Approximate the 5839 to the nearest 1:Ten 2:hundred Thousand 3:

Mathematics

1 answer:

nirvana33 [79]2 years ago

6 0

Answer:

(1): nearest ten is 5840

(2):nearest hundred is 5800

(3):nearest thousand is 6000

Step-by-step explanation:

For nearest ten we get 5840

For nearest hundred we get 5800

For nearest thousand we get 6000

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Vera_Pavlovna [14]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Suppose that X has a discrete uniform distribution on the integers 0 through 9. Determine the mean, variance, and standard devia

Anna35 [415]

With 10 integers available, $X$ has PMF

$\mathbb P(X=x)=\begin{cases}\dfrac1{10}&\text{for }0\le x\le9,x\in\mathbb Z\\\\0&\text{otherwise}\end{cases}$

We're interested in the statistics of the new random variable $Y=5X$ . To do this, we need to know the PMF for $Y$ . This isn't too hard to find.

$\mathbb P(Y=y)=\mathbb P(5X=y)=\mathbb P\left(X=\dfrac y5\right)$

Since the PMF for $X$ gives a value of $\dfrac1{10}$ whenever $x$ is an integer between 0 and 9, it follows that $\dfrac y5$ must also be an integer for the PMF to give the identical value of $\dfrac1{10}$ . This means

$\mathbb P(Y=y)=\begin{cases}\dfrac1{10}&\text{for }y\in\{0,5,\ldots,45\}\\\\0&\text{otherwise}\end{cases}$

Now the mean (expectation) is

$\mathbb E(Y)=\displaystyle\sum_yy\mathbb P(Y=y)=\frac1{10}\sum_{y\in\{0,\ldots,45\}}y$
$\mathbb E(Y)=\dfrac{225}{10}=22.5$

The variance would be

$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb V(X)=\displaystyle\sum_yy^2\mathbb P(Y=y)-\mathbb E(Y)^2$
$\mathbb V(X)=\displaystyle\frac1{10}\sum_{y\in\{0,\ldots,45\}}y^2-\left(\frac{225}{10}\right)^2$
$\mathbb V(X)=\dfrac{7125}{10}-\dfrac{50625}{100}=206.25$

The standard deviation is the square root of the variance, so you have

$\sqrt{\mathbb V(X)}=\sqrt{206.25}\approx14.3614$