This is only a sample of the book. For complete book contact us Statistics EXERCISE 14.2

This is only a sample of the book. For complete book contact us... 14. Statistics EXERCISE 14.2 1 Mark Questions 1. Find the range of the data : 22, ...
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14. Statistics EXERCISE 14.2 1 Mark Questions 1. Find the range of the data : 22, 25, 20, 32, 36, 28, 40, 45, 35, 38 (2016-1ZXWG1F) Sol. Range = 45 – 20 = 25. 2. In a frequency distribution, the mid value of a class-interval is 10 and the width of the class-interval is 6. Find the upper limit of the class-interval. (2016-3QDAGER) RELIABLE

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6 Sol. Upper limit of the class-interval isENTERPRISES 10 + = 13. 2 3. What is the class size of the class-intervals 2.5 – 5.5, 5.5 – 8.5, ... ?

(2016-06PPXTO, FTYNFCY)

Sol. Class size is 5.5 – 2.5 = 8.5 – 5.5 = 3.0. 4. If the class marks of a frequency distribution are 9.5, 16.5, 23.5, 30.5, then find the class-interval corresponding to the class mark 16.5. (2016-8GGBPWN) Sol. Class width = 16.5 – 9.5 = 7.0. 7 .0 ⎞ ⎛ 7 .0 ⎞ ⎛ So, class-interval corresponding to the class mark 16.5 is ⎜ 16.5 − ⎟⎠ − ⎜⎝ 16.5 + ⎟, ⎝ 2 ⎠ 2 i.e., 13 – 20. 5. Convert the following discontinuous class-intervals into continuous class-intervals :

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Class-intervals DOWNLO E E 1–6 7 – 12 RELIABLE 13 – 18 ENTERPRISES 19 – 24

(2016-767HL9B, ON3GWF5)

Sol. The required continuous class-intervals are as shown below :

Class-intervals 0.5 – 6.5 6.5 – 12.5 12.5 – 18.5 18.5 – 24.5

2 Marks Questions 6. The marks obtained by 25 students in an examination are given below : 370, 290, 318, 410, 378, 480, 375, 315, 225, 288, 325, 355, 402, 382, 178, 253, 154, 306, 360, 328, 440, 425, 380, 198, 230. Form a continuous frequency table with one class-interval as 200 – 250. Copyrighted Material

Sol. The required frequency table is as shown below :

Marks

Frequency

150 – 200

3

200 – 250

2

250 – 300

3

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5 7

400 – 450 4 RELIABLE

ENTERPRISES 450 – 500 1

3 Marks Questions 7. The blood groups of 30 students of class IX are recorded as follows : A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. Represent this data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students ? (2016-66H6XR9) Sol. The required frequency table is as shown below : Blood group

A

B

O

AB

Total

Frequency

9

6

12

3

30

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Most common blood group is O and rarest group is AB. OWNblood LO E D18 players 8. The following are the runs madeEby in one day cricket match : 79, 28, 45, 99, 3, 46, 8, 0, 3, 7, 24, 73, 122, 46, 27, 16, 7, 3 Form a frequency table for above RELIABLE data with equal class-intervals one of these being 0–25 (excluding 25). (2016-EU9N39A, NHHDUQG) ENTERPRISES Sol. The required frequency table is as shown below : Runs

0 − 25

25 − 50

50 − 75

75 − 100

100 − 125

Total

Frequency

9

5

1

2

1

18

4 Marks Question 9. The following data gives the weights (in grams) of 30 oranges picked from a basket : 106 107 76 109 187 95 125 92 70 139 128 100 88 84 99 113 204 141 136 123 90 115 110 97 90 107 75 80 118 82 Construct a grouped frequency distribution table taking class size equal to 20 in such a way that the mid-value of the first class is 70. From the frequency table, find the number of oranges (i) weighing more than 180 grams. Copyrighted Material

(ii) less than 100 grams. Sol. Frequency table is as given below :

(2016-XTZ3TDV, DM07VQU, 4U5QKWO)

Class [Weight (in g)] 60 – 80

Number of oranges (Frequency) 3

80 – 100 100 – 120 120 – 140 140 – 160

10 9 5 1

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160 – 180 180 – 200

0 ENTERPRISES 1 1 30

200 – 220 Total

(i) From the table, number of oranges weighing more than 180 grams is 1 + 1 = 2. (ii) Number of oranges weighing less than 100 grams is 3 + 10 = 13.

PRACTICE EXERCISE

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1 Mark Questions 1. The points scored by a basketball team in a series of matches are follows : (2016-8GWMLTF) 17, 7, 10, 25, 5, 10, 18, 10 and 24. Find the range. 2. Find the range of two digits numbers. (2016-9XH7QH5) 3. Find the class size of the class-intervals 0–4, 5–9, 10–14, ..... (2016-19T8NWO) W DO NLO EE 4. Find the upper limit of the first class-interval, if class-intervals are 15 – 19, 19 – 24, 24–29. (2016-127ZIS9) 5. The class-marks of a continuous RELIABLE distribution are 2.05, 2.15, 2.25, 2.35, 2.45, 2.55 and 2.65. Find the class-interval corresponding to 2.25 ? (2016-FH419PT) ENTERPRISES 6. If the class marks of a frequency distribution are 29.5, 36.5, 43.5, 50.5, then find the class-interval corresponding to the class mark 36.5. (2016-HE0TO1B, LVQGPXO) 3 Marks Questions 7. To solve a problem, the time taken (in seconds) by the students are as follows : 37, 31, 27, 18, 59, 45, 54, 40, 32, 43, 39, 49, 45, 40, 59, 53, 52, 50, 40, 59, 60, 23, 21, 20, 22. Construct a frequency table with equal class width and one of the class-intervals as 25–32, (32 not included) for the above. (2016-KZJQDPR, R1TGR0F, UK7EWQE) ANSWERS 1. 20

2. 89

3. 5

4. 19

5. 2.20 – 2.30

6. 33 – 40

EXERCISE 14.3 Copyrighted Material

EXERCISE 14.4 1 Mark Questions 1. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18. (2016-WORADH) Sol. Mode is 14 (four times). 2. What is the mean of the first five prime numbers ? (2016-IKDNFT4, AXXWX8R) 2 + 3 + 5 + 7 + 11 28 = = 5.6 . 5 5 3. Find the median of all prime numbers between 20 and 35. (2016-0DBZTDH, KZJQDPR) Sol. Prime numbers between 20 and 35 are 23, 29 and 31. RELIABLE They are arranged in ascending order and three in number. ENTERPRISES ⎛ 3 + 1⎞ So, median will be ⎜ th number, i.e., second number. ⎝ 2 ⎟⎠ So, median is 29. 4. In a data, 14 numbers are arranged in ascending order. If the 9th entry is increased by 5, what will be the corresponding effect on the median ? (2016-1V1LZ5S, M44RVVV)

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Sol. Mean =

Sol.

14 ⎛ 14 ⎞ th number + ⎜ + 1⎟ th number ⎝ ⎠ 2 2 Median = 2

7th number + 8th number 2 Therefore, there will be no effect on median by increasing the 9th observation by 5. 5. The mean age of three students is 28 years. If the ratio of their ages is 2 : 3 : 2, find DOWNL their ages. (2016-2ITFHFV, 9H00N2T, K57594Q)

=

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O EE Mean age of three students = 28 years So, sum of their ages = 28 × 3 = 84 years RELIABLE They are in the ratio 2 : 3 : 2. ENTERPRISES 84 × 2 So, age of the first student = years = 24 years (2 + 3 + 2) 84 × 3 Age of the second student = = 36 years (2 + 3 + 2) 84 × 2 Age of the third student = = 24 years (2 + 3 + 2) 6. Find the median of first seven whole numbers. (2016-2LXZFM6) Sol. First seven whole numbers in ascending order are : 0, 1, 2, 3, 4, 5, 6 (7 + 1) Median = th number i.e., 4th number 2 =3 7. The mean of 15 observations is 16. If each observation in the data is decreased by 25%, what will be the new mean of the data ? (2016-3QDAGER, FH419PT, QJ2VEEN) Sol.

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Sol. Mean of 15 observations is 16. So, their sum = 15 × 16 = 240 Each observation is decreased by 25%. So, their sum will be decreased by 240 × So,

25 = 60. 100

new sum = 240 – 60 = 180 180 Hence, new mean = = 12 15 ⎛ 16 − 12 ⎞ × 100%, i.e., by 25%. i.e., the mean decreased by ⎜ ⎝ 16 ⎟⎠ 10. If the mode of the data 15, 18, 14, RELIABLE 15, 15, 18, 14, 17, 18, k is 15, then find the value ENTERPRISES of k. (2016-LOGBSPN, NIFYJCT, ENECZAO) Sol. In the data, 15 occurs three times and also 18 occurs three times. So, for mode to be 15, k must be equal to 15.

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2 Marks Questions 11. Find the mean of all the prime numbers lying between 20 and 30. (2016-QUG63Y8, 1V1LZ5S, VSX6IHI; 2014-P7TY4GX)

23 + 29 52 = = 26 . 2 2 12. The observations in ascending order are given below : 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 (2016-VAER5MI, ZZDR93, WORADH; 2014-A0FG9I8) If the median of the data is 63, find the unknown observations. Sol. Number of observations are 10. N DO⎛W 10 LO ⎞ 10 EE + 1⎟ th observations, i.e., of 5th and 6th So, median is the mean of th and ⎜ ⎝ 2 ⎠ 2 observations. x+ x+2 = 63 So, we have : ENTERPRISES 2 ⇒ 2x + 2 = 126 ⇒ 2x = 124 124 ⇒ x = = 62 2 So, unknown observations are 62 and 62 + 2, i.e., 62 and 64. 13. If the mean of 5 observations x, x + 4, x + 8, x + 12 and x + 16 is 13, then find the observations. (2016-GL3UFU4, PS63PP3, OGGOG53; 2014-2H5HDHE) Sol. x + (x + 4) + (x + 8) + (x + 12) + (x + 16) = 13 × 5 ⇒ 5x + 40 = 65 ⇒ 5x = 65 − 40 ⇒ 5x = 25 25 ⇒ x = = 5. 5

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Sol. Mean =

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So, observations are 5, 5 + 4, 5 + 8, 5 + 12 and 5 + 16, i.e., they are 5, 9, 13, 17 and 21.

3 Marks Questions 16. The mean of 10 numbers is 55. If one number is included, their mean becomes 60. Find the included number. (2016-0W8BOM5) Sol. Mean of 10 numbers = 55 So, their sum = 55 × 10 = 550 Mean of 11 numbers = 60 So, their sum = 60 × 11 = 660 Hence, the included number = 660 – 550 = 110. RELIABLE 23. There are 20 students in a drama class. The mean age of 12 students is 18 years ENTERPRISES and the mean age of remaining 8 students is 23 years. Find the mean age of all (2016-301W2FH, D1KMNCJ, T3NBB1W) students in the drama class. Sol. Mean age of 12 students = 18 years So, sum of the ages of 12 students = 18 × 12 = 216 years Mean age of remaining 8 students = 23 years So, sum of the ages of these 8 students = 23 × 8 = 184 years Therefore, sum of the ages of all 20 students (216 + 184) years = 400 years

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400 = 20 years 20 24. The mean monthly salary of 10 members of a group is ` 1445. One more member whose monthly salary is ` 1500 has joined the group. Find the mean monthly salary of the 11 members. (2016-A13FYXN, DOWNLO FH419PT, IDUIU1Z, NIFYJCT, SC8YTO3) E E Sol. Mean salary of 10 members = ` 1445 So, sum of salaries of 10 members = ` 1445 × 10 = ` 14450 ENTERPRISES Sum of the salaries of 11 members = ` 14450 + ` 1500 = ` 15950

mean ages of all 20 students =

Hence,

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Hence,

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‘` 15950 11 = ` 1450

mean of salaries of 11 members =

PRACTICE EXERCISE 1 Mark Questions 1. Find the mean of first six odd numbers. (2016-2LXZFM6) 2. Find the median of first 8 natural numbers. (2016-7C1UGUM, QJ2VEEN, QWS2GLV) 3. Find the mean of all factors of 30. (2016-QWS2GLV, D1KMNCJ, 127ZIS9, SC8YTO3) 4. When a number x is added to the following set of numbers 5, 7, 8, 12, 6, 10 and 11, the mean is 9. Calculate the value of x. (2016-FRZ8DOS, ADPHMFN, 9BP8039) 5. Mean of 20 observations is 75. If each observation is divided by 3, find the new mean. (2016-FTYNFCY, N5QIERJ, XTZ3TDV) Copyrighted Material

6. If the median of the observations : x, x + 3, x + 5, x + 7, x + 10 is 9, find the fourth observation. (2016-R1TGR0F, XTZ3TDV)

2 Marks Questions 7. Find the median and mode of the data : 42, 52, 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 60 8. Find the mean of first 10 whole numbers. (2016-YLZ6HSN; 2014-B47H3GT; 2012-45028)

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3 Marks Questions 9. The mean of 15 numbers is 18. If one number is included, their mean is 19. Find the (2016-1ZXWG1F, 12L7U55) included number. 10. Find the mean and median of all the composite numbers between 90 and 100. RELIABLE

ENTERPRISES

(2016-06PPXTO, IULZCIP)

11. The mean of 10 numbers is 55. If one number is excluded, their mean becomes 50. Find the excluded number. (2016-FZYHWLA, L0GBSPN, QJ2VEEN, NHHDUQG, SNYEV4O) 12. The weights (in kg) of 15 students are 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, and 30. Find the median of the data. If the weight 44 kg is replaced by 46 kg and 35 kg is replaced by 37 kg, find the new median. (2016-JCDON1C, SMU7XKM) 13. If the mean of the following distribution is 6, find the value of p. x

2

4

6

10

p+5

f

3

2

3

1

2

(2016-BCMSAT)

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4 Marks Question 14. A class consists of 50 students out of which 30 are girls. The mean marks scored by DOWofNLboys O is 71. Determine the mean score of girls in a test is 73 (out of 100) and that E E the whole class. (2016-ZZDR93)

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ENTERPRISES

1. 6 2. 4.5 3. 9 4. 13 5. 25 6. 11 7. Median = 62, Mode = 52 9. 34 10. Mean = 94.75, Median = 94.5 11. 100 12. 35 kg; 36 kg 13. p = 7 14. 72.2

8. 4.5

VALUE BASED QUESTIONS 1. The mean monthly salary of 7 officers of a company is ` 25,000. Three more officers in the same cadre join the company with salaries ` 40,000, ` 45,000 and ` 55,000. Find the mean monthly salary of the 10 officers. Do you approve the policy of management to appoint the new officers on a much higher salaries ? Give your views. Sol. Mean of salaries of 7 officers = ` 25000 So, total salary of 7 officers = ` 25000 × 7 = ` 175000 Copyrighted Material

Sum of the salaries of 3 new officers = ` 40000 + ` 45000 + ` 55000 = ` 140000 So, sum of the salaries of 10 officers = ` 175000 + 140000 = ` 315000 315000 Their mean salary = ` = ` 31500 10 No. It is not proper to appoint new officers on much higher salaries. For this purpose, the existing workers should not be ignored. New officers should be first provided a proper training, before giving them higher salaries.

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2. Find the mean salary of 60 workers of a factory from the following table : Salary (in `) RELIABLE Number of workers 3000 16 ENTERPRISES 4000 12 5000 10 6000 8 7000 6 8000 4 9000 3 10000 1 Total 60 What types of measures can be taken in a factory for the welfare of the workers ? Number of workers (f) 16 DOWNLO E 12 E 10 8 RELIABLE 6 ENTERPRISES 4 3 1 60

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Salary (in `) (x) 3000 4000 5000 6000 7000 8000 9000 10000 Total

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fx 48000 48000 50000 48000 42000 32000 27000 10000 395000

395000 = ` 6583 (approx.) 60 Measures : (i) Proper health care (ii) Provision of children education (iii) Rewarding for good work and encouraging workers to participate in the management of the factory. So, mean = `

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PRACTICE EXERCISE 1. To increase the number of computer users in the lower income group, a state distributed free laptops to students during a year as given below : Months

No. of students

Jan-March

75

Book July-Sept.e

April-June

Oct.-Dec.

150 170

106 RELIABLE

(a) Draw a bar graph to representENTERPRISES the above data. (b) What values of the state are depicted ? (2016-XVIELLZ, GG4NB5C) 2. An organisation donates clothes to an orphanage every month. According to past record, the number of clothes donated by him are as follows : 6, 7, 21, 16, 18, 29, 6, 5, 4, 21, 18, 5, 5, 20, 29, 7, 6, 4, 21, 7, 18, 21, 20, 14, 7, 1, 8 (a) Prepare a grouped frequency distribution table and hence draw a histogram. (b) What value is depicted by the organisation ? (2016-KOWLDHT, 8GGBPWN)

ANSWERS 1. (b) Value : Duty of state : to help lower income group students in different ways. 2. (b) Value : Organisation is kind towards people living in orphanages.

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O FORMATIVE ASSESSMENT EE

REMEDIAL WORKSHEET

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1. In a mathematics test, 15 students appeared. Their marks (out of 100) are ENTERPRISES recorded as under : 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60 Find the median marks. Sol. Step 1. Arrange the given data in descending/ascending order Step 2. n = __________________ (odd) ⎛ n + 1⎞ ⎟ th observation 2 ⎠

Step 3. Median = Value of ⎜ ⎝

= _________________________________________________ = _________________________________________________ = _________________________________________________ Sol. Step 1. Arranging in ascending order; 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98, Step 2. n = 15 (odd) ⎛ n + 1⎞

th observation Step 3. Median = Value of ⎜ ⎝ 2 ⎟⎠ Copyrighted Material

⎛ 15 + 1 ⎞ ⎟ th observation 2 ⎠

= Value of ⎜ ⎝

= Value of 8th observation = 52

M.C.Q. WORKSHEET 1. The range of the data 14, 27, 29, 61, 45, 15, 9, 18 is (a) 61 (b) 52 (c) 47 (d) 53 2. The class mark of the class 120-150 is (a) 120 (b) 130 (c) 135 (d) 150 RELIABLE 3. The class mark of a class is 10 and its class width is 6. The lower limit of ENTERPRISES the class is (a) 5 (b) 7 (c) 8 (d) 10 4. In a frequency distribution, the class width is 4 and the lower limit of first class is 10. If there are six classes, the upper limit of last class is (a) 22 (b) 26 (c) 30 (d) 34 5. The class marks of a distribution are 15, 20, 25, ......., 45. The class corresponding to class mark 45 is (a) 12.5 – 17.5 (b) 22.5 – 27.5 (c) 42.5 – 47.5 (d) None of these

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ORAL ASSESSMENT SHEET

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1. The mid-point of a class is called _________________. 2. Data collected by the experimenter himself is called _______________ data. 3. The difference between maximum and NL minimum observations in the data DOW O E is called ________________. E 4. Cumulative frequency of a class is the sum total of all frequencies ________ that class. RELIABLE 5. Are the class limits and true class limits different? If yes, explain the ENTERPRISES difference. 6. The sum total of all observations divided by their number is called _________ of the data. 7. The mode of a group of observations is that value of the variable which has _________ frequency. 8. The ___________ is the middle most observation in the data, when they are arranged in increasing/decreasing order. 9. The mean of first ten natural numbers is ______________. 10. The median of first 9 natural numbers is ______________. 11. If each observation in the data is increased by ‘a’, then their ______ is also increased by ‘a’. 12. The sum of deviations of the data (observations) from the mean is _____.

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