When rounded to the nearest thousand,i become 3,000.what numbers can i be

Mathematics

1 answer:

cupoosta [38]2 years ago

4 0

I believe it's any number between 2,501 and 3,499.

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6 0

3 years ago

Consider a population list x with μx=10 and SDx = 1. A second population list, y, with μy=10 and SDy=2, is added to the first li

Basile [38]

Answer:

The combined standard deviation is 1.58114.

Step-by-step explanation:

The formula to compute the combined standard deviations of two different data sets is:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}$

Here $\mu_{c}$ is the combined mean given by:

$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}$

It is provided that the sample size is same for both the data sets, i.e. $n_{X} = n_{Y}=n$

Compute the combined mean as follows:

$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}\\=\frac{(n\times10)+(n\times10)}{n+n}}\\=\frac{20n}{2n}\\ =10$

Compute the combined standard deviation as follows:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}}\\=\sqrt{\frac{(n\times1^{2})+(n\times2^{2})+(n(10-10))+(n(10-10))}{n+n}}\\=\sqrt{\frac{n+4n}{2n} } \\=\sqrt{\frac{5n}{2n} } \\=\sqrt{\frac{5}{2}} \\=1.58114$