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# Linear ineqns. and statistics

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### Linear ineqns. and statistics

1. 1. Linear InequalityDefinition of Linear Inequality A Linear Inequality involves a linear expression in two variables by using any of the relational symbols such as <, >, ≤ or ≥More about Linear Inequality A linear inequality divides a plane into two parts. If the boundary line is solid, then the linear inequality must be either ≥ or ≤. If the boundary line is dotted, then the linear inequality must be either > or <.Example of Linear Inequality As the boundary line in the above graph is a solid line, the inequality must be either ≥ or ≤. Since the region below the line is shaded, the inequality should be ‘≤’. We can notice that the line y = - 2x + 4 is included in the graph; therefore, the inequality is y ≤ - 2x + 4. Any point in the shaded plane is a solution and even the points that fall on the line are also solutions to the inequality. 4x + 6y ≤ 12, x + 6 ≥ 14, 2x - 6y < 12 + 2x, 9y < 12 + 2x are the examples of linear inequalities.
2. 2. Properties of Linear Inequalities : Property 1 of Linear Inequalities :Adding (or Subtracting) the same number to both sidesof an Algebra Inequality does not change the order ofthe inequality sign ( i.e., >, or <).i.e., if a < b then a + c < b + c and a - c < b - cSimilarly, if a > b then a + c > b + c and a - c > b - cfor any three numbers a, b, c. Property 2 of Linear Inequalities :Multiplying (or Dividing) both sides of an AlgebraInequality by the same positive number does not changethe order of the inequality sign ( i.e., >, or <).For any three numbers a, b, c where c > 0,(i) if a < b then ac < bc and a⁄c < b⁄c(ii) if a > b then ac > bc and a⁄c > b⁄c Property 3 of Linear Inequalities :If three numbers are related in such a way that the firstis less (greater) than the second and the second is less(greater) than the third, then the first is less (greater)than the third.This is called transitive property. Property 4 of Linear Inequalities :If a and b are of the same sign and a < b (a > b), then 1⁄a> 1⁄b (1⁄a < 1⁄b).
3. 3. If reciprocals are taken to quantities of the same sign onboth sides of an inequality, then the order of theinequality is changed.Additional properties of modulus functionHere, we enumerate some more properties of modulus of a realvariable i.e. modulus of a real number.Let x,y and z be real variables. Then : |−x| = |x| |x−y| = 0 ⇔ x=y |x+y ≤ |x|+|y| |x−y| ≥ ||x|−|y|| |xy| = |x| |y| | | = , |y| ≠ 0Modulus |x-y| represents distance of x from y. Also, we know thatsum of two sides of a triangle is greater than third side.Combining these two facts, we write a general property formodulus involving real numbers as : |x−y| < |x−z| + |z−y|Square functionThere is striking similarity between modulus and square function.Both functions evaluate to non-negative values.
4. 4. y= |x| ;y≥0 y = x2 ;y≥0Their plots are similar. Besides, they behave almost alike toequalities and inequalities. We shall not discuss each of the casesas done for the modulus function, but with a specific number (4 or-4). We shall enumerate each of the possibilities, which can beeasily understood in the background of discussion for modulusfunction.1: Equality x2 = 4 ⇒ x = ±2x2 = −4 ⇒ No solution2: Inequality with non-negative numberA. Less than or less than equal to x2 ≤ 4 ⇒ −2 ≤ x ≤ 2B. Greater than or greater than equal to x2 ≥ 4 ⇒ x ≤ −2 or x ≥ 2 ⇒ (−∞,−2)∪(2,∞)3: Inequality with negative numberA. Less than or less than equal to x2 ≤ −4 ⇒No solutionB. Greater than or greater than equal to x2 ≥ −4 ⇒ Always true
5. 5. **SOME IMPOTANT RESULTS:1. Let r be a positive real number & a be a fixed realnumber. Then, (i) |x-a| < r ⇔ a-r< x <a+r i.e., x∊ (a-r,a+r)(ii) |x-a| ≤ r ⇔ a-r≤ x ≤a+r i.e., x∊ [a-r, a+r](iii) |x-a| > r ⇔ x <a-r, or x > a+r(iv) |x-a| ≥ r ⇔ x ≤a-r , or x≥ a+r2. Let a, b be positive real numbers. Then (i) a<|x|<b ⇔ x ∊ (-b, -a)U (a,b)(ii) a≤|x|≤b ⇔ x ∊ [-b, -a]U [a,b](iii) a≤|x-c|≤b ⇔ x ∊ [-b+c, -a+c]U [a+c,b+c](iv) a<|x-c|<b ⇔ x ∊ (-b+c, -a+c)U (a+c,b+c)3. (i) |x| ≤ a ⇔ -a ≤x≤a , |x| ≥a ⇔ x ≥a or x≤ -a , a >0 (ii) If ab >0 or a/b >0 (b≠0) and a, b ∊ R; then a & b are ofsame sign, i.e., either both are positive or negative.(iii) ) If ab < 0 or a/b < 0 (b≠0) and a, b ∊ R; then a & b areof opposite sign.**EXAMPLES:1. <I ⇨ <0 , reduce into two casesCase1. X-2 ≥0 , so < 0 ⇨ x>-3 but x≥2, hence x≥2 –Case2. X-2≤ 0 , so < 0⇨ > 0 ⇨ 2x+1, x+3 are ofsame sign , so we get x > -1/2 & x >-3 [when both +ve] X < -1/2 , x< -3 [both are –ve] Solutions are all real numbers less than -3 or ≥ 2 or -1/2<x≤ 2.2. |x-2| ≥ 5 By above result , we get x ≤ 2-5 , or x ≥ 2+5Solution is x ∊ (-∞, -3) U (7, ∞)3. 1 ≤ |x-2|≤ 3 By above result , we get x ∊ [-3+2, -1+2]U [1+2, 3+2]
6. 6. Solving Linear Inequations using the Properties :To solve linear inequation, collect terms containingunknown quantity (variable) on left side and constantson the right side.Then reduce the coefficient of the unknown quantity tounity.While doing this, remember the properties of the algebrainequalities. Solved Example 1 : Linear Inequalities Check whether x = 6 is a solution of 9x + 1 > 7x + 5 Solution:When x = 6, the L.H.S. of the inequation = 9x + 1 = 9(6) + 1 = 55and the R.H.S. of the inequation = 7x + 5 = 7(6) + 1 = 43 We know 55 > 43. So x = 6 satisfies the given inequation. ∴ x = 6 is a solution of 9x + 1 > 7x + 5. Solved Example 2 : Linear Inequalities If the domain of the variable is N (the natural number set), solve the inequation (x + 2) ≤ (3x - 4) Solution: (x + 2) ≤ (3x - 4) ⇒ x - 3x ≤ -4 -2 ⇒ -2x ≤ -6 Dividing throughout by -3(negative number division causes reversing of inequality sign), we get
7. 7. ⇒ x ≥ 3. Ans. Since the domain is the set of Natural numbers, x = { 3, 4, 5, 6, ..............}. Ans. Solve x + 3 < 0. If theyd given me "x + 3 = 0", Id have known how to solve: I would have subtracted 3 from both sides. I can do the same thing here: Then the solution is: x < –3 notation format pronunciationinequality x < –3 x is less than minus three i) the set of all x, such that x is a i) {x | x is a real number, x < real –3} number and x is less than minus set three ...or: ii) all x such that x is less than ii) {x | x < –3} minus three the interval from minus infinity to interval minus three either of the following graphs: graph Solve x – 4 > 0.
8. 8. If theyd given me "x – 4 = 0", then I would have solved by adding four to each side. I can do the same here: Then the solution is: x >4Just as before, this solution can be presented in any of the four followingways: notation format pronunciation inequality x>4 x is greater than or equal to four i) {x | x is a real number, x i) the set of all x, such that > 4} x is a real number, and set x is greater than or equal to four ...or: ii) all x such that x is ii) {x | x > 4} greater than or equal to four the interval from four to infinity, interval inclusive of four either of the following graphs: graph
9. 9. Graph the following inequality: y > 2x - 1Graph the following inequality: 2y 4x + 6Rewrite the inequality first.y < 2x + 3
10. 10. Solve the following system: 2x – 3y < 12 x + 5y < 20 x>0 Just as with solving single linear inequalities, it is usually best to solve as many of the inequalities as possible for "y" on one side. Solving the first two inequalities, I get the rearranged system: y > ( 2/3 )x – 4 y < ( – 1/5 )x + 4 x > 0 Copyrigap The last inequality is a common "real life" constraint: only allowing x to be positive. The line "x = 0" is just the y-axis, and I want the right-hand side. I need to remember to dash the line in, because this isnt an "or equal to" inequality, so the boundary (the line) isnt included in the solution: The "solution" of the system is the region where all the inequalities are happy; that is, the solution is where all the inequalities work, the region where all three individual solution regions overlap. In this case, the solution is the shaded part in the middle:
11. 11. el Solve the following system: 2x – y > –3 4x + y < 5 As usual, I first want to solve these inequalities for "y". I get the rearranged system: y < 2x + 3 y < –4x + 5 The solution is the lower region, where the two individual solutions overlap.The kind of solution displayed in the above example is called"unbounded", because it continues forever in at least one direction(in this case, forever downward). Solve the following system: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved x – y < –2 x–y>2 First I solve for y, and get the equivalent system: y>x+2 y<x–2
12. 12. Then I graph the first inequality: ...and then the second: But there is no place where the individual solutions overlap. (Note that the lines y = x + 2 and y = x – 2 never intersect, being parallel lines with different y-intercepts.) Since there is no intersection, there is no solution.LINEAR INEQUATIONS (Ncert)Question 16:Solve the given inequality for real x:{[Answer ( - ∞, 2] }Question 20:Solve the given inequality and show the graph of the solution onnumber line:{ Answer x ≥ -2/7 -1 0 1 }
13. 13. Question 23:Find all pairs of consecutive odd positive integers both of which aresmaller than 10 such that their sum is more than 11.{ Answer (5, 7), (7, 9) }Question 24:Find all pairs of consecutive even positive integers, both of whichare larger than 5 such that their sum is less than 23. {Answer (6, 8), (8,10), (10, 12) }Question 4:Solve the given inequality graphically in two-dimensional plane: y +8 ≥ 2xQuestion 5:Solve the given inequality graphically in two-dimensional plane: x –y≤2Question 9:Solve the given inequality graphically in two-dimensional plane: y <–2Question 8:Solve the following system of inequalities graphically: x + y ≤ 9, y >x, x ≥ 0Question 10:
14. 14. Solve the following system of inequalities graphically: 3x + 4y ≤ 60,x + 3y ≤ 30, x ≥ 0, y ≥ 0Question 13Solve the following system of inequalities graphically:4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0Question 15:Solve the following system of inequalities graphically: x + 2y ≤ 10, x+ y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0Question 2:[misc.]Solve the inequality 6 ≤ –3(2x – 4) < 12Question 6:Solve the inequalityQuestion 11:A solution is to be kept between 68°F and 77°F. What is the range intemperature in degree Celsius (C) if the Celsius/Fahrenheit (F)conversion formula is given byQuestion 12:A solution of 8% boric acid is to be diluted by adding a 2% boric acidsolution to it. The resulting mixture is to be more than 4% but lessthan 6% boric acid. If we have 640 litres of the 8% solution, howmany litres of the 2% solution will have to be added?
15. 15. Question 13:How many litres of water will have to be added to 1125 litres of the45% solution of acid so that the resulting mixture will contain morethan 25% but less than 30% acid contentQuestion 14:IQ of a person is given by the formulaWhere MA is mental age and CA is chronological age. If 80 ≤ IQ ≤140 for a group of 12 years old children, find the range of theirmental age.ANSWERS Y y+8=2xAnswer: 4 0 xAnswer 5 y x – y=2 0 x
17. 17. Answer 13 Y x=3 y=2x X 0 4x+3y=60Answer 15 y y=x X 0 x+y=1 x+2y=10
18. 18. Answers (misc.)2. (0, 1] , 6. [1, 11/3] , 11. B/w 200C and 250 C , 12. [let x be the number of litres of 2% boric acid solution.Thentotal mixture = (640+x) litres.∴ 2% of x+8% of (640) > 4% of (640+x)⇨x<1280And 2% of x+8% of (640)< 6% of(640+x) ⇨x>320.More than 320 litres but less than 1280 litres.13. [ let x litres of water be added , then (1125+x)25% <(1125x45)/100 ⇨ x<900And (1125+x)30% > 1125x45% ⇨ x>5625.More than 562.5 litres but less than 900 litres.14. 9.6 ≤ M.A. ≤ 16.8 .
19. 19. Statistics1. Statistics deals with collection presentation, analysisand interpretation of the data.2. Data can be either ungrouped or grouped. Further,grouped data could be categorized into:(a) Discrete frequency distribution,(b) Continuous frequency distribution.3. Data can be represented in the form of tables or in theform of graphs. Common graphical forms are: Barcharts,pie diagrams, histograms, frequency polygonsogives, etc.4. First order of comparison for the given data is themeasures of central tendencies. Commonly usedmeasures are (i) Arithmetic mean (ii) Median(iii) Mode.5. Arithmetic mean or simply mean is the sum of allobservations divided by the number of observations. Itcannot be determined graphically. Arithmetic mean isnot a suitable measure in case of extreme values in thedata.6. Median is the measure which divides the data in twoequal parts. The median is the middle term when thedata is sorted.Incase of odd observations the middle observation ismedian. In case of even observations the median is theaverage of the two middle observations.7. Median can be determined graphically. It does nottake into account all the observations.8. The mode is the most frequently occurringobservation. For a frequency distribution mode may ormay not be defined uniquely.
20. 20. Get the9. Measures of central tendencies namely mean, medianand mode provide us with a single value which is therepresentative of the entire data. These three measurestry to condense the entire data into a single centralvalue10. Central tendencies indicate the general magnitude ofthe data.11. Two frequency distributions may have same centralvalue but still they have different spread or they vary intheir variation from central position. So it is important tostudy how the other observations are scattered aroundthis central position.12. Two distributions with same mean can have differentspread as shown below.13. Variability or dispersion captures the spread of data.Dispersion helps us to differentiate the data when themeasures of central tendency are the same.14. Like ‘measures of central tendency’ gives a singlevalue to describe the magnitude of data. Measures ofdispersion gives a single value to describe variability.15. The dispersion or scatter of a dataset can bemeasured from two perspectives:(i) Taking the order of the observations into consideration, two measures are: (a) Range (b) Quartile Deviation(ii)Taking the distance of each observation from thecentral position, yields two measures, (a) Meandeviation, (b) Variance and Standard deviation16. Range is the difference between the highest and thelowest observation in the given data.Get the
21. 21. The greater the range is for a data, its observations arefar more scattered than the one whose range is smaller.17. The range at best gives a rough idea of thevariability or scatter.18. Quartile divides the data into 4 parts. There arethree quartiles namely Q1 Q2 Q3 and Q2 is the median only.19. The quartile deviation is one-half of the differencebetween the upper quartile and the lower quartile.20. If x1, x2, … xn are the set of points and point a is themean of the data. Then the quantity xi –a is called thedeviation of xi from mean a. Then the sum of thedeviations from mean is always zero.21.In order to capture average variation we must get ridof the negative signs of deviations. There are tworemediesRemedy I: take the Absolute values of the deviations.Remedy II: take the squares of the deviation.22. Mean of the absolute deviations about a gives the‘mean deviation about a’, where a is the mean. It isdenoted as M.D. (a). Therefore,M.D.(a) = Sum of absolute values of deviations from themean a divided by the number of observations. Meandeviation can be calculated about median or mode or anyother observations.
22. 22. 23. Merits of mean deviation (1) It utilizes all theobservations of the set. (2) It is least affected by theextreme values. (3) It is simple to calculate andunderstand.24. Mean deviation is the least when calculated aboutthe median.If the variations between the values is very high, thenthe median will not be an appropriate central tendencyrepresentative.26. Measure of variation based on taking the squares ofthe deviation is called the variance.27. Let the observations are x1, x2, x3,..,xnlet mean = xSquares of deviations: 2 d()=−iixxCase 1: The sum di is zero. This will imply that allobservations are equal to the mean x bar.Case 2: The sum di is relatively small. This will implythat there is a lower degree of dispersion. And casethreeCase 3: The sum di is large. There seems to be a highdegree of dispersion.28. Variance is given by the mean of squared deviations.If variance is small the data points are clustering aroundmean otherwise they are spread across.29. Standard deviation is simply expressed as thepositive square root of variance of the given data set.Standard deviation of the set of observations does notchange if a non-zero constant is added or subtractedfrom each observations.30. Variance takes into account the square of thedeviations.Hence, the unit of variance is in square units ofobservations.For standard deviation, its units are the same as that ofthe observations. That’s the reason why standarddeviation is preferred over variance.31. Standard deviation can help us compare two sets ofobservations by describing the variation from the"average" which is the mean. Its widely used in
23. 23. comparing the performance of two data sets. Such astwo cricket matches or two stocks.In Finance it is used to access the risk associated with a particular mutualfund.33. A measure of variability which is independent of theunits is called as coefficient of variation. It is denoted asC.V.It is given by the ratio of σ the standard deviation andthe mean x bar of the data.34. It is useful for comparing data sets with differentunits, and wildly varying means. But mean should be nonzero. If mean is zero or even if it is close to zero theCoefficient of Variation fails to help.35. Coefficient of Variation-a dimensionless constantthat helps compares the variability of two observationswith same or different units.Get the Standard Deviation and Variance Deviation just means how far from the normal Standard Deviation The Standard Deviation is a measure of how spread out numbers are. Its symbol is σ (the greek letter sigma) The formula is easy: it is the square root of the Variance. So now you ask, "What is the Variance?" Variance The Variance is defined as: The average of the squared differences from the Mean. To calculate the variance follow these steps:
24. 24. Work out the Mean (the simple average of the numbers) Then for each number: subtract the Mean and then square the result (the squared difference). Then work out the average of those squared differences. (Why Square?) Example You and your friends have just measured the heights of your dogs (in millimeters):The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm. Find out the Mean, the Variance, and the Standard Deviation. Your first step is to find the Mean: Answer: 600 + 470 + 170 + 430 + 300 1970 Mean = = = 394 5 5 so the mean (average) height is 394 mm. Lets plot this on the chart:
25. 25. Now, we calculate each dogs difference from the Mean:To calculate the Variance, take each difference, square it, and then average the result: 2 2062 + 762 + (-224)2 + 362 + (-94)2 108,520Variance: σ = = = 21,704 5 5 So, the Variance is 21,704. And the Standard Deviation is just the square root of Variance, so: Standard Deviation: σ = √21,704 = 147.32... = 147 (to the nearest mm)And the good thing about the Standard Deviation is that it is useful. Nowwe can show which heights are within one Standard Deviation (147mm) of the Mean: So, using the Standard Deviation we have a "standard" way of knowing what is normal, and what is extra large or extra small. Rottweillers are tall dogs. And Dachsunds are a bit short ... but dont tell them! Now try the Standard Deviation Calculator.
26. 26. *Note: Why square ? Squaring each difference makes them all positive numbers (to avoid negatives reducing the Variance) And it also makes the bigger differences stand out. For example 1002=10,000 is a lot bigger than 502=2,500.But squaring them makes the final answer really big, and so un-squaringthe Variance (by taking the square root) makes the Standard Deviation a much more useful number. Accuracy and Precision They mean slightly different things! Accuracy Accuracy is how close a measured value is to the actual (true) value. Precision Precision is how close the measured values are to each other. Examples of Precision and Accuracy: Low Accuracy High Accuracy High Accuracy High Precision Low Precision High PrecisionSo, if you are playing soccer and you always hit the left goal post instead of scoring, then you are not accurate, but you are precise!Class Vocabulary
27. 27. bell-shaped curve A common type of histogram characterized by a high center, tapered sides, and bell- flared edges. A bell-shaped curve reflects conditions that exhibit natural variation.data A collection of numbers or facts that is used as a basis for making conclusions.fraction A numerical expression representing a part of a larger whole. A fraction can be converted to a decimal by dividing the upper number, or numerator, by the lower number, or denominator.histogram A visual graph that shows the frequency of a range of variables.horizontal scale The portion of a histogram that lists the range of variables.mean The average of a numerical set. It is found by dividing the sum of a set of numbers by the number of members in the group.median The value of a numerical set that equally divides the number of values that are larger and smaller. For example, in a set containing nine numbers, the median would be the fifth number.mode The value of a numerical set that appears with the greatest frequency.natural variation Variation resulting from sources that are normal and expected. Natural variation is predictable over time.percentage A numerical expression that includes a percent sign, with 100 assumed as the denominator.population mean The mean of a numerical set that includes all the numbers within the entire group.probability The likelihood that a particular event will happen in the future. Probability can be expressed as a fraction, ratio, or
28. 28. percentage.process A set of activities that uses resources to transform inputs into outputs. Essentially, a process describes the way "things get done."random sampling The process of collecting and analyzing only a small representative portion of a larger group. Each item must have the same likelihood of being selected.range The difference between the smallest and the largest values within a numerical set.ratio A numerical expression representing a part of a larger whole or proportion. A ratio consists of two numbers separated by a colon.sample mean A mean of a numerical set that includes an average of only a portion of the numbers within a group.standard deviation A number representing the degree of variation within a numerical set.statistics The science of collecting, summarizing, and analyzing numerical data. Statistics makes it possible to predict the likelihood of events.unnatural variation Variation resulting from one or more sources that involve a fundamental change in a process. Unnatural variation is undesirable.variation A difference between two or more similar things.vertical scale The portion of a histogram that indicates the frequency of each variable.
29. 29. Question 2:Find the mean deviation about the mean for the data38, 70, 48, 40, 42, 55, 63, 46, 54, 44Question 4:Find the mean deviation about the median for the data36, 72, 46, 42, 60, 45, 53, 46, 51, 49Question 6:Find the mean deviation about the mean for the dataxi 10 30 50 70 90fi 4 24 28 16 8Question 7:Find the mean deviation about the median for the data.xi 5 7 9 10 12 15fi 8 6 2 2 2 6Question 8:Find the mean deviation about the median for the dataxi 15 21 27 30 35fi 3 5 6 7 8Question 9:
30. 30. Find the mean deviation about the mean for the data.Income per day Number of persons 0-100 4 100-200 8 200-300 9 300-400 10 400-500 7 500-600 5 600-700 4 700-800 3Question 10:Find the mean deviation about the mean for the dataHeight in cms Number of boys 95-105 9 105-115 13 115-125 26 125-135 30 135-145 12 145-155 10Question 11:Find the mean deviation about median for the following data:
31. 31. Marks Number of girls 0-10 610-20 820-30 1430-40 1640-50 450-60 2Question 3:Find the mean and variance for the first 10 multiples of 3Question 6:Find the mean and standard deviation using short-cut method.xi 60 61 62 63 64 65 66 67 68fi 2 1 12 29 25 12 10 4 5Question 1:From the data given below state which group is more variable, A orB? Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80Group A 9 17 32 33 40 10 9Group B 10 20 30 25 43 15 7Question 2:
32. 32. From the prices of shares X and Y below, find out which is morestable in value:X 35 54 52 53 56 58 52 50 51 49Y 108 107 105 105 106 107 104 103 104 101Question 4:The following is the record of goals scored by team A in a footballsession:No. of goals scored 0 1 2 3 4 No. of matches 1 9 7 5 3For the team B, mean number of goals scored per match was 2 witha standarddeviation 1.25 goals. Find which team may be considered moreconsistent?Question 5:The sum and sum of squares corresponding to length x (in cm) andweight y(in gm) of 50 plant products are given below:Which is more varying, the length or weight?Question 1:The mean and variance of eight observations are 9 and 9.25,respectively. If six of the observations are 6, 7, 10, 12, 12 and 13,find the remaining two observations.
33. 33. Question 2:The mean and variance of 7 observations are 8 and 16, respectively.If five of the observations are 2, 4, 10, 12 and 14. Find theremaining two observations.Question 3:The mean and standard deviation of six observations are 8 and 4,respectively. If each observation is multiplied by 3, find the newmean and new standard deviation of the resulting observations.Question 4:Given that is the mean and σ2 is the variance of n observations x1,x2 … xn. Prove that the mean and variance of the observations ax1,ax2, ax3 …axn are and a2 σ2, respectively (a ≠ 0).Question 5:The mean and standard deviation of 20 observations are found tobe 10 and 2, respectively. On rechecking, it was found that anobservation 8 was incorrect. Calculate the correct mean andstandard deviation in each of the following cases:(i) If wrong item is omitted. (ii) If it is replaced by 12.Question 6:The mean and standard deviation of marks obtained by 50 studentsof a class in three subjects, Mathematics, Physics and Chemistry aregiven below: Subject Mathematics Physics Chemistry Mean 42 32 40.9
34. 34. Standard deviation 12 15 20Which of the three subjects shows the highest variability in marksand which shows the lowest?Question 8:The mean and standard deviation of a group of 100 observationswere found to be 20 and 3, respectively. Later on it was found thatthree observations were incorrect, which were recorded as 21, 21and 18. Find the mean and standard deviation if the incorrectobservations are omitted.Answers2. Mean = 50, Mean deviation about mean = = 8.44. Median(M) = (13+14)/2=13.5, Mean deviation about (M) =2.336. Mean = ( )/ = 4000/80 = 50, Mean deviation about mean= =1280/80 =167. Median(M) = 7, Mean deviation = = 84/26 = 3.238. Median = 30, Mean deviation= 148/29 = 5.19. Mean( ) = a + h , a=assumed mean , = )/ ,where = - a)/h, h is class size. = 350 + [4x100/50 ]= 350+8 = 358Mean deviation = 7896/50 = 157.9210. Mean = 130+(-47x10/100) = 130 – 4.7 = 125.3
35. 35. Mean deviation = 1128.8/100 = 11.28811. Median = l + xh = 20+ (11x10)/14 = 390/14 = 27.8 Mean deviation = 517.8/50 = 10.3563. Mean = (3+6+9+........+30)/10 = 165/10 =16.5,variance=740.25/10= 74.0256. Mean = 64, Standard deviation ( ) = Variance = [ when frequency is not given]Variance = [ when frequency is given] Variance = =NShort-cut methodVariance = [N ],We can use any one of above formulas (when frequency is given)Variance = 2.86, Standard deviation ( ) = 1.691. assumed mean = 45, mean for A = 45 – 6x150/150= 39Variance for A is 227.84 and mean for B is 39 , variance for B=243.842. = 51, variance for x = 35 = 105 , variance fo y = 4 , ∴ price of share y will be more stable∵ the variance of y is less than that of x.
36. 36. 4. C.V. (coefficient of variation) = (standard deviation)x100/Mean== 100, ≠ 0 [ used fo comparing the variability or dispersionof two series, series having greater C.V. is said to be more variablethan other )Mean = 50/25 = 2,variance= 30/25=1.2, s.d.= 1.1,C.V.(A) = 1.2x100/2=55%, C.V.(B) = 1.25X100/2=62.5%5. Mean of length = 4.24 , the mean of weight = 5.22Variance of length = 0.079, standard deviation of length = Variance of weight = 1.904, hence weight is more varying.(misc.) 1. [Same as example]Mean =9 = (60+x+y)/8 ⇨ x+y = 12...........(i)Variance = 9.25 = = 74 = 210+ – 216 , by solving it with (i) ,we get, x=4 & y=8 .2. Answer is x =6, y = 8 [ same as Q. 1]3. Mean = 8 = ⇨ = 48 , on multiplying each term by 3Mean = = = 24, variance= =9.4²=144.4. [same as Q. 3] USE formula of variance as5. (i) = 200, since 8 has been omitted then correctmean = (200-8)/19 = 10.11, since incorrect variance are 10 & 4 , so
37. 37. = (4+100)20 = 104x20 = 2080 2correct variance = = –( == –(10.11)2 = (2080 – 64)/19 – (10.11)2 = 1440/361, S.D.=1.99(ii) Mean = = 200/20+4/20 = 10.2Since incorrect mean & variance are 10 , 4 , so , = 20x104 = 2080 , i = 1 to 19Correct variance = –(10.2)2 as 8 has been replaced by 12.(2016+144)/20 – 104.04 = 3.96, S.D. = 1.996. C.V.(Maths.=28.57%, C.V.( Phy.) = 46.87%, C.V.(Chem.)=48.89%Lowest coefficient of variation of maths. And highest coefficient ofvariation is of Chem., so chem.. shows highest variability & maths.Shows lowest variability.8. = 2000, since wrong items are 21, 21 & 18 has been omittedthen correct mean is (2000 – 60)/97 = 20, since incorrect mean &variance are 20 & 9 –(20)2 = 9 ⇨ = (400+9)x100 = 40900Correct mean = –(20)2 = (40900 – 1206)/97 - 400 = 9.21S.D. = 3.1 (Approx.)