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Answer in Statistics and Probability for Kyle #91564

Author:
January 24th, 2023

Let a1 be the mark for the first test, a2 be the mark for the second test and so on up to a5. For a5 to be the smallest, the average of all scores must be 70: (a1+a2+a3+a4+a5)/5=70.Because the average of the first four estimates 68: (a1+a2+a3+a4)/4=68 => a1+a2+a3+a4=272 substitute the sum in the previous equation: (272

+a5)/5=70 => 272+a5=350 => a5=78. Answer: 78

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