Higher Secondary - First year STATISTICS Practical Book th_statistics_practicals.indd 07-09-08 8:00:9
Introduction Statistical tools are important for us in daily life. They are used in the analysis of data pertaining to various activities such as production, consumption, distribution, banking and insurance, trade, transport, etc. Practical work also gives students many opportunities to use their minds to adopt suitable statistical tools and methods on various types of analysis for the given sample data. Objectives It facilitates comparison with similar data. Tabulation of data Compares the tabular data with diagrammatic representation of data Represents the data in a graph Presents the data in suitable diagrams Distinguishes diagrammatic and graphical representation of data Calculates the mathematical averages and the positional averages Computes quartiles, Deciles, Percentiles and interprets Measures the spread or dispersion Understands the theorems on probability and applies in problems Measures the Skewness Fittings Binomial and poisson distribution Instructions To Students Students must attend all the practical classes. They must also remember that there is a great degree of co-ordination between theory problems and practical problems. The following are some of the items that they must bring to the Practical Classes. Practical observation note book Practical record Pencil sharpener Eraser A measuring scale Graph sheets Compass and protractor Calculator th standard Statistics Practical Come prepared with theory part of the practical subject. They should submit the practical records periodically for correction and evaluation. They must maintain strict discipline and silence in the statistical laboratory. They should write the date and experiment number in their observation note books. th_statistics_practicals.indd 07-09-08 8:00:9
Students must answer any three out of five questions from the following topics. Formation of Frequency Table. Diagrammatic Representation of data. Graphical Representation of data 4. Measures of Central Tendency 5. Measures of Dispersion 6. Measures of Skewness 7. Simple problem in probability 8. Probability Mass Function and Probability Density Function 9. Computation of Mean and Variance for Random Variables 0. Fitting Binomial Distribution. Fitting Poisson Distribution MODEL QUESTION PAPER I Marks 5 Duration: ½ hrs. Answer any three of the following. 5 5. Construct a stem and leaf plot for the following and find range, Median and Mode.., 0.7, 0.9,.44,.0, 0.88, 0.99, 0.7, 0.9, 0.98,., 0.79,.4,.9,.08, 0.94,.06,.,.0.. Administration of a school wished to initiate suitable preventive measures against breakage of equipments in its chemistry laboratory. Information collected about breakage of equipment occurred during the year 07 in the laboratory are given below. Equipments Burette Conical flask Test tube Pipette No. of Breakages 45 75 50 0 Draw pareto diagram for the above data. Which equipment requires more attention in order to reduce breakages?. Find the Q, Q, D 7, of the frequency distribution given below. Marks in Statistics < 0 0-0 0-0 0-40 40-50 50-60 60-70 > 70 Number of students 8 0 0 8 4 4. Given three identical boxes I, II and III containing two coins. In Box I both coins are gold coin, in Box II both are silver coin, in Box III there is one gold and one silver coin. A person chooses a box at random and takes out a coin if the coin is gold what is the probability that the other coin in the box is also gold? 5. A set of 4 coins are tossed 64 times. The number of occurrences of head is tabulated as follows. Number of Heads 0 4 Number of times 5 7 6 Fit a Binomial distribution for the foresaid data and find the expected frequencies. th_statistics_practicals.indd 07-09-08 8:00:9
ANSWERS FOR MODEL QUESTION PAPER - I. Aim: To construct a stem and leaf plot and finding the values of range, Median and Mode. Formula: Calculation: Range : L S Median : Mid value of the data Mode : Term occurring most frequently Write the numbers in ascending order 0.7, 0.7, 0.79, 0.88, 0.9, 0.9, 0.94, 0.98, 0.99,.0,.0,.06,.08,.,.,.4,.9,.,.9,.44. Stem Leaf 0.7,, 9 0.8 8 0.9,, 4, 8,9.0,, 6, 8.,, 4, 9.. 9.4 4 Range: Median: [From the stem and leaf plot, the lowest and highest value can be found out easily] Highest value (L).44, Lowest value (S) 0.7 Range L - S.44 0.7 0.7 th 0 item + th item 0. + 0. 04. 0. Mode: 0.9 Result: Range 0.7 Median.0 Mode 0.9 4 th_statistics_practicals.indd 4 07-09-08 8:00:4
. Aim: To draw a pareto diagram. Given: Formula: Calculation: Equipment Data To find percentage value total 00 Arrange the given data in descending order of frequency. No. of Breakages (f) Test tube 50 Conical flask 75 Burette 45 Pipette 0 Number of Breakages in percentage Total 00 00 Cumulative percentage 50 00 50 00 50 75 00 5 00 75 45 00 5 00 90 0 00 0 00 00 No of breakages in the chemistry laboratory 00 70 40 0 Y 75% 90% 00% Y axis : unit 0 breakages No. breakages 80 50 0 90 60 50% Tt Test tube Cf Conical flask Bu Burette Pi Pipette 0 0 Tt Cf Bu Pi Equipment X Result: 50% of breakage is due to test tube and 5% is due to conical flask. i. Pareto diagram is drawn ii. The school administration has to focus more attention on reducing the breakages of test tubes and conical flasks. 5 th_statistics_practicals.indd 5 07-09-08 8:00:46
. Aim: To find Q, Q and D 7 Formula: Q l+ N 4 m c f Q N m l+ 4 c f D Calculations: 7 l+ 7N m 0 c f To find Q and Q Marks in Statistics No. of Students f Cumulative frequency Below 0 8 8 0 0 0 (m) 0-0 (l) 0 f 40 0 40 7 40 50 0 0 (m) 50-60 (l) 8 f 0 60 70 4 Above 70 4 46 N 46 6. 5 4 4 N m Q l+ 4 c f l 0, N 6. 5, f 0 m 0 c 0 4 6. 5 0 Q 0 + 0 0 0 + 6. 5 0 + 8.5 8.5 N 46 65. 09. 5 4 4 6 th_statistics_practicals.indd 6 07-09-08 8:00:58
Q l+ N m 4 c f l 50, N 4 09. 5, f 8 m 0 c 0 09. 5 0 Q 50 + 0 8 50 + 75. 8 0 50 + 68. 568. 7N 7 46 0. 0 0 7N m D7 l+ 0 c f l 50, 7N 0 0., f 8 m 0 c 0 0. 0 D 7 50 + 0 8 50 + 0. 8 0 50 + 0 8 50 + 07. 507. Result: Q 8.5, Q 5.68, D 7 50.7 4. Aim: To find the probability by using Baye s theorem Formula: P E A Calculation: P( E ) P( A E ) i P E P A E i i E, E and E be the events that the boxes I, II, III are chosen respectively. P ( E) P ( E) P ( E ) Let A be the event that the gold coin is drawn P ( A E ) 0 P ( A E ) 0 P ( A E ) 7 th_statistics_practicals.indd 7 07-09-08 8:0:8
P ( E A) + 0 + 6 + 6 6 + 6 Result: Probability that the second coin in the box is gold is /. 5. Aim: To fit a Binomial distribution Formula: (i) x f x N (ii) x p, q p n (iii) x n x p ( x) nc p q, x 0,... n (iv) (v) F ( 0) N p( 0) x n x p F( x+ ) F x x + q Calculation: n 4 6 (i) x 64 (ii). 5 x f f x 0 0 5 5 46 7 5 4 6 4 Total 64 6 x p n. 5 4 0. 55 05. q p 05. 047. 8 th_statistics_practicals.indd 8 07-09-08 8:0:
(iii) x 4 x x p x 4c 05. 047., x 0,... 4 ( 0) 4 c ( 05. ) ( 0. 47) (iv) F 0 N p 0 (v) p 0 0 4 0 ( 047. ) 0. 05 F ( 0) 64 0. 05. n x p F( x+ ) F x x + q When x 0 4 0 05 F( 0+ ) F 0 0+. 047. 4.. F() 4.464 4 When x F ( + ) 4 4 05 F +. 047 (). 4 464.. F 454. 5 When x 4 05 F ( + ) F +. 047. 45.. F 847. 8 When x 4 05. F ( + ) F 047. 847 4.. F ( 4) 5. 5 + Result: (i) (ii) The fitted binomial distribution is x 4 x x P X x 4c 0. 5 047., x 0,,, The Expected frequencies are x 0 4 Observed frequencies 5 7 6 Expected frequencies 4 5 8 5 9 th_statistics_practicals.indd 9 07-09-08 8:0:5
MODEL QUESTION PAPER II Marks 5 Duration: ½ hrs. Answer any three of the following. 5 5. The number of hours spent by a School Student on Various activities on a working day, is given below. Construct a pie chart using the angle measurement. Activity Sleep School Play Homework Others Number of hours 8 6 4 Draw a pie chart to represent the above information.. The following is the distribution of marks obtained by 09 Students in a Subject in an institution. Find the Geometric mean. Marks 4-8 8 - - 6 6 0 0-4 4-8 8 - - 6 6-40 No of Students 6 0 8 0 5 0 6. The wholesale price of a commodity for seven consecutive days in a month is a follows. Days 4 5 6 7 Commodity/Price/Quintal 40 60 70 45 55 86 64 Find varance and standard deviation 4. A number is selected randomly from the digits through 9. Consider the events. A {, 4, 6, 8, 9}, B {, 4, 8, 9}, C {, 5, 8, 9} Find (i) P (A/B) (ii) P (A/C) (iii) P (B/C) (iv) P (B/A) 5. The p.d.f of a continuous random variable X is given by x, 0 < x < f ( x) find its mean and variance. 0, elsewhere 0 th_statistics_practicals.indd 0 07-09-08 8:0:5
ANSWERS FOR MODEL QUESTION PAPER II. Aim: Construct a pie chart using the angle measurement Formula: Calculation: The central angle of a Component is [value of the Component/Total value] 60 0 Activity Duration in hours Central angle Sleep 8 School 6 Play Homework Others 4 8 0 0 60 0 4 6 0 0 60 90 4 0 0 60 45 4 0 0 60 45 4 4 0 0 60 60 4 Total 4 60 0 Homework 45 Play 45 Sleep 0 Others 60 School 90 th_statistics_practicals.indd 07-09-08 8:0:58
. Aim: To find the Geometric mean. Formula: Calculation: GM. Anti log n i f i N log xi Marks Midpoint (x i ) f i log x i f i log x i 4 8 6 6 0.778 4.669 8 0 0.0000 0.0000 6 4 8.46 0.698 6 0 5 0.55 7.6590 0 4 5.44 0.60 4 8 6.450 6.800 8 0 0.477 4.770-6 4 6.55 9.890 6-40 8.5798.596 Total N 09 7.96 GM. Anti log n i fi log xi N 7. 96 Anti log 09 GM. 8. 4 Anti log [. 587] Result: Geometric mean marks of 09 students in a subject is 8.4. Aim: To find the variance and standard deviation Formula: d Variance d n n Standard deviation var iance th_statistics_practicals.indd 07-09-08 8:0:04
Calculation: Standard deviation Observations d x - A d 40-5 5 60 5 5 70 5 5 45-0 00 55 0 0 86 96 64 9 8 d Variance d n n 67 5 7 7 5 5 06 var iance 6 4. 5 Result: Variance 06 Standard deviation 4.5 5 67 4. Aim: To find the conditional probability Formula: P A B P A C P B C P B A P( B) ( C) PC ( C) PC ( A) P( A) P A B P A P B P B th_statistics_practicals.indd 07-09-08 8:0:
Calculation: A {, 4, 6, 8, 9}, B {, 4, 8, 9}, C {, 5, 8, 9} A B{ 4, 8, 9} A C { 8, 9} B C 5 4 P( A) P( B) 9 9 P( A B) P( A C) 9 9 P B C The probability for the occurrence of A given that B has occurred is P A P A B ( B) 9 P( B) 4 9 P( A B) 4 The probability for the occurrence A given that C has occurred is P A P A C ( C) 9 PC 4 9 P( A C) 4 Similarly the Conditional Probability of B given C P B P B C ( C) 9 PC 4 9 P( B C) The Conditional probability of B given A is P B P B A ( A) 9 P( A) 5 9 P( B A) 5 Result: P( A B) 4 P( A C) 4 P( B C) P( B A) 5 5. Aim: To find mean and variance Formula: E X x f x dx E X x f x dx Variance X E X E X 4 th_statistics_practicals.indd 4 07-09-08 8:0:7
Calculation: x E( X) x dx 0 x 0 x dx 0 8 0 4 E( X ) 0 x x dx x dx 0 4 x 4 6 4 Variance X E X E X 0 4 6 9 Result: 9 Mean E (X) 4 Variance (X) /9 5 th_statistics_practicals.indd 5 07-09-08 8:0:5
MODEL QUESTION PAPER III STATISTICS PRACTICAL Marks 5 Duration: ½ hrs. Answer any three of the following. 5 5. Construct a bi-variate frequency distribution table for the following data of twenty students. Marks in Economics Marks in Statistics 5 7 0 4 0 8 5 0 6 8 6 5 7 9 5 0 0 0 4 4 0 0 4. Find the measures of central tendencies for the following data. Wages ( ) 60-70 50-60 40-50 0-40 0-0 No of labourers 5 0 0 5. Draw Box - Whisker plot for the following, 5, 0,,, 0,, 9, 7, 7, 6 4. Two coins are tossed one by one. First throw is considered as X and second throw is considered as Y. The joint probability distribution of X and Y is given by X Y 0 0.5 0.5 0 0.5 0.5 Verify E ( XY) E( X) E( Y ) 5. The following mistakes per page were observed in a book Number of mistakes (per page) 0 4 Number of pages 90 9 5 0 Fit a poisson distribution and estimate the expected frequencies. 6 th_statistics_practicals.indd 6 07-09-08 8:0:54
. Aim: ANSWERS MODEL QUESTION PAPER III To construct a bi-variate frequency distribution. Calculation Let X denote marks in Economics Y denote marks in Statistics Eco/Stat 0 - - 4 4-6 6-8 8-0 0 - - 4 8-0 0 - l ll ll l l - 4 l l l lll lll l 4-6 ll l Frequency Distribution Table X/Y 0 - - 4 4-6 6-8 8-0 0 - - 4 Total 8-0 0-7 - 4 0 4-6 Total 5 4 4 4 0. Aim: To find the measures of central tendencies Formula: Calculation n fi xi i x N, x i is the midpoint of the class interval. Median l + N m C f f f0 Mode l + f f f 0 C Wages No of labourers (f) Mid value x f x 0 0 5 5 5 0 40 0 5 50 40 50 0 45 900 50 60 5 55 75 60 70 65 95 N 4 fx845 A.M f x x N 845 4 4. 9 7 th_statistics_practicals.indd 7 07-09-08 8:0:0
Median No of labourers Cumulative Wages (f) frequency 0 0 5 5 0 40 0 5 40 50 0 5 50 60 5 40 60 70 4 N 4 N 4. 5 f 0, m 5, l 40, N. 5, C 0 l + Median Mode N m C f. 5 5 40 + 0 0 65 40 +. 0 0 65 40 + 50 40+.5 4.5 Wages No of labourers (f) 0 0 5 0 40 0 40 50 0 50 60 5 60 70 f 0, f 0, l 40, f 5, c 0 0 f f0 Mode l + f f f 0 c 8 th_statistics_practicals.indd 8 07-09-08 8:0:0
40 0 0 + 0 0 5 0 0 40 + 40 5 0 40 + 00 5 40 + 4 44 Result : Measures of Central tendencies Arithmetic mean x 4. 9 Median 4.5 Mode 44. Aim: Draw box- whisker plot. Formula Calculation : Q Q Q n + 4 th n + th n + 4 item th item item First arrange in the ascending order,5,0,,,6,7,7,9,0, n Q n + 4 th item Q + 4 4 th rd item Q 0 th item item 9 th_statistics_practicals.indd 9 07-09-08 8:0:
Q n + th th + th 6 item Q 6 th. item item item Q n + 4 + 4 th 9 th item Q 9. th th item item item Minimum Value and Maximum Value L Q 0 Q 6 Q 9 H L Q 0 Q 6 Q 9 H L Q 0 Q 9 H Result: Q 6 Box- Whisker plot is drawn using Five number summary. 0 th_statistics_practicals.indd 0 07-09-08 8:0:4
4. Aim To Verify E( xy) E( x) E( y) Formula x y P E xy xy j i i j ij XY 0 Total 0.5 0.5 0.5 0 0.5 0.5 0.5 Total 0.5 0.5 Calculation: E ( x) E y x P i i i yp j j j A random variable x, y can table the values o and E( xy) P ( xi yj) x y i j i j x0.5+0+0.5+0+0.5+0x0.5 0.5 E x i x p x0.5 +0+0-5 0.5 E y j i j i yp X0.5+0X0.50.5 E( xy) E( y) 05. 05.. 5 E( xy) E( x) E( y) Result: E( xy) E( x) E( y) is verified 5. Aim To fit a Poisson distribution and estimate the expected frequencies Formula i) x fx ii) x iii) P( x) f e x x! th_statistics_practicals.indd 07-09-08 8:0:
o iv) P e ( 0) e 0! v) F 0 N P 0 vi) F x ( + ) F( x) x + i) Mean x f fx 0 0 90 90 9 8 5 5 4 0 0 Total 5 4 iii) P x x e x! fx x f 4 5 x e 044. 044. ( 044. ) x! x ii) x 044. 044. 0 e ( 044. ) 044. iv) P( 0) e 0. 6440. 0! v) F 0 N P 0 5 0. 6440 09. 4. vi) F x ( + ) F( x) x + F F F F Result: F 0 044. 0 () ( + ) + 09. 4 9. 5 044. F( ) 9. 5 0. 7 + + 044. F( ) 0. 7. 97 + + 044. 4 F( ) 97. 0. + + Fitted Poisson distribution is 0. 44 x e 044. P( x x), x 0,,... x! Expected Frequencies are given below x 0 4 Total Observed frequency Expected frequency 90 9 5 0 5 0 9 0 0 5 th_statistics_practicals.indd 07-09-08 8:0:44
MODEL QUESTION PAPER IV STATISTICS PRACTICAL Marks 5 Duration: ½ hrs. Answer any three of the following. 5 5. Draw more than ogive curve for the following data showing the marks secured by the students of class XI in a school. Marks No of students 0-0 0-0 0-0 4 9 0-40 0 40-50 8 50-60 5 60-70 70-80 a) Estimate the total number of students who secured marks more than. b) Find the median of the data. (i) From the known data, mean 7.5, mode 8 and variance.69 then find the karl peasson coefficient of skewness. (ii) If Q 40 Q 50,Q 60 find Bocoleys Coefficient of skewness. ) A Question Paper contains section A with 5 Questions and section B with 7 Questions. A Student is required to attempted 8 Questions in all, Selecting at least from each section. In how many ways can a student select the Question? 4) A discrete random variable has the following distribution function. x 0 4 5 6 7 8 P( x) a a 5a 7a 9a a a 5a 7a Find (i) a (ii) P( x< ) (iii) P( x ) 5 (iv) P < x< 7 5. Student of a class were given an aptitude test. Their marks were found to he normally distributed with mean 60 and the standard deviation 5. What Percentage of students scored (i) more than 60 marks (ii) less than 56 marks (iii) between 45 and 65 marks. th_statistics_practicals.indd 07-09-08 8:0:48
ANSWERS MODEL QUESTION PAPER IV. Aim To find the median of the data by using ogive curve Calculation. More then ogive More than ogive Less than ogive Marks less than No of students Marks greater than No of students 0 0 4 0 6 0 4 0 5 0 7 40 5 0 8 50 60 8 40 50 8 0 70 4 60 5 80 4 70 80 70 60 No of students 50 40 0 Median 0 0 x x 6 0 0 0 0 40 50 60 70 80 Marks Result: a) The total number of students who secured marks more than 6 students b) Median 6 4 th_statistics_practicals.indd 4 07-09-08 8:0:48
. Aim To find the coefficient of skewness Given data (i) mean 7.5, Mode 8, variance.69 (ii) Q 40 Q 50 Q 60 Formula: (i) Karl Pearson coefficient of skewness Mean mod e S tan darddeviation Q + Q Q (i) Bowler s Coefficient of skewness Q Q Calculation: (i) Karl Pearson coefficient of skweness Standard deviation var iance 69.. 75. 8 Coeff.of skweness. 065. 05.. ii) Bowley s coefficient of skewness 40 + 60 50 60 40 00 00 0 0 Given distribution is symmetric Result: (i) Karl Pearson Coefficient of skewness -0.5 (ii) Bowley s coefficient of skewness 0 and hence the distribution is symmetric. 5 th_statistics_practicals.indd 5 07-09-08 8:0:5
. Aim To find the number of ways selecting the question Given data Section No. of questions A 5 B 7 Calculation: To select at least questions from each section to take totally 8 questions. Section Number of Selections A (5) 4 B(7) 5 4 Combinations 5 4 7 6 5C 7C5 0 0 5C 7 5 7 5 7 6 5 4 C4 C C 75 7 5 5C5 7C 5 Total number of selection is 0+ 75+540 Result: Number of ways selection 8 questions from section A and B is 40. 4. Aim To find a and their probabilities Calculation Since P( x x) is probability mars function P x x P( x 0)+ P( x ) P( x )+ P( x )+ P( x 4)+ P( x 5)+ P( x 6)+ P( x 7)+ P( x 8) a+ a + 5a + 7a + 9a + a + a + 5a + 7a 8a a 8 (ii) P( x< ) P( x o) + p( X )+ p X a+ a + 5a 6 th_statistics_practicals.indd 6 07-09-08 8:0:57
9 8 (iii) P( x 5) P( x 5) + P( x 6)+ P( x 7) + P x 8 a + a 5a + 7a 56 8 ( ) + ( ) + ( ) (iv) P < x< 7 P x 4 P x 5 P x 6 Result 9a + a + a 8 (i) a 8 (iv) P x 7 < < 8 (ii) P( x< ) 9 8 (iii) P( x 5) 56 8 5. Aim : To find the percentage of students scoring more than / less than particular marks using normal distribution Given data 60, 5 Formula (i) x (ii) Standard normal table Calculation: (i) more than 60 marks 60 60 P( x> 60) Pz ( > ) 5 p( z > 0) P 0 < z < 0.5000 Percentage of students scoring more than 60 marks 0.5 00 50 % 7 th_statistics_practicals.indd 7 07-09-08 8:04:0
- z 0 (ii) Less than 56 marks P x< 56 56 60 Pz ( < ) 5 P z < 08. P( < z < 0) P 0. 8< z < 0 0. 5000 P( 0< z < 08. ) 05. 0. 88 0.9 Percentage of students scored less than 56 marks 0.9x00.9% - z -8 z 0 (ii) Between 45 and 65 marks 45 60 65 60 P( 45 < x< 65) P < z< 5 5 P < z < P( < z < 0)+ P 0< z < 0.4986 +0.4 0.899 Percentage of students scored marks between 45 and 65 is 0.899x008.99% - z z 0 z Result (i) Percentage of students scored more than 60 marks 50% (ii) Percentage of students scored less than 56 marks.9% (iii) Percentage of students scored marks between 45 and 65 is 8.99% 8 th_statistics_practicals.indd 8 07-09-08 8:04:08