In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by
- 4
- 3
- 5
- 6
Solution:
Let \({x_1}\) be the least number
\({x_{10}}\) be the largest number
Given \(\frac{{{x_2} + {x_3} + \ldots + {x_{10}}}}{9} = 47\)
\({x_2} + {x_3} + \cdots {x_9} + {x_{10}} = 423 \to (1)\)
\(\frac{{{x_1} + {x_2} \cdots + {x_9}}}{9} = 42\)
\({x_1} + {x_2} + \cdots {x_9} = 378\quad \to (2)\)
\((1) - (2) = {x_{10}} - {x_1} = 45\)
Sum of 10 observations
\({x_1} + {x_2} + {x_3} + \cdots {x_{10}} = 423 + {x_1}\)
Since the minimum value of \({x_{10}}\) is 47, the minimum value of \({x_1}\) is 2, minimum average \( = \frac{{423 + 2}}{{10}} = 42.5\)
The maximum value of \({x_1}\) is 42 ,
Maximum average \( = \frac{{423 + 42}}{{10}} = 46.5\)
Required difference =4
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