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IrinaVladis [17]
3 years ago
14

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

have an equal mean. If the two sets of numbers are combined, find the variance.​
Mathematics
1 answer:
Alenkinab [10]3 years ago
7 0

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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Step-by-step explanation:

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The probability that the mean of the sample is greater than $325,000

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                           = 0.5 - A(2.413)

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<u><em>Final answer:-</em></u>

The probability that the mean of the sample is greater than $325,000

P( X > 3,25,000) = P( Z >2.413) = 0.008

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