Question

An engineer wants to measure a voltage $v$. Being aware of random error in each measurement, the engineer takes $n$ measurements and reports the average of the $n$ measurements as the estimated value of $v$. That is, assume that the $i$-th measurement, $Y_i$, is $Y_i = v + X_i$ where $X_i$ is the error in the $i$-th measurement. Assuming that the $X_i$'s are IID random variables with $E[X_i] = 0$ and $Var[X_i] = 0.0025$ volts$^2$, the engineer reports the average of the measurements as $M_n = \frac{Y_1 + Y_2 + ... + Y_n}{n}$ Determine the minimum number of measurements the engineer needs to make so that she is 95\% sure that the reported value is within 0.01 volts of the true value of $v$, i.e., find the value of $n$ such that $P[v - 0.01 \le M_n \le v + 0.01] \ge 0.95$ Round your answer to the nearest integer.

          An engineer wants to measure a voltage $v$. Being aware of random error in each measurement, the engineer takes $n$ measurements and reports the average of the $n$ measurements as the estimated value of $v$. That is, assume that the $i$-th measurement, $Y_i$, is
$Y_i = v + X_i$
where $X_i$ is the error in the $i$-th measurement. Assuming that the $X_i$'s are IID random variables with $E[X_i] = 0$ and $Var[X_i] = 0.0025$ volts$^2$, the engineer reports the average of the measurements as
$M_n = \frac{Y_1 + Y_2 + ... + Y_n}{n}$
Determine the minimum number of measurements the engineer needs to make so that she is 95\% sure that the reported value is within 0.01 volts of the true value of $v$, i.e., find the value of $n$ such that
$P[v - 0.01 \le M_n \le v + 0.01] \ge 0.95$
Round your answer to the nearest integer.

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An engineer wants to measure a voltage v. Being aware of random error in each measurement, the engineer takes n measurements and reports the average of the n measurements as the estimated value of v. That is, assume that the i-th measurement, Yi, is
Yi = v + Xi
where Xi is the error in the i-th measurement. Assuming that the Xi's are IID random variables with E[Xi] = 0 and Var[Xi] = 0.0025 volts^2, the engineer reports the average of the measurements as
Mn = (Y1 + Y2 + ... + Yn)/(n)
Determine the minimum number of measurements the engineer needs to make so that she is 95% sure that the reported value is within 0.01 volts of the true value of v, i.e., find the value of n such that
P[v - 0.01 ≤ Mn ≤ v + 0.01] ≥ 0.95
Round your answer to the nearest integer.

Added by Stacy B.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Sri K

01/28/2023

Step 1

0025$. Therefore, $M_n$ is also a random variable with mean $v$ and variance $\frac{0.0025}{n}$. Show more…

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An engineer wants to measure a voltage $v$. Being aware of random error in each measurement, the engineer takes $n$ measurements and reports the average of the $n$ measurements as the estimated value of $v$. That is, assume that the $i$th measurement, $Y_i$, is $$Y_i = v + X_i$$ where $X_i$ is the error in the $i$th measurement. Assuming that the $X_i$'s are IID random variables with $E[X_i] = 0$ and $$Var[X_i] = 0.0025 \text{ volts}^2$$, the engineer reports the average of the measurements as $$M_n = \frac{Y_1 + Y_2 + ... + Y_n}{n}$$ Determine the minimum number of measurements the engineer needs to make so that she is 95% sure that the reported value is within 0.01 volts of the true value of $v$, i.e., find the value of $n$ such that $$Pr[v - 0.01 \le M_n \le v + 0.01] \ge 0.95$$ Round your answer to the nearest integer.

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00:01 Hello students, today we will discuss about this question.

00:03 In this question we are given that an engineer is measuring a quantity cube.

00:09 It is assumed that there is a random error in each measurement.

00:13 So the engineer will take an measurement and report the average of the measurement as the estimated value of q.

00:21 So therefore significantly if yi is the value that is obtained in the ith method, then we can assume that yi is equal to q plus xi where xi is error in the ith measurement and the engineer reports the average of measurement that is mn is equals to y1 plus y2 plus up to yn now how many measurements does the engineer need to take until he is 95 percentage of sure that the final error is less than 0 .1 units so here we need to find the number of samples that is equals to question mark.

01:04 So first of all, we have given that yi is equals to q plus xi.

01:11 Now it is given that x are the identically and independently distributed random variable with mean is equals to zero and variance is equals to four.

01:22 So the mean and the variance of yi that can be computed as e of yi is equal to e of q plus x i e of x i so that is equals to we can write q plus e of x i so that is equals to we can write q because the mean of x i that is equals to zero and here variance of y i that is equals to variance of q plus variance of xi, so that is equals to variance of q that is zero plus variance of xi that is equals to 4, so is equals to 4.

02:01 The average of the measurement that can be calculated as the following expression, so that is mn is equal to y1 plus y2 plus up to plus yn divided by n...

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