 + 12 + 12 = 36 gives a=4a = 4.

Adım Adım Çözüm

1
Calculate the sum of all four integers from the given mean.
Sum =4×9=36= 4 \times 9 = 36.
The mean of nn numbers is the sum divided by nn.
2
Determine the two largest numbers using the mode.
Let the ordered numbers be abcda \le b \le c \le d. Since the mode is 1212 and the set contains 44 numbers with a single mode, c=12c = 12 and d=12d = 12.
For 1212 to be the mode in a 4-element set without tie, it must appear at least twice.
3
Use the median to find the second integer bb.
Median =b+c2=b+122=10    b+12=20    b=8= \frac{b + c}{2} = \frac{b + 12}{2} = 10 \implies b + 12 = 20 \implies b = 8.
The median of an even number of ordered elements is the arithmetic mean of the two middle elements.
4
Solve for the smallest integer aa.
a+b+c+d=36    a+8+12+12=36    a+32=36    a=4a + b + c + d = 36 \implies a + 8 + 12 + 12 = 36 \implies a + 32 = 36 \implies a = 4.
Subtracting the sum of the known three numbers from the total sum gives the smallest integer.

Anahtar Kavram

Using mean, median, and mode definitions simultaneously to deduce unknown values in an ungrouped dataset
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