Then expected daily demand and variance are respectivelyIn a pizza hut, the following distribution is found for the daily demand of pizzas. Then expected daily…
Then expected daily demand and variance are respectively- 7.28 and 1.52
- 1.52 and 7.28
- 7.28 and 54.52
- 7.28 and 53
Solution
Mean $=\Sigma P_i x_i=7.28$
and variance $=\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}^2}-\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=54.52-53=1.52$Asked in: MHT CET 2022 (05 Aug Shift 2)