Below-average 60271

In a given group of numbers, one number is equal to the average of all, the largest number is 7 larger than the average, the smallest is 7 smaller than the average, and most numbers in the group have below-average values.

What is the smallest number of numbers in a group?

Correct answer:

n =  5

Step-by-step explanation:

$$\begin{gathered} x_1=7+\mu \\ {x}_2=\mu -7 \\ cnt(x_i<\mu )\ge n/2 \\ \mu =\frac{\Sigma x_i}{n} \\ {x}_2<\mu \\ {x}_1>\mu \\ n=2,\mu =0 \\ n=3,x_3=\mu =0 \\ \mu =1 \\ {x}_1=\mu +7=1+7=8 \\ {x}_2=\mu -7=1-7=-6 \\ {x}_3=(\mu \cdot 3-x_1-x_2)/1=(1\cdot 3-8-(-6))/1=1 \\ {\mu }_3=\frac{x_1+x_2+x_3}{3}=\frac{8+(-6)+1}{3}=1 \\ {y}_1=\mu +7=1+7=8 \\ {y}_2=\mu -7=1-7=-6 \\ {y}_3=(\mu \cdot 4-y_1-y_2)/2-1=(1\cdot 4-8-(-6))/2-1=0 \\ {y}_4=(\mu \cdot 4-y_1-y_2-y_3)/1=(1\cdot 4-8-(-6)-0)/1=2 \\ {\mu }_4=\frac{y_1+y_2+y_3+y_4}{4}=\frac{8+(-6)+0+2}{4}=1 \\ cnt(y<\mu )=2 \\ cnt(y\ge \mu )=1+1 \\ {z}_1=\mu +7=1+7=8 \\ {z}_2=\mu -7=1-7=-6 \\ {z}_3=(\mu \cdot 5-z_1-z_2)/3-1=(1\cdot 5-8-(-6))/3-1=0 \\ {z}_4=z_3=0 \\ {z}_5=(\mu \cdot 5-z_1-z_2-z_3-z_5)/1=(1\cdot 5-8-(-6)-0-3)/1=3 \\ {\mu }_5=\frac{z_1+z_2+z_3+z_4+z_5}{5}=\frac{8+(-6)+0+0+3}{5}=1 \\ cnt(z<\mu )=3 \\ cnt(z\ge \mu )=1+1 \\ n=5 \end{gathered}$$

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