## NCERT Solutions Class 11th Maths Chapter 15: Statistics## Exercise 15.1
1. 4, 7, 8, 9, 10, 12, 13, 17
Mean of the given data = x = 1/8 × = 80/8 = 10 The respective deviations from mean for the given data will be: 4 - 10 = 7 - 10 = 8 - 10 = 9 - 10 = 10 - 10 = 12 - 10 = 13 - 10 = 17 - 10 =
Absolute values of the deviations: 6, 3, 2, 1, 0, 2, 3, 7
Mean deviation for the given data = Sum of deviations/Number of observations = 24/8 = 3 2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data = x = 1/10 × = 500/10 = 50 The respective deviations from mean for the given data will be: 38 - 50 = 70 - 50 = 48 - 50 = 40 - 50 = 42 - 50 = 55 - 50 = 63 - 50 = 46 - 50 = 54 - 50 = 44 - 50 =
Absolute values of the deviations: 12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Mean deviation for the given data = Sum of deviations/Number of observations = 84/10 = 8.4
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Arrange the given data in ascending order. 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 Number of observations = n = 12 Median of the given data = ((n/2) (n/2) (n/2 + 1) Median = (13 + 14)/2 = 13.5 The respective deviations from median for the given data will be: 10 - 13.5 = 11 - 13.5 = 11 - 13.5 = 12 - 13.5 = 13 - 13.5 = 13 - 13.5 = 14 - 13.5 = 16 - 13.5 = 16 - 13.5 = 17 - 13.5 = 17 - 13.5 = 18 - 13.5 = Absolute value of respective deviations from median: 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Mean deviation = MD = 1/12 × = 28/12 = 2.333 4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Arrange the given data in ascending order. 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 Number of observations = n = 10 Median of the given data = ((n/2) (n/2) (n/2 + 1) Median = (46 + 49)/2 = 47.5 The respective deviations from median for the given data will be: 36 - 47.5 = - 42 - 47.5 = - 45 - 47.5 = 46 - 47.5 = 46 - 47.5 = 49 - 47.5 = 51 - 47.5 = 53 - 47.5 = 60 - 47.5 = 72 - 47.5 = Absolute value of respective deviations from median: 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Mean deviation = MD = 1/10 × = 70/10 = 7
Number of observations = N =
Mean = x = 1/N × Sum of absolute deviations from mean = Mean deviation about the mean = 1/N ×
Number of observations = N =
Mean = x = 1/N × Sum of absolute deviations from mean = Mean deviation about the mean = 1/N ×
Number of observations = N = 26 (even) (N/2) (N/2 + 1) Median = ((N/2) = (7 + 7)/2 = 14/2 = 7
Sum of absolute deviations from median = Mean deviation about the median = 1/N ×
Number of observations = N = 29 (odd) The cumulative frequency greater than N/2 = 14.5 is 21 and the corresponding observation for it is 30 15 16 Median = (15 = (30 + 30)/2 = 60/2 = 30
Sum of absolute deviations from median = Mean deviation about the median = 1/N ×
Number of observations = N =
Mean = x = 1/N × Sum of absolute deviations from mean = Mean deviation about the mean = 1/N ×
Number of observations = N =
Mean = x = 1/N × Sum of absolute deviations from mean = Mean deviation about the mean = 1/N × 11. Find the mean deviation about median for the following data :
Number of observations = N = 50 (N/2) The class interval that contains the 25 l = 20, h = 10 Median = l + h × (N/2 - c)/f = 20 + 10 × (25 - 14)/14 = 20 + 10 × 11/14 = (280 + 110)/14 = 390/14 = 27.85
Sum of absolute deviations from median = Mean deviation about the median = 1/N × 12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:
The data needs to be converted into continuous frequency distribution. Therefore, subtract 0.5 from the lower limit and add 0.5 to the upper limit in each class interval.
Number of observations = N = 100 (N/2) The class interval that contains the 50 l = 35.5, h = 5 Median = l + h × (N/2 - c)/f = 35.5 + 5 × (50 - 37)/26 = 35.5 + 5 × 13/26 = 35.5 + 2.5 = 38
Sum of absolute deviations from median = Mean deviation about the median = 1/N × ## Exercise 15.2
1. 6, 7, 10, 12, 13, 4, 8, 12
Mean = x = ( = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8 = 72/8 = 9
Variance = σ = 1/8 × 74 = 9.4 Hence, Mean of the data = 9 Variance of the data = 9.4 2. First n natural numbers
Mean = x = ( = (n(n + 1)/2)/n = (n + 1)/2 Variance = σ = 1/n × = 1/n × [ = 1/n × [n(n + 1)(2n + 1)/6 + n(n + 1) = (n + 1)(2n + 1)/6 + (n + 1) = [2(n + 1)(2n + 1) + 3(n + 1) = [2(2n = [2(2n = [4n = (n Hence, Mean of the data = (n + 1)/2 Variance of the data = (n 3. First 10 multiples of 3
The first 10 multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Mean = x = ( = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10 = 165/10 = 16.5
Variance = σ = 1/10 × 742.5 = 74.25 Hence, Mean of the data = 16.5 Variance of the data = 74.25
N =
Mean = x = 1/N × Variance = σ = 1/40 × 1736 = 43.4 Hence, Mean of the data = 19 Variance of the data = 43.4
N =
Mean = x = 1/N × Variance = σ = 1/22 × 640 = 29.09 Hence, Mean of the data = 100 Variance of the data = 29.09 6. Find the mean and standard deviation using short-cut method.
Let assumed mean = A = 64. We have h = 1.
Mean = x = A + h × ( = 64 + 1 × (0)/100 = 64 Variance = σ = (1 = 28600/10000 = 2.86 Therefore, standard deviation of the given data = σ = √2.86 = 1.69 Hence, Mean of the data = 64 Standard deviation of the data = 1.69
N =
Mean = x = 1/N × Variance = σ = 1/30 × 68280 = 2276 Hence, Mean of the data = 107 Variance of the data = 2276
N =
Mean = x = 1/N × Variance = σ = 1/50 × 6600 = 132 Hence, Mean of the data = 27 Variance of the data = 132 9. Find the mean, variance and standard deviation using short-cut method
Let assumed mean = A = (90 + 95)/2 = 92.5. We have h = 5.
Mean = x = A + h × ( = 92.5 + 5 × (6)/60 = 92.5 + 0.5 = 93 Variance = σ = (1 = (15240 - 36)/3600 = 15204/3600 = 105.583 Therefore, standard deviation of the given data = σ = √105.583 = 10.27 Hence, Mean of the data = 93 Variance of the data = 105.58 Standard deviation of the data = 10.27 10. The diameters of circles (in mm) drawn in a design are given below:
The data needs to be converted into continuous frequency distribution. Therefore, subtract 0.5 from the lower limit and add 0.5 to the upper limit in each class interval.
Let assumed mean = A = (40.5 + 44.5)/2 = 42.5. We have h = 4.
Mean = x = A + h × ( = 42.5 + 4 × (25)/100 = 42.5 + 1 = 43.5 Variance = σ = (4 = 16(19900 - 625)/10000 = 19275/625 = 30.84 Therefore, standard deviation of the given data = σ = √30.84 = 5.55 Hence, Mean of the data = 43.5 Standard deviation of the data = 5.55 ## Exercise 15.31. From the data given below state which group is more variable, A or B?
Group A: Let assumed mean = A = 45. We have h = 10.
Mean = x = A + h × ( = 45 + 10 × (-6)/150 = 45 - 0.4 = 44.6 Variance = σ = (10 = 100(51300 - 36)/22500 = 51264/225 = 227.84 Standard deviation = σ = √227.84 = 15.09 Coefficient of Variance for Group A = σ/x × 100 = 15.09/44.6 × 100 = 33.83 Group B: Let assumed mean = A = 45. We have h = 10.
Mean = x = A + h × ( = 45 + 10 × (-6)/150 = 45 - 0.4 = 44.6 Variance = σ = (10 = 100(54900 - 36)/22500 = 54864/225 = 243.84 Standard deviation = σ = √243.84 = 15.61 Coefficient of Variance for Group A = σ/x × 100 = 15.61/44.6 × 100 = 35 Therefore, C.V. is higher for Group B. Hence, Group B is more variable. 2. From the prices of shares X and Y below, find out which is more stable in value:
Number of terms = n = 10 Mean for X = x = ∑ x = 510/10 = 51 Variance for X = 1/n = 1/10 = 1/100 × [263600 - 260100] = 3500/100 = 35 Standard deviation = σ = √35 = 5.91 Coefficient of Variance = σ/x × 100 = 5.91/51 × 100 = 11.58 Mean for Y = y̅ = ∑ y = 1050/10 = 105 Variance for Y = 1/n = 1/10 = 1/100 × [1102900 - 1102500] = 400/100 = 4 Standard deviation = σ = √4 = 2 Coefficient of Variance = σ/y̅ × 100 = 2/105 × 100 = 1.904 C.V. of X is higher. Hence, Y is more stable. 3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
**Which firm A or B pays larger amount as monthly wages?****Which firm, A or B, shows greater variability in individual wages?**
Number of wage earners in firm A = 586 Total amount paid as monthly wages in firm A = 5253 × 586 = Rs 3078258 Mean monthly wages of firm B = Rs 5253 Number of wage earners in firm B = 648 Total amount paid as monthly wages in firm B = 5253 × 648 = Rs 3403944 Therefore, firm B pays a larger amount as monthly wages
Standard deviation of firm A = √100 = 10 Coefficient of Variance for firm A = 10/5253 × 100 = 0.19 Variance of firm B = 121 Standard deviation of firm B = √121 = 11 Coefficient of Variance for firm B = 11/5253 × 100 = 0.20 C.V. of B is higher. Hence, firm B has greater variability in individual wages. 4. The following is the record of goals scored by team A in a football session:
Mean for team A = x = ∑ f Variance = 1/n = 1/25 = [3250 - 2500]/625 = 750/625 = 1.2 Standard Deviation for team A = σ = √1.2 = 1.09 Coefficient of variation for team A = σ/x × 100 = 1.09/2 × 100 = 54.5 Mean for team B = 2 Standard deviation for team B = 1.25 Coefficient of variation for team B = 1.25/2 × 100 = 62.5 C.V. of team B is greater than team A. Hence, Team A is more consistent. 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:
Mean for length = x = ∑ x Variance in length = 1/n = 1/50 = [45140 - 44944]/2500 = 196/2500 = 0.0784 Standard deviation for length = σ = √0.0784 = 0.28 Coefficient of variance in length = σ/x × 100 = 0.28/4.24 × 100 = 6.60 Mean for weight = y̅ = ∑ y Variance in weight = 1/n = 1/50 = [72880 - 68121]/2500 = 4759/2500 = 1.9036 Standard deviation for weight = σ = √1.9036 = 1.37 Coefficient of variance in weight = σ/x × 100 = 1.37/5.22 × 100 = 26.24 C.V. of weight is greater than C.V. of height. Hence, weight is more varying. ## Miscellaneous Exercise1. 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.
Let the remaining two observations to be x and y respectively. Then, the observations are: 6, 7, 10, 12, 12, 13, x, y Mean = x = (6 + 7 + 10 + 12 + 12 + 13 + x + y)/8 9 = (60 + x + y)/8 72 = 60 + x + y
Variance = 1/n × 9.25 = 1/8 × [(6 - 9) 74 = (-3) 74 = 9 + 4 + 1 + 9 + 9 + 16 + 81 + 81 + x 74 = 210 + x 74 = 210 + x x
Now, square both sides in x + y = 12 x 80 + 2xy = 144 2xy = 64 Subtract 2xy = 64 from x x (x - y)
CASE I: x - y = 4 x = 4 + y x + y = 12 4 + 2y = 12 2y = 8
So, x = 4 + 4
CASE II: x - y = -4 x = -4 + y x + y = 12 -4 + 2y = 12 2y = 16
So, x = -4 + 8
Hence, the remaining two observations are 4 and 8. 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.
Let the observations be x Mean = x = (x 8 = (x Let the resulting observations after multiplying each observation by 3 be: y New mean = y̅ = (y = (3x = 3(x = 3(x) = 3(8) = 24 Original Standard Deviation = σ = √(1/n × 4 = √(1/6 × Square both sides 16 = 1/6 × Now, x 16 = 1/6 × 16 = 1/6 × 1/3 16 × 9 = 1/6 × 144 = 1/6 × New standard deviation = √(1/6 × Hence, new mean is 24 and new standard deviation is 12. 4. Given that x is the mean and σ 2 is the variance of n observations x
Mean = x = (x x = (x Let each new observation be denoted by y y x New mean = y̅ = (ax = a(x = ax Variance = σ σ σ a New variance = 1/n Hence, proved. 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:
Incorrect mean = 1/n × 10 = 1/20 × 200 = Incorrect sum of observations = 200 Correct sum of observations = 200 - 8 = 192 Number of observations after 8 is omitted = N = 20 - 1 = 19 Correct mean = 1/19 × 192 = 10.11 Incorrect standard deviation = √(1/n × 2 = √(1/n × Square both sides 4 = 1/20 × 104 = 1/20 × 2080 = Incorrect sum of squares of observations = 2080 Correct sum of squares of observations = 2080 - 8 Correct Standard deviation = √(1/N × = √(1/19 × 2016 - 102.1) = √(106.1 - 102.01) = √4.09 = 2.02
Correct sum of observations = 200 - 8 + 12 = 204 Number of observations = N = 20 Correct mean = 1/20 × 204 = 10.2 Incorrect sum of squares of observations = 2080 Correct sum of squares of observations = 2080 - 8 Correct Standard deviation = √(1/N × = √(1/20 × 2160 - 104.04) = √(108 - 104.04) = √3.96 = 1.98 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:
Coefficient of Variance in Mathematics = Standard Deviation/Mean × 100 = 12/42 × 100 = 28.57 Coefficient of Variance in Physics = 15/32 × 100 = 46.87 Coefficient of Variance in Chemistry = 20/40.9 × 100 = 48.89 C.V. of marks in Chemistry is the highest while that of Mathematics is lowest Hence, chemistry has the highest variability in marks while Mathematics shows the lowest. 7. 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.
Number of observations = 100 Incorrect mean = 1/n × 20 = 1/100 × 2000 = Incorrect sum of observations = 2000 Correct sum of observations = 2000 - 21 - 21 - 18 = 1940 Number of observations after incorrect observations are omitted = N = 100 - 3 = 97 Correct mean = 1/97 × 1940 = 20 Incorrect standard deviation = √(1/n × 3 = √(1/n × Square both sides 9 = 1/100 × 409 = 1/100 × 40900 = Incorrect sum of squares of observations = 40900 Correct sum of squares of observations = 40900 - 21 Correct Standard deviation = √(1/N × = √(1/97 × 39694 - 400) = √(409.21 - 400) = √9.21 = 3.036 |