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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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5x-3=19? Answer for Brainliest
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Answer:

x = 4.4

Step-by-step explanation:

I'm going to assume you want to solve for x so here we go.

You need to work backwards for this equation, and whatever you do to the LHS, you do to the RHS.

First, you need to remove the minus 3, which means that on both sides, you add 3. Adding three on the LHS makes the -3 disappear, and adding 3 on the RHS makes the 19 go to a 22.

Your equation is now 5x=22.

Since 5x means 5 × x, to get rid of it, you need to divide 5x by 5. Doing it to the LHS will make the five disappear, and doing it to the RHS will make it go to 22 ÷ 5 which equals 4.4

Therefore, x = 4.4

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there are 5 walnut trees currently in the park park workers will plant 8 more walnut trees today how many walnut trees will the
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Step-by-step explanation:

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2 years ago
Can someone help<br><br> Write an equivalent expression for<br> 63.62.65.6.
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5 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimeters,
tiny-mole [99]

Answer:

Probability that the sample mean would be greater than 141.4 millimetres is 0.3594.

Step-by-step explanation:

We are given that Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimetres, and a standard deviation of 7.

A random sample of 39 steel bolts is selected.

Let \bar X = <u><em>sample mean diameter</em></u>

The z score probability distribution for sample mean is given by;

                            Z  =  \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} }  } }  ~ N(0,1)

where, \mu = population mean diameter = 141 millimetres

           \sigma = standard deviation = 7 millimetres

           n = sample of steel bolts = 39

Now, Percentage the sample mean would be greater than 141.4 millimetres is given by = P(\bar X > 141.4 millimetres)

      P(\bar X > 141.4) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} }  } } > \frac{141.4-141}{\frac{7}{\sqrt{39} }  } } ) = P(Z > 0.36) = 1 - P(Z \leq 0.36)

                                                            = 1 - 0.6406 = <u>0.3594</u>

The above probability is calculated by looking at the value of x = 0.36 in the z table which has an area of 0.6406.

8 0
3 years ago
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