Linear ineqns. and statistics

Linear Inequality

Definition 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.

Properties of Linear Inequalities :

Property 1 of Linear Inequalities :

Adding (or Subtracting) the same number to both sides of an Algebra Inequality does not change the order of the inequality sign ( i.e., '>', or '<'). i.e., if a b then a + c > b + c and a - c > b - c for any three numbers a, b, c.


Multiplying (or Dividing) both sides of an Algebra Inequality by the same positive number does not change the 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


If three numbers are related in such a way that the first is 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.


If a and b are of the same sign and a b), then 1⁄a > 1⁄b (1⁄a < 1⁄b). If reciprocals are taken to quantities of the same sign on

both sides of an inequality, then the order of the inequality is changed.

Additional properties of modulus function

Here, we enumerate some more properties of modulus of a real variable 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|

| xy | = ¿ x∨ ¿

¿ y∨¿¿¿ , |y| ≠ 0

Modulus |x-y| represents distance of x from y. Also, we know that sum of two sides of a triangle is greater than third side. Combining these two facts, we write a general property for modulus involving real numbers as :

|x−y| < |x−z| + |z−y|

Square function

There is striking similarity between modulus and square function. Both functions evaluate to non-negative values.

y= |x| ;y≥0

y = x2 ;y≥0

Their plots are similar. Besides, they behave almost alike to equalities and inequalities. We shall not discuss each of the cases as done for the modulus function, but with a specific number (4 or -4). We shall enumerate each of the possibilities, which can be easily understood in the background of discussion for modulus function.

1: Equality

x2 = 4 ⇒ x = ±2

x2 = −4 ⇒ No solution

2: Inequality with non-negative number

A. Less than or less than equal to

x2 ≤ 4 ⇒ −2 ≤ x ≤ 2

B. Greater than or greater than equal to

x2 ≥ 4 ⇒ x ≤ −2 or x ≥ 2 ⇒ (−∞,−2)∪(2,∞)

3: Inequality with negative number

A. Less than or less than equal to

x2 ≤ −4 ⇒No solution

B. Greater than or greater than equal to

x2 ≥ −4 ⇒ Always true

**SOME IMPOTANT RESULTS:1. Let r be a positive real number & a be a fixed real number. 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|0 (ii) If ab >0 or a/b >0 (b≠0) and a, b ∊ R; then a & b are of same 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 are of opposite sign.

**EXAMPLES: 1. ¿ X−2∨ ¿

X+3¿ -3 but x≥2, hence

x≥2

Case2. X-2≤ 0 , so – 2 x−1

x+3 < 0⇨ 2 x+1x+3 > 0 ⇨ 2x+1,

x+3 are of

same 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+5 Solution 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]

Solving Linear Inequations using the Properties :

To solve linear inequation, collect terms containing unknown quantity (variable) on left side and constants on the right side. Then reduce the coefficient of the unknown quantity to unity. While doing this, remember the properties of the algebra inequalities.

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 = 55 and 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 ⇒ x ≥ 3. Ans. Since the domain is the set of Natural numbers,

x = { 3, 4, 5, 6, ..............}. Ans.

Solve x + 3 < 0.

If they'd given me "x + 3 = 0", I'd 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 pronunciation

inequality x < –3 x is less than minus three

set

i) {x | x is a real number, x < –3}

...or:

ii) {x | x < –3}

i) the set of all x, such that x is a real

number and x is less than minus three

ii) all x such that x is less than minus three

intervalthe interval from minus infinity to

minus three

graph either of the following graphs:

Solve x – 4 > 0.

If they'd 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 > 4

Just as before, this solution can be presented in any of the four following ways:

notation format pronunciation

inequality x > 4 x is greater than or equal to four

set

i) {x | x is a real number, x > 4}

...or:

ii) {x | x > 4}

i) the set of all x, such that

x is a real number, and

x is greater than or equal to four

ii) all x such that x is greater than or equal to four

intervalthe interval from four to infinity,

inclusive of four

graph

either of the following graphs:

Graph the following inequality:

y > 2x - 1

Graph the following inequality:

2y 4x + 6

Rewrite the inequality first.y < 2x + 3

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 Copyrig

ap

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 isn't an "or equal to" inequality, so the boundary (the line) isn't 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:

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

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 on

number line:

{ Answer x ≥ -2/7 -1 0 1 }

Question 23:

Find all pairs of consecutive odd positive integers both of which are smaller 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 which are 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 ≥ 2x

Question 5:

Solve the given inequality graphically in two-dimensional plane: x – y ≤ 2

Question 9:

Solve the given inequality graphically in two-dimensional plane: y < –2

Question 8:

Solve the following system of inequalities graphically: x + y ≤ 9, y > x, x ≥ 0

Question 10:

Solve the following system of inequalities graphically: 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

Question 13

Solve the following system of inequalities graphically:

4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

Question 15:

Solve the following system of inequalities graphically: x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

Question 2:[misc.]

Solve the inequality 6 ≤ –3(2x – 4) < 12

Question 6:

Solve the inequality

Question 11:

A solution is to be kept between 68°F and 77°F. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F)

conversion formula is given by

Question 12:

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Question 13:

How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content

Question 14:

IQ of a person is given by the formula

Where 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 their mental age.

ANSWERS

Y

y+8=2x

Answer: 4 0 x

Answer 5 y x – y=2

0 x

Answer 9 y

0 x

y=-2

Answer 8

x+y=9 y=x

Answer 10

3x+4y=60

0 x+3y=30 x

Answer 13

Y x=3 y=2x

X 0

4x+3y=60

Answer 15 y y=x

X 0

x+y=1 x+2y=10

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.Then total mixture = (640+x) litres.∴ 2% of x+8% of (640) > 4% of (640+x)⇨x<1280

And 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<900

And (1125+x)30% > 1125x45% ⇨ x>5625.

More than 562.5 litres but less than 900 litres.

14. 9.6 ≤ M.A. ≤ 16.8 .

Statistics1. Statistics deals with collection presentation, analysis and 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 the form of graphs. Common graphical forms are: Bar charts,pie diagrams, histograms, frequency polygons ogives, etc. 4. First order of comparison for the given data is the measures of central tendencies. Commonly used measures are (i) Arithmetic mean (ii) Median(iii) Mode. 5. Arithmetic mean or simply mean is the sum of all observations divided by the number of observations. It cannot be determined graphically. Arithmetic mean is

not a suitable measure in case of extreme values in the data. 6. Median is the measure which divides the data in two equal parts. The median is the middle term when the data is sorted. Incase of odd observations the middle observation is median. In case of even observations the median is the average of the two middle observations. 7. Median can be determined graphically. It does not take into account all the observations. 8. The mode is the most frequently occurring observation. For a frequency distribution mode may or may not be defined uniquely. Get the

9. Measures of central tendencies namely mean, median and mode provide us with a single value which is the representative of the entire data. These three measures try to condense the entire data into a single central value 10. Central tendencies indicate the general magnitude of the data. 11. Two frequency distributions may have same central value but still they have different spread or they vary in their variation from central position. So it is important to study how the other observations are scattered around this central position. 12. Two distributions with same mean can have different spread as shown below. 13. Variability or dispersion captures the spread of data. Dispersion helps us to differentiate the data when the measures of central tendency are the same. 14. Like ‘measures of central tendency’ gives a single value to describe the magnitude of data. Measures of dispersion gives a single value to describe variability. 15. The dispersion or scatter of a dataset can be measured 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 the central position, yields two measures, (a) Mean deviation, (b) Variance and Standard deviation

16. Range is the difference between the highest and the lowest observation in the given data. Get the

The greater the range is for a data, its observations are far more scattered than the one whose range is smaller. 17. The range at best gives a rough idea of the variability or scatter. 18. Quartile divides the data into 4 parts. There are three quartiles namely Q1 Q2 Q3 and Q2 is the median

only.

19. The quartile deviation is one-half of the difference between the upper quartile and the lower quartile. 20. If x1, x2, … xn are the set of points and point a is the

mean of the data. Then the quantity xi –a is called the

deviation of xi from mean a. Then the sum of the deviations from mean is always zero. 21.In order to capture average variation we must get rid of the negative signs of deviations. There are two remedies

Remedy 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 is denoted as M.D. (a). Therefore, M.D.(a) = Sum of absolute values of deviations from the mean 'a ' divided by the number of observations. Mean deviation can be calculated about median or mode or any other observations. 23. Merits of mean deviation (1) It utilizes all the observations of the set. (2) It is least affected by the extreme values. (3) It is simple to calculate and understand. 24. Mean deviation is the least when calculated about the median. If the variations between the values is very high, then the median will not be an appropriate central tendency representative. 26. Measure of variation based on taking the squares of the deviation is called the variance. 27. Let the observations are x1, x2, x3,..,xn let mean = x Squares of deviations: 2 d()=−iixx Case 1: The sum di is zero. This will imply that all observations are equal to the mean x bar. Case 2: The sum di is relatively small. This will imply that there is a lower degree of dispersion. And case three Case 3: The sum di is large. There seems to be a high degree of dispersion. 28. Variance is given by the mean of squared deviations. If variance is small the data points are clustering around mean otherwise they are spread across. 29. Standard deviation is simply expressed as the positive square root of variance of the given data set. Standard deviation of the set of observations does not change if a non-zero constant is added or subtracted from each observations. 30. Variance takes into account the square of the deviations. Hence, the unit of variance is in square units of observations.

For standard deviation, its units are the same as that of the observations. That’s the reason why standard deviation is preferred over variance. 31. Standard deviation can help us compare two sets of observations by describing the variation from the "average" which is the mean. Its widely used in comparing the performance of two data sets. Such as two cricket matches or two stocks. In Finance it is used to access the risk associated with a particular mutual fund.

33. A measure of variability which is independent of the units is called as coefficient of variation. It is denoted as C.V. It is given by the ratio of σ the standard deviation and the mean x bar of the data. 34. It is useful for comparing data sets with different units, and wildly varying means. But mean should be non zero. If mean is zero or even if it is close to zero the Coefficient of Variation fails to help. 35. Coefficient of Variation-a dimensionless constant that helps compares the variability of two observations with 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:

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:

Mean = 600 + 470 + 170 + 430 + 300

= 1970

= 3945 5

so the mean (average) height is 394 mm. Let's plot this on the chart:

Now, we calculate each dogs difference from the Mean:

To calculate the Variance, take each difference, square it, and then average the result:

Variance: σ2 = 2062 + 762 + (-224)2 + 362 + (-94)2

= 108,520

= 21,7045 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. Now we 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 don't tell them!

Now try the Standard Deviation Calculator.

*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-squaring the 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 AccuracyHigh Precision

High AccuracyLow Precision

High AccuracyHigh Precision

So, 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

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

Class Vocabulary

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 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.

Class Vocabulary

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.

Question 2:

Find the mean deviation about the mean for the data

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Question 4:

Find the mean deviation about the median for the data

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Question 6:


xi 10 30 50 70 90

fi 4 24 28 16 8

Question 7:

Find the mean deviation about the median for the data.

xi 5 7 9 10 12 15

fi 8 6 2 2 2 6

Question 8:

Find the mean deviation about the median for the data

xi 15 21 27 30 35

fi 3 5 6 7 8

Question 9:

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 3

Question 10:


Height in cms Number of boys

95-105 9

105-115 13

115-125 26

125-135 30

135-145 12

145-155 10

Question 11:

Find the mean deviation about median for the following data:

Marks Number of girls

0-10 6

10-20 8

20-30 14

30-40 16

40-50 4

50-60 2

Question 3:

Find the mean and variance for the first 10 multiples of 3

Question 6:

Find the mean and standard deviation using short-cut method.

Question 1:

From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Group A 9 17 32 33 40 10 9

Group B 10 20 30 25 43 15 7

Question 2:

xi 60 61 62 63 64 65 66 67 68

fi 2 1 12 29 25 12 10 4 5

From the prices of shares X and Y below, find out which is more stable in value:

X 35 54 52 53 56 58 52 50 51 49

Y 108 107 105 105 106 107 104 103 104 101

Question 4:

The following is the record of goals scored by team A in a football session:

No. of goals scored 0 1 2 3 4

No. of matches 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard

deviation 1.25 goals. Find which team may be considered more consistent?

Question 5:

The sum and sum of squares corresponding to length x (in cm) and weight 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.

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 the remaining 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 new mean 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 to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard 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 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Subject Mathematics Physics Chemistry

Mean 42 32 40.9

Standard deviation 12 15 20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Question 8:

The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Answers

2. Mean(x¿)¿ = 50, Mean deviation about mean =∑ x i−¿ x∨¿n

¿ = 8.4

4. Median(M) = (13+14)/2=13.5, Mean deviation about (M) =2.33

6. Mean = ( ∑ f i xi )/ ∑ f i = 4000/80 = 50,

Mean deviation about mean= ∑ f i ¿¿¿¿ =1280/80 =16

7. Median(M) = 7, Mean deviation = ∑f i(x¿¿ i−M )

∑ f i

¿ = 84/26 = 3.23

8. Median = 30, Mean deviation= 148/29 = 5.1

9. Mean(x) = a + hu , a=assumed mean , u =∑ f i ui )/ ∑ f i ,where

ui =¿¿- a)/h, h is class size.

x = 350 + [4x100/50 ]= 350+8 = 358

Mean deviation = 7896/50 = 157.92

10. Mean = 130+(-47x10/100) = 130 – 4.7 = 125.3

Mean deviation = 1128.8/100 = 11.288

11. Median = l + N2

−cf

f xh = 20+ (11x10)/14 = 390/14 = 27.8

Mean deviation = 517.8/50 = 10.356

3. Mean = (3+6+9+........+30)/10 = 165/10 = 16.5,variance=740.25/10= 74.025

6. Mean = 64, Standard deviation (σ) = √variance

Variance = ∑ (x¿¿ i−x )2

n¿ [ when frequency is not given]

Variance = ∑ f i(x¿¿ i−x )2

f i

¿ [ when frequency is given]

Variance = ¿ ,∑ f i = N

Short-cut method

Variance = h ²N ² [N∑ f i ui

2−¿¿ ] ,

We can use any one of above formulas (when frequency is given)

Variance = 2.86, Standard deviation (σ) = 1.69

1. assumed mean = 45, mean for A = 45 – 6x150/150= 39

Variance for A is 227.84 and mean for B is 39 , variance for B =243.84

2. x = 51, variance for x = 35

y = 105 , variance fo y = 4 , ∴ price of share y will be more stable∵ the variance of y is less than that of x.

4. C.V. (coefficient of variation) = (standard deviation)x100/Mean=

= σx × 100, x ≠ 0 [ used fo comparing the variability or dispersion of

two series, series having greater C.V. is said to be more variable than 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 = 5.22

Variance of length = 0.079, standard deviation of length = √ .079

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 = ∑ (x¿¿ i−9)2

8¿ = 48+x2+ y2−216

8

74 = 210+ x2+ y2 – 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 = ∑ xi

6 ⇨ ∑ x i = 48 , on multiplying each term by 3

Mean = ∑ 3 xi

6 = 3∑ xi

6 = 24, variance= 3²

∑ (x¿¿ i−8)2

6¿ =9.4²=144.

4. [same as Q. 3] USE formula of variance as ∑ (ax¿¿i−a xn)2

n¿

5. (i) ∑ x i = 200, since 8 has been omitted then correct

mean = (200-8)/19 = 10.11, since incorrect variance are 10 & 4 , so

∑ (x¿¿ i¿) ² ¿¿ = (4+100)20 = 104x20 = 2080

correct variance = ∑ (x¿¿ i−x )2

n¿ = ∑ xi

2

n –( x ¿¿ 2 = ∑ (x¿¿ i−10.11)2

19¿

= ∑ xi2

19 –(10.11)2 = (2080 – 64)/19 – (10.11)2 = 1440/361, S.D.=1.99

(ii) Mean = ∑ x i−8+12

20 = 200/20+4/20 = 10.2

Since incorrect mean & variance are 10 , 4 , so ,

∑ (x¿¿ i¿) ² ¿¿ = 20x104 = 2080 , i = 1 to 19

Correct variance = ∑ x i2+12²

20 –(10.2)2 as 8 has been replaced by 12.

(2016+144)/20 – 104.04 = 3.96, S.D. = 1.99

6. C.V.(Maths.=28.57%, C.V.( Phy.) = 46.87%, C.V.(Chem.)=48.89%

Lowest coefficient of variation of maths. And highest coefficient of variation is of Chem., so chem.. shows highest variability & maths. Shows lowest variability.

8. ∑ x i = 2000, since wrong items are 21, 21 & 18 has been omitted then correct mean is (2000 – 60)/97 = 20, since incorrect mean & variance are 20 & 9

∑ xi2

100 –(20)2 = 9 ⇨ ∑ (x¿¿ i¿) ² ¿¿ = (400+9)x100 = 40900

Correct mean = ∑ xi2

97 –(20)2 = (40900 – 1206)/97 - 400 = 9.21

S.D. = 3.1 (Approx.)

Technology

Linear ineqns. and statistics