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notsponge [240]

3 years ago

13

There were 8 pizzas at the pizza party for two families. The Houston family at 2 2/3 pizzas. The Ozark ate 2 5/6 pizzas. What is

the whole number in the closest estimate of much pizza was left

1 answer:

xz_007 [3.2K]3 years ago

6 0

2 plus 2 equals 4 so then they almost finished two pizzas right there so you have 5 2/3

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Answer:

$E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8$

Now we can find the second moment with this formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4$

The variance would be given by:

$Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96$

And the deviation would be:

$Sd(X) =\sqrt{6.96}= 2.638$

Step-by-step explanation:

For this case we have the following distribution given:

X 1 2 7

P(X) 1/5 2/5 2/5

We need to begin finding the mean with this formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8$

Now we can find the second moment with this formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4$

The variance would be given by:

$Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96$

And the deviation would be:

$Sd(X) =\sqrt{6.96}= 2.638$

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3 years ago

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Elan Coil [88]

Answer:

Yes

Step-by-step explanation:

4 0

3 years ago

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