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larisa [96]
4 years ago
11

PLZ!!!!HELP!!!What is the absolute deviation for 15 in the data set?

Mathematics
2 answers:
choli [55]4 years ago
6 0

Answer: the absolute deviation for 15 is 10

Step-by-step explanation:

Brut [27]4 years ago
5 0
20 + 20 + 30 + 15 + 40 = 125
125/5= 25
25-15 = 10
Absolute deviation for 15 is 10
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1 minute = 0.25km

Step-by-step explanation:

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How can i solve this
lana66690 [7]
X+3-2x+5=10
+2x +2x
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Rufina [12.5K]

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timofeeve [1]
Answer : 12.56 inches²
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2 years ago
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Suppose a certain type of fertilizer has an expected yield per acre of mu 1 with variance sigma 2, whereas the expected yield fo
mart [117]

Answer:

See the proof below.

Step-by-step explanation:

For this case we just need to apply properties of expected value. We know that the estimator is given by:

S^2_p= \frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2}

And we want to proof that E(S^2_p)= \sigma^2

So we can begin with this:

E(S^2_p)= E(\frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2})

And we can distribute the expected value into the temrs like this:

E(S^2_p)= \frac{(n_1 -1) E(S^2_1) +(n_2 -1) E(S^2_2)}{n_1 +n_2 -2}

And we know that the expected value for the estimator of the variance s is \sigma, or in other way E(s) = \sigma so if we apply this property here we have:

E(S^2_p)= \frac{(n_1 -1 )\sigma^2_1 +(n_2 -1) \sigma^2_2}{n_1 +n_2 -2}

And we know that \sigma^2_1 = \sigma^2_2 = \sigma^2 so using this we can take common factor like this:

E(S^2_p)= \frac{(n_1 -1) +(n_2 -1)}{n_1 +n_2 -2} \sigma^2 =\sigma^2

And then we see that the pooled variance is an unbiased estimator for the population variance when we have two population with the same variance.

8 0
3 years ago
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