Mathematics: A test instrument needs to be calibrated periodically to prevent measurement errors. After some time of use without calibration, it is known that the probability density function of the measurement error is f(x)=1-0.5x for 0 x 2 millimeters. Determine the mean, variance and standard deviation of the random variable.

A test instrument needs to be calibrated periodically

Ответ:

gabbym39077

22.01.2022

Mathematics

$E(X) =\frac{x^2}{2} -\frac{0.5}{3}x^3 \Big|_0^2 = (\frac{2^2}{2}- \frac{0.5}{3}*(2^2)) -0 = 0.667 = \frac{2}{3}$

For the variance we need to calculate first the second moment given by:

$E(X^2) = \int_{0}^2 x^2 (1-0.5x) dx=\int_{0}^2 x^2-0.5x^3 dx$

And after solve the integral we got:

$E(X^2) = \frac{x^3}{3} -\frac{0.5}{4} x^4 \Big|_0^2 = \frac{2^3}{3} -\frac{0.5}{4} 2^4 = \frac{8}{3} -2 = \frac{2}{3}$

And for this case the variance would be:

$Var(X) = E(X^2) -[E(X)]^2 = \frac{2}{3} -(2/3)^2 = \frac{2}{9}$

And the deviation would be:

$Sd(X) = \sqrt{\frac{2}{9}}= \frac{\sqrt{2}}{3}$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

For this case w ehave the following density function:

f(x) = 1 -0.5x , 0 < x< 2

We can determine the mean with the following integral:

$E(X) = \int_{0}^2 x(1-0.5x) dx$

And if we solve the integral we got:

$E(X) =\frac{x^2}{2} -\frac{0.5}{3}x^3 \Big|_0^2 = (\frac{2^2}{2}- \frac{0.5}{3}*(2^2)) -0 = 0.667 = \frac{2}{3}$

For the variance we need to calculate first the second moment given by:

$E(X^2) = \int_{0}^2 x^2 (1-0.5x) dx=\int_{0}^2 x^2-0.5x^3 dx$

And after solve the integral we got:

$E(X^2) = \frac{x^3}{3} -\frac{0.5}{4} x^4 \Big|_0^2 = \frac{2^3}{3} -\frac{0.5}{4} 2^4 = \frac{8}{3} -2 = \frac{2}{3}$

And for this case the variance would be:

$Var(X) = E(X^2) -[E(X)]^2 = \frac{2}{3} -(2/3)^2 = \frac{2}{9}$

And the deviation would be:

$Sd(X) = \sqrt{\frac{2}{9}}= \frac{\sqrt{2}}{3}$

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Ответ:

19deleonl

17.01.2021

Mathematics

Therefore,

Priya's way is correct to solve the equation.

Other person Strategyis Equation is

23+ x -23=4-23

Step-by-step explanation:

Given:

Jada and Priya are trying to solve the equation,

23+ x=4

For Jada way of calculation :

I think we should multiply each side by 32 because that is the reciprocal of 23.

Also Reciprocal of 23 is $\dfrac{1}{23}$

We have

$32\times (23+x)=32\times 4$ ...Which is INCORRECT

For Priya way of calculation :

I think we should add -23 to each side because that is the opposite of 23.

We have

23+ x +(-23)=4+(-23)

plus minus= minus , (+ - = - )

23 will get cancelled and she will get "x"

$23+ x -23=4-23\\\\x=-19$ ...Which is CORRECT

Hence Priya's way is correct to solve the equation.

The other Person strategy to solve will be:

I think we should subtract 23 to each side because that is the opposite of 23.

We have

$23+ x -23=4-23\\\\x=-19$ ...Which is also CORRECT.

Therefore,

Priya's way is correct to solve the equation.

Other person Strategyis Equation is

23+ x -23=4-23

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