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lions [1.4K]
3 years ago
11

Nine voltage measurements in a circuit have produced the following values: 12.09, 12.05, 11.94, 11.99, 12.12, 12.05, 12.02, 11.9

5, 12.14. If the standard deviation for the above data set is 0.07V, determine whether any values can be rejected. (5 Points)
Mathematics
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

we can not reject any value

Step-by-step explanation: From data we can test the highest and the lowest value to evaluate if one of these values are out of certain confidence Interval

If we established CI = 95 % then  α = 5 % and  α/2 = 0,025

From data we find the mean of the values

μ₀  = 12,03   and σ = 0,07

From z table we find z score for 0,025 is  z(c) = ± 1,96

So limits of our CI are:

12,03 + 1,96 =  13,99

12,03 - 1,96 = 10,07

And all our values are within  ( 10,07 , 13,99)

So we can not reject any value

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The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo
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Answer:

Follows are the solution to the given points:

Step-by-step explanation:

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\to f_X (x) \ \ xe^{-x} \ , \ x>0

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M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx

integrating the values by parts:

u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0}  -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\  

        = \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\

Therefore, the moment value generating by the function is =\frac{1}{(t-1)^2}

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E(X^n)=?

Using formula: E(X^n)= M^{n}_{X}(0)

form point (a):

\to M_{X}(t)=\frac{1}{(t-1)^2}

Differentiating the value with respect of t

M'_{X}(t)=\frac{-2}{(t-1)^3}

when t=0

M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\

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