Please answer Q3.11 The following measurements were taken with an analogue meter

Question

Please answer Q3.11 The following measurements were taken with an analogue meter of the current flowing in a circuit (the circuit was in steady state and therefore, although the measurements varied due to random errors, the current flowing was actually constant): 21.5mA, 22.1 mA, 21.3mA, 21.7mA, 22.0mA, 22.2mA, 2I.S mA, 21.4 mA, 21.9 mA, 22.1mA Calculate the mean value, the deviations from the moan and the standard deviation. The measurements in a data set are subject to random errors but it is known that the data set fits a Gaussian distribution. Use standard Gaussian tables to determine the percentage of measurements that lie within the boundaries of plusminus 1.5sigma. where sigma is the standard deviation of the measurements. The thickness of a set of gaskets varies because of random manufacturing disturbances but the thickness values measured belong to a Gaussian distribution. If the mean thickness is 3 mm and the standard deviation is 0.25. calculate the percentage of gaskets that have a thickness greater than 2.5 mm. Determine the values of v_2 - v_1 that will saturate an op amp for each of the following condition sets: A = 10^4. +V_cc = 20V. - V_cc = -20V A = 10^7 +V_cc = 20V. - V_cc = -20V A = 4 times 10^5. + V_cc = 16 V. - V_cc = - 18 V A = 2 times 10^6. + V_cc = 18 V. - V_cc = -16 V Determine the value of v_2 - v_1 that will produce positive saturation of an op amp for each of the following condition sets: A = 10^4. + V_cc = 20V_cA_c = 0.1.v_c = 8V A = 10^7. + V_cc = 20V.A_c = 0.1.v_c = 8V A = 4 times 10^5. + V_cc = 16 V. A_c = 0.08. v_c = 6V A = 2 times 10^6. +V_cc = 18V.A_c = 0.12.v_c = 5V For each of the following condition sets, determine the CMMR and CMR required to limit the common-mode error to a maximum of 2% of the output when v_c = 0 V A = 10^4. v_2 - v_1 = 1.0 mV. maximum v_c = 8 V A = 10^7. v_2 - v_1 = 0.8 mu V. maximum v_c = 8V A = 4 times 10^5. v_2 - v_1 = 20 mu V. maximum u_c = 6 V A = 2 times 10^6. v_2 - v_1 = 5 mu V. maximum v_c = 5 V A = 2 times 10^5. v_2 - v_1 = 80 mu V. maximum v_c = 5 V

Explanation / Answer

In the question you asked about Mean value, deviation from the mean and Standard deviation.

Mean is defined as sum of samples divided by number of sample.

In the question 10 readings are given, therefore

Mean= Sum of readings/no of readings;

Sum of readings=21.5+22.1+21.3+21.7+22.0+22.2+21.8+21.4+21.9+22.1=218 mA

No of readings=10

There Mean=218/10=21.8 mA.

Deviation from mean is difference between actual reading and mean.

Deviation calculation is shown in table below.

Standard Deviation:

To calculate standard deviation first calculate square of deviation. Then sum the values.

After summation divide the sum by number of reading and then take square root.

From the above table we can get deviation.

Standard deviation = square root(Sum of square of deviation/No of readings)

Sum of square of deviation = (-0.3)2 +(0.3)2 +(-0.5)2 +(-0.1)2 +(0.2)2 + (0.4)2 +(0)2 +(-0.4)2 + (0.1)2 +(0.3)2 = 0.9

No of readings=10

Standard deviation =sqrt(0.9/10)=0.3

Sl no Reading (mA) Mean (mA) Deviation (mA) (Reading-Mean) 1 21.5 21.8 -0.3 2 22.1 21.8 +0.3 3 21.3 21.8 -0.5 4 21.7 21.8 -0.1 5 22.0 21.8 +0.2 6 22.2 21.8 +0.4 7 21.8 21.8 0 8 21.4 21.8 -0.4 9 21.9 21.8 +0.1 10 22.1 21.8 +0.3

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