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In a pizza hut, the following distribution is found for the daily demand of pizzas.
Then expected daily demand and variance are respectively
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Then expected daily demand and variance are respectively
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The correct answer is:
7.28 and 1.52
Mean $=\Sigma P_i x_i=7.28$
and variance $=\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}^2}-\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=54.52-53=1.52$
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