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dolphi86 [110]
3 years ago
10

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

tion of the random variable equals 5x=y
Mathematics
1 answer:
Anna35 [415]3 years ago
8 0
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
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you have to cross out both 150 and -150, take 45 and subtract 150 which will give you -105. 
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take -35 and divide by both sides. Cross out the two -35's. Take -105/-35 which your answer is x = 3. Hopes this helps. :)

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