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aleksklad [387]

2 years ago

7

WILL MARK BRANLIEST the first one plz

1 answer:

ruslelena [56]2 years ago

3 0

Answer:

A,D,E, and F

Step-by-step explanation:

They all equal -1/4.

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Set A of six numbers has a standard deviation of 3 and set B of four numbers has a standard deviation of 5. Both sets of numbers

Alenkinab [10]

Given:

$\sigma_A=3$

$n_A=6$

$\sigma_B=5$

$n_B=4$

$\overline{x}_A=\overline{x}_B$

To find:

The variance. of combined set.

Solution:

Formula for variance is

$\sigma^2=\dfrac{\sum (x_i-\overline{x})^2}{n}$ ...(i)

Using (i), we get

$\sigma_A^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{n_A}$

$(3)^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$9=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$54=\sum (x_i-\overline{x}_A)^2$

Similarly,

$\sigma_B^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{n_B}$

$(5)^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$25=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$100=\sum (x_i-\overline{x}_B)^2$

Now, after combining both sets, we get

$\sigma^2=\dfrac{\sum (x_i-\overline{x}_A)^2+\sum (x_i-\overline{x}_B)^2}{n_A+n_B}$

$\sigma^2=\dfrac{54+100}{6+4}$

$\sigma^2=\dfrac{154}{10}$

$\sigma^2=15.4$

Therefore, the variance of combined set is 15.4.

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