• Choice A

    Let the 9th number be 91 which is greater than 90.

    $$The\ mean\ of\ Set\ A=(62+76+76+80+82+87+94+96)\div8=\frac{653}{8}=81.625$$ $$The\ median\ of\ Set\ A=(80+82)\div2=81$$ $$The\ mean\ of\ Set\ B=\frac{653+91}{9}\approx82.66$$ $$The\ median\ of\ Set\ B=82$$ $$\Rightarrow 82.66\gt 81.625$$ $$\Rightarrow 82\gt 81$$

    The mean and the median of Set B will each be greater than the mean and the median of Set A.

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