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Class 10 Maths Chapter 13, , Exercise 13.1 Page No: 157, , Choose the correct answer from the given four options:, 1. In the formula, , For finding the mean of grouped data di’s are deviations from a of, (A) Lower limits of the classes, , (B) Upper limits of the classes, (C) Mid points of the classes, (D) Frequencies of the class marks, Solution:, (C) Mid points of the classes, Explanation:, We know,, d i = xi – a, Where,, xi are data and ‘a’ is the assumed mean, So, di are the deviations from of mid – points of the classes., Hence, the option (C) is correct, 2. While computing mean of grouped data, we assume that the frequencies are, (A) Evenly distributed over all the classes, (B) Centred at the class marks of the classes, (C) Centred at the upper limits of the classes, (D) Centred at the lower limits of the classes, Solution:, (B) Centered at the class marks of the classes, Explanation:, In computing the mean of grouped data, the frequencies are centered at the class marks of the, classes., Hence, the option (B) is correct, , 3. If xi’s are the mid points of the class intervals of grouped data, fi’s are the corresponding, frequencies and x is the mean, then (fixi – [latex]\overline{x}[/latex] ) is equal to, (A)0 (B) –1 (C) 1 (D) 2, Solution:

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(A) 0, Explanation:, Mean (x) = Sum of all the observations/ Number of observations, x = (f1x1 + f2x2 + …..+ fnxn) / f1 + f2 +……+ fn, x = Σfixi / Σfi, Σfi = n, x = Σfixi / n, n x = Σfixi …………… (1), Σ (fixi – x) = (f1x1 – x) + (f2x2 – x)+ …..+ (fnxn – x), Σ (fixi – x) = (f1x1 + f2x2 + …..+ fnxn) – (x +x +….n times), Σ (fixi – x) = Σfixi –nx, Σ(fixi – x) = nx – nx ( From eq1), Σ(fixi – x) = 0, Hence, option (A) is correct, , 4. In the formula x = a + h(fiui/fi), for finding the mean of grouped frequency distribution, ui =, (A) (xi+a)/h, , (B) h (xi – a), (C) (xi –a)/h, , (D) (a – xi)/h, Solution:, , (C) (xi –a)/h, , Explanation:, , According to the question,, , x = a + h(fiui/fi),, , Above formula is a step deviation formula., In the above formula,, xi is data values,, a is assumed mean,, h is class size,, When class size is same we simplify the calculations of the mean by computing the coded mean of, u1,u2,u3…..,, Where ui = (xi – a)/h, Hence, option (C) is correct, 5. The abscissa of the point of intersection of the less than type and of the more than type, cumulative frequency curves of a grouped data gives its, (A) mean (B) median, (C) mode (D) all the three above, Solution:, (B) Median, Explanation:

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Since, the intersection point of less than ogive and more than ogive gives the median on the, abscissa, the abscissa of the point of intersection of the less than type and of the more than type, cumulative frequency curves of a grouped data gives its, Hence, option (B) is correct, 6. For the following distribution :, Class 0-05 5-10 10-15 15-20 20-25, Frequency 10 15 12 20 9, the sum of lower limits of the median class and modal class is, (A)15 (B) 25 (C) 30 (D) 35, Solution:, (B) 25, Explanation:, , Class, , Frequency, , Cumulative, Frequency, , 0-5, , 10, , 10, , 05-Oct, , 15, , 25, , Oct-15, , 12, , 37, , 15-20, , 20, , 57, , 20-25, , 9, , 66, , From the table, N/2 = 66/2 = 33, which lies in the interval 10 – 15., Hence, lower limit of the median class is 10., The highest frequency is 20, which lies in between the interval 15 – 20., Hence, lower limit of modal class is 15., Therefore, required sum is 10 + 15 = 25., Hence, option (B) is correct, 7. Consider the following frequency distribution:, Class 0-05 6-11 12-17 18-23 24-29, Frequency 13 10 15 8 11, The upper limit of the median class is, (A)17 (B) 17.5 (C) 18 (D) 18.5, Solution:, (B) 17.5, Explanation:, According to the question,, Classes are not continuous, hence, we make the data continuous by subtracting 0.5 from lower limit, and adding 0.5 to upper limit of each class.

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Class, , Frequency, , Cumulative, Frequency, , 0.5-5.5, , 13, , 13, , 6.5-11.5, , 10, , 23, , 11.5-, , 15, , 38, , 8, , 46, , 11, , 57, , 17.5, 17.523.5, 23.529.5, According to the question,, N/2 = 57/2 = 28.5, 28.5 lies in between the interval 11.5 – 17.5., Therefore, the upper limit is 17.5., Hence, option (B) is correct, 8. For the following distribution:, Marks Number of students, Below 10 3, Below 20 12, Below 30 27, Below 40 57, Below 50 75, Below 60 80, The modal class is, (A)10-20 (B) 20-30 (C) 30-40 (D) 50-60, Solution:, , (C) 30-40, Explanation:, , Marks, , Number, , Cumulative, , of, , Frequency, , students, Below 10, , 3=3, , 3, , Oct-20, , (12 – 3) =, , 12

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9, 20-30, , (27 – 12), , 27, , = 15, 30-40, , (57 – 27), , 57, , = 30, 40-50, , (75 – 57), , 75, , = 18, 50-60, , (80 – 75), , 80, , = 5, Here, we see that the highest frequency is 30, which lies in the interval 30 – 40., Hence, option (C) is correct, 9. Consider the data :, , Class, Frequency, , 65-85, 4, , 85-105, 5, , 105-, , 125-, , 145-, , 165-, , 185-, , 125, , 145, , 165, , 185, , 205, , 13, , 20, , 14, , 7, , 4, , The difference of the upper limit of the median class and the lower limit of the modal class is, A0 (B) 19 (C) 20 (D) 38, Solution:, (C) 20, Explanation:, , Class, , Frequency, , Cumulative, Frequency, , 65-85, , 4, , 4, , 85-105, , 5, , 9, , 105-125, , 13, , 22, , 125-145, , 20, , 42, , 145-165, , 14, , 56, , 165-185, , 7, , 63, , 185-205, , 4, , 67, , Here, N/2 = 67/2 = 33.5 which lies in the interval 125 – 145., Hence, upper limit of median class is 145.

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Here, we see that the highest frequency is 20 which lies in 125 – 145., Hence, the lower limit of modal class is 125., ∴ Required difference = Upper limit of median class – Lower limit of modal class, = 145 – 125 = 20, Hence, option (C) is correct, 10. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated, below, , Class, Frequency, , 13.8-14, , 14-14.2, , 2, , 4, , 14.2-, , 14.4-, , 14.6-, , 14.4, , 14.6, , 14.8, , 5, , 71, , 48, , 14.8-15, 20, , The number of athletes who completed the race in less than 14.6 seconds is :, A11 (B) 71 (C) 82 (D) 130, Solution:, (C) 82, Explanation:, The number of athletes who completed the race in less than 14.6 second= 2 + 4 + 5 + 71 = 82, Hence, option (C) is correct, 11. Consider the following distribution :, Marks obtained Number of students, More than or equal to 0 63, More than or equal to 10 58, More than or equal to 20 55, More than or equal to 30 51, More than or equal to 40 48, More than or equal to 50 42, The frequency of the class 30-40 is, (A) 3 (B) 4 (C) 48 (D) 51, Solution:, (A) 3, Explanation:, , Marks, , Number, , Cumulative, , Obtained, , of, , Frequency, , students, 0-10, , (63 – 58), , 5, , = 5, Oct-20, , (58 – 55), , 3, , = 3, 20-30, , (55 – 51), , 4

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= 4, 30-40, , (51 – 48), , 3, , = 3, 40-50, , (48 – 42), , 6, , = 6, 50<, , 42 = 42, , 42, , Hence, frequency in the class interval 30 – 40 is 3., Hence, option (A) is correct, 12. If an event cannot occur, then its probability is, (A)1 (B) ¾ (C) ½ (D) 0, Solution:, (D) 0, Explanation:, The event which cannot occur is said to be impossible event., The probability of impossible event = zero., Hence, option (D) is correct, 13. Which of the following cannot be the probability of an event?, (A)1/3 (B) 0.1 (C) 3% (D)17/16, Solution:, (D)17/16, Explanation:, Probability of an event always lies between 0 and 1., Probability of any event cannot be more than 1 or negative as (17/16) > 1, Hence, option (D) is correct, , Exercise 13.2 Page No: 161, , 1. The median of an ungrouped data and the median calculated when the same data is grouped, are always the same. Do you think that this is a correct statement? Give reason., Solution:, In order to calculate the median of a grouped data, the formula used is based on the assumption, that the observations in the classes are uniformly distributed or equally spaced. Hence, we cannot, say that the statement “the median of an ungrouped data and the median calculated when the same, data is grouped are always the same” is always correct., , 2. In calculating the mean of grouped data, grouped in classes of equal width, we may use the, formula

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where a is the assumed mean. a must be one of the mid-points of the classes. Is the last, statement correct? Justify your answer., Solution:, No, the statement is not correct. It is not necessary that assumed mean should be the mid – point, of the class interval. a can be considered as any value which is easy to simplify it., 3. Is it true to say that the mean, mode and median of grouped data will always be different?, Justify your answer., Solution:, No, the values of mean, mode and median of grouped data can be the same as well, it depends, on the type of data given., 4. Will the median class and modal class of grouped data always be different? Justify your, answer., , Solution:, The median class and modal class of grouped data is not always different, it depends on the data, given., 5. In a family having three children, there may be no girl, one girl, two girls or three girls. So,, the probability of each is ¼. Is this correct? Justify your answer., Solution:, No it is not correct that in a family having three children, there may be no girl, one girl, two girls or, three girls, the probability of each is ¼. ., Let boys be B and girls be G, Outcomes can be BBB , GGG , BBG , BGB , GBB, GGB, GBG , BGG, Then Probability of 3 girls = 1/8, Probability of 0 girls = 1/8, Probability of 2 girls = 3/8, Probability of 1 girl = 3/8, 6. A game consists of spinning an arrow which comes to rest pointing at one of the regions (1,, 2 or 3) (Fig. 13.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons., Solution:

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Total no. of outcome = 360, p(1)= 90/360 =1/4, p(2) = 90/ 360 = 1/4, p(3) = 180/360 =1/2, Hence, it is clear that the outcome are not equal, 7. Apoorv throws two dice once and computes the product of the numbers appearing on the, dice. Peehu throws one die and squares the number that appears on it. Who has the better, chance of getting the number 36? Why?, Solution:, Apoorv throw two dice at once., Hence, the total number of outcomes = 36, Number of outcomes for getting product 36 = 1(6×6), ∴ Probability for Apoorv = 1/36, Peehu throws one die,, Hence, the total number of outcomes = 6, Number of outcomes for getting square = 36, ∴ Probability for Peehu = 6/36 = 1/6, Therefore, Peehu has a better chance of getting the number 36., , Exercise 13.3 Page No: 166, 1. Find the mean of the distribution :, Class 1-3 3-5 5-7 7-10, Frequency 9 22 27 17, Solution:, We first, find the class mark xi of each class and then proceed as follows., , Class, , Class, , Frequency, , f ixi, , Marks (xi), , (fi), , 01-Mar, , 2, , 9, , 18, , 03-May, , 4, , 22, , 88

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05-Jul, , 6, , 27, , 162, , 07-Oct, , 8.5, , 17, , 144.5, , Σfi = 75, , Σfixi =, 412.5, , Mean,, , Therefore, mean of the given distribution = 5.5., 2. Calculate the mean of the scores of 20 students in a mathematics test :, Marks 10-20 20-30 30-40 40-50 50-60, Number of students 2 4 7 6 1, Solution:, We first, find the class mark xi of each class and then proceed as follows, , Class, , Class, , Frequency, , Marks, , (fi), , f ixi, , (xi), Oct-20, , 15, , 2, , 30, , 20-30, , 25, , 4, , 100, , 30-40, , 35, , 7, , 245, , 40-50, , 45, , 6, , 270, , 50-60, , 55, , 1, , 55, , Σfi = 20, , Σfixi =, 700, , Mean,, , Therefore, mean of scores of 20 students in mathematics test = 35., 3. Calculate the mean of the following data :, Class 4 – 7 8 –11 12– 15 16 –19, Frequency 5 4 9 10, Solution:, The given data is not continuous., So, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class., , Class, 3.5 – 7.5, , Class, , Frequency, , Marks (xi), , (fi), , 5.5, , 5, , f ixi, 27.5

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7.5 – 11.5, , 9.5, , 4, , 38, , 11.5 –, , 13.5, , 9, , 121.5, , 17.5, , 10, , 175, , Σfi = 28, , Σfixi =, , 15.5, 15.5 –, 19.5, 362, Mean,, , Therefore, mean of the given data = 12.93., 4. The following table gives the number of pages written by Sarika for completing her own book, for 30 days :, Number of pages written per day 16-18 19-21 22-24 25-27 28-30, Number of days 1 3 4 9 13, Find the mean number of pages written per day., Solution:, , Class, , Mid –, , Number of, , f ixi, , Marks, , Value (xi), , days (fi), , 15.5 –, , 17, , 1, , 17, , 20, , 3, , 60, , 23, , 4, , 92, , 26, , 9, , 234, , 29, , 13, , 377, , Σfi = 30, , Σfixi =, , 18.5, 18.5 –, 21.5, 21.5 –, 24.5, 24.5 –, 27.5, 27.5 –, 30.5, 780, The given data is not continuous., Hence, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class., , Therefore, the mean of pages written per day = 26., 5. The daily income of a sample of 50 employees are tabulated as follows :, Income (in Rs) 1-200 201-400 401-600 601-800

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Number of employees 14 15 14 7, Find the mean daily income of employees., Solution:, , C.I, , xi, , di =, , Fi, , f id i, , (xi – a), 1 – 200, , 100.5, , – 200, , 14, , – 2800, , 201 –, , 300.5, , 0, , 15, , 0, , 500.5, , 200, , 15, , 2800, , 700.5, , 400, , 7, , 2800, , Σfi = 50, , Σfixi =, , 400, 401 –, 600, 601 –, 800, 2800, ∴ Assumed mean, a = 300.5 and di = (xi – a), , = 300.5 + 2800/50, = 356.5, Hence, the average daily income of employees = Rs.356.5, 6. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given, in the following table :, Number of seats 100-104 104-108 108-112 112-116 116-120, Frequency 15 20 32 18 15, Determine the mean number of seats occupied over the flights., Solution:, , Class, , Class, , Frequency, , Deviation, , Interval, , Marks, , (fi), , (di = xi –, , (xi), 100 –, , f id i, , a), , 102, , 15, , – 8, , – 120, , 106, , 20, , – 4, , – 80, , 110, , 32, , 0, , 0, , 114, , 18, , 4, , 72, , 118, , 15, , 8, , 120, , 104, 104 –, 108, 108 –, 112, 112 –, 116, 116 –

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120, N = Σfi =, , Σfidi =, , 100, , – 8, , ∴ Assumed mean, a = 110, Class width, h = 4, And total observations, N = 100, Hence, finding mean,, , = 110 + (-8/100), = 110 – 0.08, = 109.92, But we know that the seats cannot be in decimal., Therefore, the number of seats = 109., 7. The weights (in kg) of 50 wrestlers are recorded in the following table :, , Weight (in, , 100-110, , kg), Number, , 4, , 110-, , 120-, , 130-, , 140-, , 120, , 130, , 140, , 150, , 14, , 21, , 8, , 3, , of, wrestlers, Find the mean weight of the wrestlers., Solution:, , Weight, , Number, , Class, , Deviation, , (in kg), , of, , Marks (xi), , (di = xi –, , Wrestlers, , f id i, , a), , (fi), 100 – 110, , 4, , 105, , – 20, , – 80, , 110 – 120, , 14, , 115, , – 10, , – 140, , 120 – 130, , 21, , 125, , 0, , 0, , 130 – 140, , 8, , 135, , 10, , 80, , 140 – 150, , 3, , 145, , 20, , 60, , N = Σfi =, , Σfidi = –, , 50, , 80, , ∴ Assumed mean, (a) = 125, Class width, (h) = 10, and total observations, (N) = 50

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By step deviation method,, , = 125 – 16, = 123.4kg, Hence, mean weight of wrestlers = 123.4kg, 8. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and, details are tabulated as given below :, , Mileage (km/l) 10-12 12-14 14-16 16-18, Number of cars 7 12 18 13, Find the mean mileage., The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this, claim?, Solution:, , Mileage (km, , Class –, , Number of, , -1, , L ), , Marks (xi), , cars (fi), , 10 – 12, , 11, , 7, , 77, , 12 – 14, , 13, , 12, , 156, , 14 – 16, , 15, , 18, , 270, , 16 – 18, , 17, , 13, , 221, , Σfi = 50, , Σfixi =, , Total, , f ixi, , 724, Here, Σfi = 50, Σfixi = 724, , = 724/50 = 14.48, Hence, mean mileage = 14.48 km/h, No, I don’t agree with the claim because the manufacturer is claiming mileage 1.52 km/h more than, average mileage., 9. The following is the distribution of weights (in kg) of 40 persons :, Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80, Number of persons 4 4 13 5 6 5 2 1, , Construct a cumulative frequency distribution (of the less than type) table for the data above., Solution:

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Weight (in, , Cumulative, , kg), , frequency, , Less than 45, , 4, , Less than 50, , 4 + 4 = 8, , Less than 55, , 8 + 13 =, 21, , Less than 60, , 21 + 5 =, 26, , Less than 65, , 26 + 6 =, 32, , Less than 70, , 32 + 5 =, 37, , Less than 75, , 37 + 2 =, 39, , Less than 80, , 39 + 1 =, 40, , 10. The following table shows the cumulative frequency distribution of marks of 800 students in, an examination:, Marks Number of students, Below 10 10, Below 20 50, Below 30 130, Below 40 270, Below 50 440, Below 60 570, Below 70 670, Below 80 740, Below 90 780, Below 100 800, Construct a frequency distribution table for the data above., Solution:, The frequency distribution table for the given data is:

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Class, , Number of, , Interval, , students, , 0-10, , 10, , Oct-20, , 50 – 10 = 40, , 20-30, , 130 – 50 =, 80, , 30-40, , 270 – 130 =, 140, , 40-50, , 440 – 270 =, 170, , 50-60, , 570 – 440 =, 130, , 60-70, , 670 – 570 =, 100, , 70-80, , 740 – 670 =, 70, , 80-90, , 780 – 740 =, 40, , 90-100, , 1. – 780 = 20, , 11. Form the frequency distribution table from the following data :, Marks (out of 90) Number of candidates, More than or equal to 80 4, More than or equal to 70 6, More than or equal to 60 11, More than or equal to 50 17, More than or equal to 40 23, More than or equal to 30 27, More than or equal to 20 30, More than or equal to 10 32, More than or equal to 0 34, Solution:, The frequency distribution table for the given data is:, , Class, , Number of, , Interval, , students, , 0-10, , 34 – 32 =, 2

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Oct-20, , 32 – 30 =, 2, , 20-30, , 30 – 27 =, 3, , 30-40, , 27 – 23 =, 4, , 40-50, , 23 – 17 =, 6, , 50-60, , 17 – 11 =, 6, , 60-70, , 11 – 6 = 5, , 70-80, , 6 – 4 = 2, , 80-90, 12. Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in, a class:, , Height Frequency Cumulative frequency, (in cm), , 150-155 12 a, 155-160 b 25, 160-165 10 c, , 165-170 d 43, 170-175 e 48, 175-180 2 f, Total 50, , Solution:, , Height, , Cumulative, , Cumulative, , (in, , frequency, , frequency, , cm), , given, , 150 –, , Frequency, , 12, , a, , 12, , b, , 25, , 12 + b, , 10, , c, , 22 + b, , d, , 43, , 22 + b +, , 155, 155 –, 160, 160 –, 165, 165 –, 170, 170 –, , d, e, , 48, , 22 + b +

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175, , d + e, , 175 –, , 2, , f, , 24 + b +, , 180, , d + e, , Total, , 50, , On comparing last two tables, we get, a = 12, ∴ 12 + b = 25, ⇒ b = 25 – 12 = 13, 22 + b = c, ⇒ c = 22 + 13 = 35, 22 + b + d = 43, ⇒ 22 + 13 + d = 43, ⇒ d = 43 – 35 = 8, 22 + b + d + e = 48, ⇒ 22 + 13 + 8 + e = 48, ⇒ e = 48 – 43 = 5, 24 + b + d + e = f, ⇒ f = 24 + 13 + 8 + 5 = 50, 13. The following are the ages of 300 patients getting medical treatment in a hospital on a, particular day:, , Age (in years) 10-20 20-30 30-40 40-50 50-60 60-70, Number of patients 60 42 55 70 53 20, Form:, ALess than type cumulative frequency distribution., , , , More than type cumulative frequency distribution.Solution:(i)Less than type cumulative, frequency distribution of the data is given below., , (i) Less than type, Age (in year), , Number of, patients, , Less than 10, , 0, , Less than 20, , 60 + 0 = 60, , Less than 30, , 60 + 42 =, 102, , Less than 40, , 102 + 55 =, 157, , Less than 50, , 157 + 70 =

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227, Less than 60, , 227 + 53 =, 280, , Less than 70, , 280 + 20, =300, , , , (ii), , , , More than type cumulative frequency distribution of the data is given below., , (i) More than type, Age (in year), , Number of, patients, , More than or, , 60 + 42 + 55 +, , equals 10, , 70 + 53 + 20 =, 300, , More than or, , 42 + 55 + 70 +, , equals 20, , 53 + 20 = 240, , More than or, , 55 + 70 + 53 +, , equals 30, , 20 = 198, , More than or, , 70 + 53 + 20 =, , equals 40, , 143, , More than or, , 53 + 20 = 73, , equals 50, More than or, , 20, , equals 60, More than or, , 0, , equals 70, , , , 14. Given below is a cumulative frequency distribution showing the marks secured by 50, students of a class:, , , , Marks Below 20 Below 40 Below 60 Below 80 Below 100, , , , Number of students 17 22 29 37 50, , , , Form the frequency distribution table for the data., , , , Solution:

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The frequency distribution table for given data., , , , Marks, , Number of, students, , 0 – 20, , 12, , 20 – 40, , 22 – 17 = 5, , 40 – 60, , 29 – 22 = 7, , 60 – 80, , 37 – 29 = 8, , 80 – 100, , 50 – 37 = 13, , , , 15. Weekly income of 600 families is tabulated below :, , , , Weekly income Number of families, , , , (in Rs), , , , 0-1000 250, , , , 1000-2000 190, , , , 2000-3000 100, , , , 3000-4000 40, , , , 4000-5000 15, , , , 5000-6000 5, , , , Total 600, , , , Compute the median income., , , , Solution:, , Weekly, , Number, , Cumulative, , Income, , of, , frequency, , families, , (cf), , (fi), 0-1000, , 250, , 250, , 1000-, , 190, , 250 + 190, , 2000, 2000-, , = 400, 100, , 3000, 3000-, , = 540, 40, , 4000, 4000-, , 6000, , , 540 + 40, = 580, , 15, , 5000, 5000-, , 440 + 100, , 580 + 15, = 595, , 5, , 595 + 5 =, 600

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, , According to the question,, , , , n = 600, , , , ∴ n/2 = 600/2 = 300, , , , Cumulative frequency 440 lies in the interval 1000 – 2000., , , , Hence, lower median class, l = 1000, , , , f = 190,, , , , cf = 250,, , , , Class width, h = 1000, , , , And total observation n = 600, , , , , = 1000 + 5000/19, , , , = 1000 + 263.15 = 1263.15, , , , Hence, the median income is Rs.1263.15., , , , 16. The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching, centre are given as follows:, , , , Speed (km/h) 85-100 100-115 115-130 130-145, , , , Number of players 11 9 8 5, , , , Calculate the median bowling speed., , , , Solution:, , , , First we construct the cumulative frequency table, , Speed, , Number, , Cumulative, , ( in, , of, , frequency, , km/h), , players, , 85 –, , 11, , 11, , 9, , 11 + 9 =, , 100, 100 –, 115, 115 –, , 20, 8, , 130, 130 –, 145, , 20 + 8 =, 28, , 5, , 28 + 5 =, 33, , , , It is given that, n = 33, , , , ∴ n/2 = 33/2 = 16.5, , , , Hence, the median class is 100 – 115.

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, , Where, lower limit(l) = 100, , , , Frequency (f) = 9, , , , Cumulative frequency (cf) = 11, , , , And class width(h) = 15, , , , , = 100 + 82.5/9, , , , = 100 + 9.17, , , , = 109.17, , , , Hence, the median bowling speed is 109.17 km/h., , , , 17. The monthly income of 100 families are given as below :, , , , Income (in Rs) Number of families, , , , 0-5000 8, , , , 5000-10000 26, , , , 10000-15000 41, , , , 15000-20000 16, , , , 20000-25000 3, , , , 25000-30000 3, , , , 30000-35000 2, , , , 35000-40000 1, , , , Calculate the modal income., , , , Solution:, , , , According to the data given,, , , , The highest frequency = 41,, , , , 41 lies in the interval 10000 – 15000., , , , Here, l = 10000, fm = 41,f1 = 26,f2 = 16 and h = 5000, , , , , = 10000 + 15×125, , , , = 10000 + 1875, , , , = 11875, , , , Hence, the modal income = Rs.11875 per month.

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, , 18. The weight of coffee in 70 packets are shown in the following table :, , , , Weight (in g) Number of packets, , , , 200-201 12, , , , 201-202 26, , , , 202-203 20, , , , 203-204 9, , , , 204-205 2, , , , 205-206 1, , , , Determine the modal weight., , , , Solution:, , , , In the given data, the highest frequency is 26, which lies in the interval 201 – 202, , , , Here, l = 201,fm = 26,f1 = 12,f2 = 20 and (class width) h = 1, , , , , Hence, the modal weight = 201.7 g., , , , 19. Two dice are thrown at the same time. Find the probability of getting, , , , A Same number on both dice., Different numbers on both dice., , , , Solution:, , , , Two dice are thrown at the same time., , , , So, total number of possible outcomes = 36, , , , (i) Same number on both dice., , , , Possible outcomes = (1,1), (2,2), (3, 3), (4, 4), (5, 5), (6, 6)., , , , Hence, number of possible outcomes = 6, , , , Therefore, the probability of getting same number on both dice = 6/36 = 1/6, , , , (ii) Different number on both dice., , , , Hence, number of possible outcomes, , , , = 36 – Number of possible outcomes for same number on both dice, , , , = 36 – 6 = 30, , , , Therefore, the probability of getting different number on both dice = 30/36 = 5/6, , , , 20. Two dice are thrown simultaneously. What is the probability that the sum of the, numbers appearing on the dice is, , , , (i) 7? (ii) a prime number? (iii) 1?

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, , Solution:, , , , According to the question,, , , , Two dice are thrown simultaneously., , , , So, that number of possible outcomes = 36, , , , (i) Sum of the numbers appearing on the dice is 7., , , , So, the possible outcomes = (1, 6), (2,5), (3, 4), (4, 3), (5, 2), (6, 1)., , , , Hence, number of possible outcomes = 6, , , , ∴ the probability that the sum of the numbers appearing on the dice is 7 = 6/36 = 1/6, , , , (ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11., , , , So, the possible outcomes are (1, 1), (1,2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5),, (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) = (6, 5)., , , , Hence, number of possible outcomes = 15, , , , ∴ the probability that the sum of the numbers appearing on the dice is a prime number =, 15/36 = 5/12, , , , (iii) Sum of the numbers appearing on the dice is 1., , , , It is not possible, so its probability is zero., , , , ∴ the probability that the sum of the numbers appearing on the dice is 1 = 0, , , , 21. Two dice are thrown together. Find the probability that the product of the numbers on, the top of the dice is, , , , (i) 6 (ii) 12 (iii) 7, , , , Solution:, , , , Number of total outcomes = 36, , , , (i) When product of the numbers on the top of the dice = 6., , , , The possible outcomes = (1, 6), (2,3), (3, 2), (6, 1)., , , , Hence, number of possible ways = 4, , , , ∴ Probability that the product of the numbers on the top of the dice is 6= 4/36 = 1/9, , , , (ii) When product of the numbers on the top of the dice = 12., , , , The possible ways are (2, 6), (3,4), (4, 3), (6, 2)., , , , Hence, number of possible ways = 4, , , , ∴ Probability that the product of the numbers on the top of the dice is 12= 4/36 = 1/9, , , , (iii) Product of the numbers on the top of the dice cannot be 7., , , , Hence, the probability is zero., , , , ∴ Probability that the product of the numbers on the top of the dice is 7 = 0