Arithmetic Mean This represents the most probable value of the measured variable. The more readings you take, the more accurate result you will get.

: Example Readings given as x1 , x2 , x3 ,…….., xn, the mean ( x̅ ) is given is : x̅ = (x1 + x2 + …….. + xn ) / n, As an example, it might be possible to get a precise value for the number of people in a room. However, if there are 50 people moving about it , might be difficult. You might count 3 or 4 times and get a different value each time. In science there is often a wide range of acceptable values . NOTE: For several number of readings, there is not an absolute answer (x̅). This fact leads to the concept of deviation.

Deviation from the mean This represents the departure of a reading from the mean. + = d1 = x1 - x̅ - = d2 = x2 - x̅ . = dn = xn - x̅ NOTE: Notice that any deviation may have a (+) or a(-) value .However, the algebraic sum of all deviation must be zero.

Example: Consider the following readings that have been taken by an ammeter: 12.8 mA 12.2 mA 12.5 mA *Mean: (x̅)= (12.8+12.2+12.5)/3 =12.5 mA. *Deviations: d1 =12.8 - 12.5 = 0 .3 d2 = 12.2 -12.5 = -0.3 d3 = 12.5 - 12.5 = 0

Average Deviation The average deviation is an estimate of how far off the actual values are from the average value assuming that the measuring device is accurate. So, it is an indication of an instrument precision (highly precise instrument yields a low average deviation between readings). * By definition, average Deviation: (Davg): For the previous example: 3 = 0.2 mA / ( Davg = ( |0.3| + | -0.3| + |0|

Standard Deviation The standard deviation is a more accurate method of finding the error margin than the average deviation method. However, the (Davg ) is relatively easier to calculate. NOW: Finding the (Ϭ) of an entire population is unrealistic. Therefore, in most cases, the (Ϭ) is estimated by examining a random sample taken from the population. Hence : The most common measure used is :

Variance V = Ϭ2 The variance is a convenient quantity to work with. The variance is given as the mean square deviation: V = Ϭ2 The variance is a convenient quantity to work with. NOTE: In general, the standard deviation of a set of readings should be less than (10%) of the average value of readings to insure that the average value is an acceptable indication of the true value.

Example: X1 = 147.2 X2 = 147.4 X3 = 147.9 X4 = 147.8 X5 = 147.1 X6 = 147.5 X7= 147.6 X8 = 147.4 X9 = 147.6 X10 = 147.5 Solution: *Mean: = ( X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 ) / 10 = 147.5 *Average deviation: d1 = 147.2 – 147.5= -0.3 d2 = 147.4 – 147.5= -0.1 d3 = 147.9 – 147.5= 0.4 d4 = 147.8 – 147.5= 0.3 d5 = 147.1 – 147.5=-0.4 d6 = 147.5 – 147.5= 0 d7= 147.6 – 147.5= 0.1 d8 = 147.4 – 147.5= -0.1 d9= 147.6 – 147.5= 0.1 d10 = 147.5 – 147.5= 0

(d1)2 = 0. 09 (d2)2 = 0. 1 (d3)2 = 0. 16 (d4)2 = 0. 09 (d5)2 = 0 (d1)2 = 0.09 (d2)2 = 0.1 (d3)2 = 0.16 (d4)2 = 0.09 (d5)2 = 0.16 (d6)2 = 0 (d7)2 = 0.01 (d8)2 = 0.01 (d9)2 = 0.01 (d10)2 = 0 Davg = 0.2 *Standard deviation: = √(0.81/9) = 0.3 * To see if the mean value is a suitable indication or no ??? (.3/147.5) *100% =.2 %= (x̅) Ϭ / ** Since (0.2 %) is less than (10 %) the mean value is a suitable indication.

*Probability of error: Consider having (50) voltage readings recorded to the nearest (0.1 V). Value # of readings 100.0 19 99.7 1 100.1 10 99.8 4 100.2 3 99.9 12 This is known as ( Gaussian Curve ) with normal distribution.

Notes: *The largest number of readings is at the central value. *With more readings (taken at smaller increments) the curve will be smoother. *The shaper and narrower the curve is, the more definitely, one way say that the most probable value of the true readings is the central value (Mean).

*The probability of a given error will be symmetrical about the *Some observations may include small disturbing effects (random error), these errors can be of either positive (+) or negative (-) with equal probability so, that the mean value will be the true value of the measure value. *The probability of a given error will be symmetrical about the zero value. *For a normally dispersed data, approximately (68 %) of all cases lie between (+ Ϭ) and the (- Ϭ) from the mean. r= ± 0.675 Ϭ : r)) *Probable error Normalization distribution Zero Error Normal Law

NOW: The area under the curve between (-Ϭ) and (+Ϭ) represents the cases that differ from the mean by no more than standard deviation. *Based on the normal law, the error distribution curve is presented as shown here. Total area include Ϭ ± .5 .68 1 .95 2 .997 3

NOTE: The possibilities as to the form of the error distribution curve, can be states as following: 1. Small errors are more parable than larger error. 2. There is an equal probability of plus and minus errors , so that the probability of a given error will be symmetrical about the zero value. Example: A large number of nominally 100 ῼ resistor is measured , with (x̅)= 100ῼ with standard deviation Ϭ = 0.2ῼ . We know that, on the average, 68% (2/3) of all resistors have values which lie between the limits of ±0.2 ῼ of the mean. Hence, there is a two-to-one chance that any resistors selected will be within these limits.

Example: Consider the following readings: 101.2, 101.5, 101.3, 101.7, 101.3, 101.1, 101.3, 101.2, 101.0, 101.4. For these readings, the followings are obtained: * = 101.3 * Ϭ = 0.2 *Probable error = ± 0.675 Ϭ = ± 0.675 (0.2) = ± 0.1349 ῼ Hence, the value will be given as: R = 101.3 ± 0.134 ῼ (di)2 di 0.01 -0.1 0.16 0.4 0.0 0.09 -0.3 0.04 0.2 0.1 -0.2

*Limiting Error: Example (1): The accuracy of an element is guaranteed to a certain percentage of the full scale reading where the limits of these deviations from a specific values known as “Limiting Error” or “guaranteed error”. Example (1): Given a resistance with R= 500 ± 10 %. That is, 10 % * (500) = ± 50 Hence R falls between (450) and (500). Here R is guaranteed to fall between the limits of (± 50).

Example (2): A Voltmeter (0-150V) has a guaranteed accuracy of (1 %) of the full scale value. What is the limiting error when reading 83 V? Originally, (1 %) of (150) is (1.5V). Hence, for reading 83, the value is represented by 83 ± 1.5 and thus, the Limiting error is Limiting error = (1.5/83) * 100% = 1.8 % Hence, it is concluded that taking measurements close to the full scale value decreases the limiting error percent.