In a pizza hut,the following distribution is | vedclass

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(A) The expected daily demand (mean) is given by $E(X) = \Sigma P_i x_i$.$E(X) = (5 	imes 0.07) + (6 	imes 0.2) + (7 	imes 0.3) + (8 	imes 0.3) +

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$7.28$ and $1.52$

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(A) The expected daily demand (mean) is given by $E(X) = \Sigma P_i x_i$.$E(X) = (5 	imes 0.07) + (6 	imes 0.2) + (7 	imes 0.3) + (8 	imes 0.3) + (9 	imes 0.07) + (10 	imes 0.06)$$E(X) = 0.35 + 1.2 + 2.1 + 2.4 + 0.63 + 0.6 = 7.28$.The variance is given by $Var(X) = \Sigma P_i x_i^2 - (E(X))^2$.First,calculate $\Sigma P_i x_i^2$:$\Sigma P_i x_i^2 = (25 	imes 0.07) + (36 	imes 0.2) + (49 	imes 0.3) + (64 	imes 0.3) + (81 	imes 0.07) + (100 	imes 0.06)$$\Sigma P_i x_i^2 = 1.75 + 7.2 + 14.7 + 19.2 + 5.67 + 6.0 = 54.52$.Now,calculate variance:$Var(X) = 54.52 - (7.28)^2$$Var(X) = 54.52 - 52.9984 \approx 1.52$.Thus,the expected demand is $7.28$ and the variance is $1.52$.

No. of Pizzas $(x_i)$	$5$	$6$	$7$	$8$	$9$	$10$
Probability $(P_i)$	$0.07$	$0.2$	$0.3$	$0.3$	$0.07$	$0.06$

In a pizza hut,the following distribution is found for the daily demand of pizzas. Then the expected daily demand and variance are respectively:
No. of Pizzas $(x_i)$ $5$ $6$ $7$ $8$ $9$ $10$
Probability $(P_i)$ $0.07$ $0.2$ $0.3$ $0.3$ $0.07$ $0.06$

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In a pizza hut,the following distribution is found for the daily demand of pizzas. Then the expected daily demand and variance are respectively:No. of Pizzas $(x_i)$$5$$6$$7$$8$$9$$10$Probability $(P_i)$$0.07$$0.2$$0.3$$0.3$$0.07$$0.06$