Mathematics-29.Unit-21-Measures of Central Tendency and Dispersion

STATISTICS One of the most important objectives of statistical analysis is to get one siangle value that describes the characteristics of entire mass of unwieldy data such a value is called central value or the average. It is a single value, which represents a group of values. Classification of Data: (i) Discrete series (ii) Continuous series TYPE OF AVERAGES (a) Mean (i) Arithmetic (ii) Weighted arithmetic mean (iii) Geometric Mean (iv) Weighted Geometric Mean (v) Harmonic Mean (vi) Weighted Harmonic Mean (b) Median (c) Mode The Arithmetic Mean: The arithmetic mean of a statistical data is defined as the quotient of the sum of all the values of the variable by the total number of items. It is denoted by A.M. For an individual series (a) A.M. = (b) A.M. = A + , where d = x – A, A is the assumed mean For a frequency distribution, (a) A.M = (Mean of grouped data) (b) Short Cut Method A.M. = A + , where d = x – A, where A is the assumed mean. (c) Step Deviation Method A.M. = A + h, where A is the assumed mean and u = . Weighted Arithmetic Mean: If w1, w2, w3, …, wn are the weights assigned to the values x1, x2, x3, …, xn respectively, then the weighted average is defined as: Weighted Arithmetic Mean = . Example -1: A group of 10 items has mean 6. If the mean of 4 of these items is 7.5, then the mean of remaining items is (A) 6.5 (B) 5.5 (C) 4.5 (D) 5.0 Solution: Sum of all the 10 items = 10 × 6 = 60 Sum of four of these items 4 × 7.5 = 30 ∴ sum of the remaining six items = 60 –30 = 30 hence, the mean of remaining six items = = 5 Exercise 1: The mean of a set of observations is . If each observation is divided by α, α ≠ 0 and then is increased by 10, then the mean of the new set is (A) (B) (C) (D) Median: Median is defined as the middle most or the central value of the variables in a set of observations, when the observations are arranged either in ascending or in descending order of their magnitudes. It divides the arranged series in two equal parts. Median is a position average, whereas, the arithmetic mean is the calculated average. When a series consists of an even number of terms, median is the arithmetic mean of the two central items. It is generally denoted by M. Case I: When n is odd. In this case th value is the median i.e. Case II: When n is even. In this case there are two middle terms th and . The median is the average of those two terms, i.e. th term Case III: When the series is continuous. In this case the data is given in the form of a frequency table with class-interval, etc., and the following formula is used to calculate the Median. M = L + , where L = lower limit of the class in which the median lies n = total number of frequencies, i.e., n = Σf. f = frequency of the class in which the median lies C = cumulative frequency of the class preceding the median class i = width of the class-interval of the class in which the median lies. Quartile: Just as the median divides a set of observations (when arranged in ascending or descending order of magnitudes), into two equal parts, similarly quartile divides the observations into four equal parts. The value of the item midway, between the first item and the median is known as first or lower quartile and is denoted be Q1. The value of the item midway between the last item and the median is known Third or Upper Quartile and is denoted Q3. The median is known as the Second Quartile and denoted by Q2. Case I: For ungrouped data. Q1 = th item, Q3 = the item. Case II: For a frequency distribution Q1 = , , where L = lower limit of the class in which a particular quartile lies, f = Frequency of the class-interval in which a particular quartile lies i = Class-interval of the class in which a particular quartile lies, C = Cumulative frequency of the class preceding the class in which the particular quartile lies In general, Qh = L + h = 1, 2, 3, 4 Deciles: The values which divide the series when arranged in ascending or descending order of magnitudes in ten equal parts are called deciles. There are 9 deciles in all, which are denoted by D1, D2, D3, D4, D5, D6, D7, D8, D9 respectively. Case I: For ungrouped data. Case II: When the series is grouped. , where L, C, f, i have their usual meanings and h = 1, 2, 3, 4, 5, 6, 7, 8, 9. Percentiles: The values of the variables which divide the series, when arranged in ascending or descending order, into 100 equal parts are called percentiles. There are 99 percentiles denoted by P1, P2, P3, P4,…,P99 respectively. Case I: When the series is ungrouped. The percentiles are calculated by the following formula: h = 1, 2, …, 99 Case II: When the series is grouped. xi, h = 1, 2, …, 99, where L, C, f, i have their usual meanings. Mode: Mode is defined as that value in a series which occurs most frequently. In a frequency distribution mode is that variate which has the maximum frequency. Continuous Frequency Distribution: i) Modal Class: It is that class in grouped frequency distribution in which the mode lies. Mode = , where L = the lower limit of the modal class i = the width of the modal class f1 = the frequency of the class preceding modal class fm = the frequency of the modal class f2 = the frequency of the class succeeding modal class. Sometimes it so happened that the above formula fails to give the mode. In this case, the modal value lies in a class other than the one containing maximum frequency. In such cases we take the help of the following formula: Mode = , where L, f1, f2, i have usual meanings. Asymmetrical Distribution: A distribution in which mean, median and mode coincide is called symmetrical distribution. If the distribution is moderately asymmetrical, then mean, median and mode are connected by the formula. Mode = 3 Median – 2Mean Example -2: The quartile deviation of the daily wages (in Rs) of 7 persons given below: 12, 7, 15, 10, 17, 19, 25 is (A) 14.5 (B) 5 (C) 9 (D) 4.5 Solution: Arranging the given observations in ascending order, we have 7, 10, 12, 15, 17, 19, 25. So, Q1 = 2nd observation = 10 and Q3 = 6th observation = 19 Hence Q.D = = 4.5 Geometric Mean: If x1, x2, x3, …, xn are n values of a variable x, none of them being zero, then the Geometric mean G is defined as G = (x1, x2, x3, …, xn )1/n. Geometric Mean for Frequency Distribution: Geometric mean of n values x1, x2, x3, …, xn of a variable x, occurring with frequency f1, f2, f3, …, fn respectively is given by or G = antilog Harmonic Mean: The harmonic mean of n items x1, x2, x3,…, xn is defined as: Harmonic Mean = Harmonic Mean of Frequency Distribution: Let x1, x2, x3, …, xn be n items which occur with frequencies f1, f2, f3, …, fn respectively. Then their Harmonic Mean is given by Harmonic Mean = Relation between Arithmetic Mean, Geometric Mean and Harmonic Mean: The arithmetic mean (A. M.), Geometric mean (G.M.) and Harmonic Mean (H.M.) for a given set of observations of a series are related as under: A. M ≥ G.M ≥ H.M MEASURES OF DISPERSION Dispersion means scatterness. The degree to which numerical data tend to spread about an average value is called the dispersion of the data. There are four measures of dispersion. Range: Range = L – S, where L = largest value; S = smallest value. Coefficient of Range = Example -3: The range of the following set of observations 2, 3, 5, 9, 8, 7, 6, 5, 7, 4, 3, is (A) 11 (B) 7 (C) 5.5 (D) 6 Solution: (B) Quartile Deviation: Quartile deviation = Coefficient of Quartile Deviation = Mean Deviation: It is the average of the modulus of the deviations of the observations in a series taken form mean or median. Methods for Calculation of Mean Deviation: Case I: For Ungrouped Data. In this case the mean deviation is given by the formula Mean Deviation = M.D. = , where ‘d’ stands for the deviation from the mean or median and |d| is always positive whether d itself is positive or negative and n is the total number of items. Case II: For Grouped data. Let x1, x2, x3, …, xn occur with frequencies f1, f2, f3, ,fn respectively and let Σf = n and M can be either Mean or Median, then the mean deviation is given by the formula. Mean Deviation = Where d = |x – M| and Σf = n. Coefficient of Mean Deviation = or = (In case the deviations are taken from mean) Standard Deviation: The Positive square root of the average of squared deviations of all observations taken from their mean is called standard deviation. It is generally denoted by the Greek alphabet  or s. Variance: The square of the standard deviation is called variance and is denoted by σ2. Coefficient of Standard Deviation: It is the ratio of the standard deviation to it’s A.M. i.e., Coefficient of standard deviation = . Standard Deviation for Ungrouped Data: Direct method: In case of individual series, the standard deviation can be obtained by the formula. [First Form] where d = x - and x = value of the variable or observation, = arithmetic mean, n = total number of observations. Short-cut Method: This method is applied to calculate Standard deviation, when the mean of the data comes out to be a fraction. In that case it is very difficult and tedious to find the deviations of all observations from the mean by the earlier method. The formula used is [Second Form] where d = x – A, A = assumed mean, n = total number of observations. Standard deviation for grouped data: It is calculated by the following formula. [Third Form] where is A.M., x is the size of the item, and f is the corresponding frequency in the case of discrete series. But when the mean has a fractional value, then the following formula is applied to calculate S.D. [Fourth Form] where d = x – A, A = assumed mean, n = Σf = total frequency. Standard deviation in continuous series: Direct Method. The standard deviation in the case of continuous series is obtained by the following formula. [Fifth Form] where x = mid-value, = A.M., f = frequency, n = total frequency. Combined Standard Deviation: Let σ1 and σ2 be the S.D. of the two groups containing n1 and n2 items respectively. Let be their respective A.M. Let x and σ be the A.M. and S.D. of the combined group respectively. Then . , where . Variance = , or Variance = σ2 ∴ σ = Variance = × i2 (Continuous Series) Standard Deviation of n Natural Number: σ = Imperical Relation: ● Mean deviation = ● Semi-intererquatile range = Example -4: The S.D of 7 scores 1, 2, 3, 4, 5, 6, 7 is (A) 4 (B) 2 (C) (D) none of these Solution: S.D of first n natural number is For n = 7 this value = Example -5: The variance of 2, 4, 6, 8, 10 is (A) 8 (B) (C) 6 (D) none of these Solution: Variance of 1, 2, 3, 4, 5, is ( variance of first n natural number is (n2 –1)/12, when each item is doubled (i.e. 2, 4, 5, 8, 10) variance is multiplied by 22 = 4. Required variance = 4 × 2 = 8) Symmetric and Skew-symmetric In a symmetrical distribution, Mean, Median, Mode coincide. Here, frequencies are symmetrically distributed on both sides of some central value. A distribution which is not symmetrical, is called skew-symmetrical. In a moderately skew-symmetric distribution, Mean - Mode = 3 (Mean - Median) In a positively skew-symmetric distribution, the value of mean is maximum and that of mode is least, and the median lies between the two. In a negatively skew-symmetric distribution, the value of mode is maximum and that of mean is least, and the median lies between the two. In a negatively skew-symmetric distribution, the value of mode is maximum and that of mean is least, and the median lies between the two. Absolute Measures of skewness are Relative Measures of Skewness are (i) Karl Pearson's coefficient of skewness. . It is lies between –1 and 1. (ii) Bowley's coefficient of skewness . It also lies between –1 and 1. (iii) Kelley's cofficient of skewness AREA RESULT For a symmetrical distribution (normal curve), (i) the interval contains 68.27% items. (ii) the interval contains 95.45% items. (iii) the interval contains 99.74% items. OBJECTIVE ASSIGNMENT 1. Co-efficient of variation of two series are 75% are 90% and their standard deviations 15 and 18. Their mean is: (A) 10 (B) 20 (C) 30 (D) 40 Solution: co-efficient of variance = ⇒ for what series 75 = ⇒ = 20 for second series 90 = ⇒ = 20 Thus both series have same mean i.e. 20. 2. If a variable x takes values xi such that , for i = 1, 2,....., n, then (A) (B) (C) (D) Sol. Since S.D. ≤ range = b – a ∴ var(x) ≤ (b – a)2 So, (d) is the answer. 3. A.M. of C0, 2 C1, 3 C2, …, (n + 1) Cn, where (1 + x)n= C0 + C1x+ C2x2 +…+ Cnxn is equal to: (A) ((B) (C) (D) Solution: (1 + x)n = C0 + C1x + C2x2 + ..... + Cnxn Multiplying with x, x (1 + x)n = C0x + C1x2 + C2x3 + ..... + Cnxn+1 Differentiating w. r. t. x, nx (1 + x)n−1 + (1 + x)n = C0 + 2C1x + 3C2x2 + ..... + (n + 1)Cnxn Putting x = 1, n (2n−1) + 2n = C0 + 2C1 +3 C2 + ...... + (n + 1) Cn so, A.M. = = . 4. If a variable takes values 0, 1, 2, …, n with frequencies 1, nC1, nC2, …, nCn then the A.M. is (A) n (B) (2n/n) (C) n+1 (D) (n/2) Solution: = 5. If G is the G.M. of the product of r sets of observations with geometric means G1, G2, …, Gr respectively, then G is equal to (A) log G1 + log G2 + …+ log Gn (B) G1 . G2 … Gn (C) log G1. log G2 … logGn (D) none of these Solution: Taking X as the product of variates X1, X2, …, Xr corresponding to r sets of observations, i.e. , X = X1 . X2 …Xr we have ⇒ ⇒ ⇒ log G = log G1 + log G2 + … + log Gr ⇒ G = G1 G2 … Gr 6. The mean deviation about the median of the series of batsman in ten innings 54, 38, 42, 44, 46, 48, 54, 55, 56, 76 is (A) 8.6 (B) 7.6 (C) 8 (D) none of these Solution: The data is : 34, 38, 42, 44, 46, 48, 54, 55, 56, 76. Median = the mean of 5th and 6th term = Now x1’s ≥ M are 48, 54, 55, 63, 70 x1’s < M are 34, 38, 42, 44, 46 ∴ n1 = 5 and n2 = 5 ∴ s1 = 48 + 54 + 55 + 63 + 78 = 290 s2 = 34 + 38 + 42 + 46 = 204. Mean deviation = 7. For a set of 100 observations, taking assumed mean as 4, the sum of the deviations is –11 cm, and the sum of the squares of these deviations is 275 cm2. The coefficient of variation is (A) 41.13% (B) 40.13% (C) 42.13% (D) None of these. Solution: and  = Coefficient of Variation = = 41.13%. 8. S.D of a data is 6. When each observation is increased by 1, then the S.D of new data is (A) 5 (B) 7 (C) 6 (D) 8 Solution: S.D (and variance) of a data is not changed when each observation is increased (or decreased) by the same constant. 9. The mean of the set of numbers x1, x2, x3, ......xn is , then the mean of the numbers xi + 2i, 1 ≤ i ≤ n, is (A) + 2n (B) + n + 1 (C) + 2 (D) none of these Solution: ⇒ and hence = = + (n + 1) 10. If a variable takes the discrete value  + 4,  –7/2,  –5/2,  –3,  –2,  + 1/2,  –1/2,  + 5, ( > 0), then the median is (A)  –5/4 (B)  –1/2 (C)  –2 (D)  + 5/4 Solution: Arranging the data, we have –7/2, –3, –5/2, –2, –1/2,  + 1/2,  + 4,  +5 Median is 1/2(4th observation + 5th observation) = 1/2( –2 +  –1/2) =  –5/4 11. If in a moderately asymmetrical distribution, mode and mean of the data are 6 and 9 respectively, then median is (A) 8 (B) 7 (C) 6 (D) 5 Solution: 6 = 3median –2  9 ⇒ 3 median = 24 ⇒ median = 8 12. The mean height of 29 male workers is 71 cms and 31 female workers is 48 cms. Find the combined mean height of all 60 workers in the factory. Sol. Combined mean height = = 59.1 cm 13. The geometric mean of 6 observations was calculated as 13. It was later observed that one of the observation was recorded as 28 instead of 36. The correct geometric mean is (A) (B) 3 (C) 13 (D) 13 Sol. Let x1, x2 ………, x6 are the observations and x1 = 28 ⇒ 28. x2 ……… x6 = 136 ⇒ x2 ……… x6 = Now correct observation is 36 instead of 28 ⇒ 36. x2………x6 =  36 So, now geometric mean = 13 14. The algebraic sum of deviations of a set of values from the arithmetic mean is: (A) 0 (B) 1 (C) depends on mean (D) none of these Sol. Let set of values be x1, x2, ……… xn and their mean be ⇒ deviation of values from mean = ⇒ sum of deviations = = = = 0 15. The mean of observations is 4.4 and the variance is 8.24. If three of the five observations are 1, 2 and 6, find the other two observations. (A) 9, 4 (B) 2, 11 (C) 7, 6 (D) 5, 8 Sol. Let other two observation be a and b ⇒ a + b + 1 + 2 + 6 = 5  4.4 ⇒ a + b = 13 … (1) 2 = (a2 + b2 + 1 + 4 + 36) − (4.4)2 = 8.24 ⇒ a2 + b2 = 97 … (2) from (1) and (2) we get, a = 9, b = 4 16. If a and b are two positive numbers, then (A) AM ≤ GM ≥ HM (B) AM ≥ GM ≤ HM (C) AM ≤ GM ≤ HM (D) AM ≥ GM ≥ HM Sol. A.M. of two positive numbers = G.M. of two positive numbers = H.M. of two positive numbers = Case I: AM ≥ GM Let AM < GM ⇒ ⇒ a + b − 2 < 0 ⇒ ( < 0 which is not possible ⇒ AM ≥ GM Case II: GM ≥ HM Let GM < HM ⇒ < ⇒ a + b < 2 ⇒ < ⇒ AM < GM which again not possible so from case I and case II AM ≥ GM ≥ HM 17. The mean annual salaries paid to 1000 employees of a company was Rs. 3400. The mean annual salaries paid to male and female employees were Rs. 200 and Rs. 4200 respectively. The percentage of males and females employed by the company are : (A) 50, 50 (B) 40, 60 (C) 70, 30 (D) 20, 80 Sol. Let number of female workers be nf and number of male workers be nm ⇒ = 3400 ⇒ 200 nm = 4200 (1000 − nm) = 34  105 ⇒ nm = 200 and nf = 800 ⇒ 20% male workers. 80% female workers 18. The number of children in 25 families of a locality are recorded as follows: 3, 1, 4, 0, 2, 1, 1, 2, 3, 3, 2, 2, 2, 5, 0, 1, 4, 1, 2, 1, 2, 3, 0, 1, 4. Then mean number of children per family is (A) 2 (B) 3 (C) 4 (D) 5 Sol. Mean = Ans. (A) 19. The mean of 12 numbers is 24. If 5 is added in every number, the new mean is (A) 25 (B) 29 (C) 84 (D) none of these Sol. = 24 Now, = 24 + 5 = 29 20. The A.M. of n numbers of a series is . If the sum of first (n-1) terms is k, then the nth number is (A) (B) (C) (D) Sol. Ans. (B) 21. If a variate X is expressed as a linear function of two variates U and V in the form X = aU +bV, then mean of X is (A) (B) (C) (D) none of these Sol. xi = aui + bvi ⇒ = a + b 22. The arithmetic means of a set of observations is . If each observation is divided by  and then it is increased by 12, then the mean of the new series is (A) (B) ( + 12)/ (C) (D) Sol. (A) 23. If a variable takes values 0, 1, 2,……,n with frequencies qn, (n)qn–1p, . Where p + q =1, then the mean is (A) nq (B) np (C) n(p+q) (D) none of these Sol. Mean of binomial distributions = np 24. The weighted mean of first n natural numbers whose weights are equal is given by (A) (2n+1)/2 (B) (n+1)/2 (C) (n-1)/2 (D) (2n-1)/n Sol. Ans. (B) 25. If the arithmetic mean of two numbers is 10 and their geometric mean is 8, their Harmonic mean is (A) 6.2 (B) 6.4 (C) 6.3 (D) 5.4 Sol. a + b = 20, ab = 64 clearly a = 4, b = 16 26. If G1,G2 are the geometric means of two series of observations and G is the GM of the ratios of the corresponding observations then G is equal to (A) (G1/G2) (B) log G1–logG2 (C) log G1/log G2 (D) log (G1,G2) Sol. Let x1, x2, ………, xn and y1, y2, ………, yn be two series of observation ⇒ G1n = x1. x2. x3 ……… . xn, G2n = y1. y2. y3 ……… . yn Now, = Gn (given) ⇒ = Gn ⇒ G = 27. Which one of the following is not a measure of location (A) median (B) mean (C) mode (D) Q1 Sol. Ans. (B) (C) 28. The sum of squares of deviations from mean is (A) least (B) zero (C) maximum (D) none of these Sol. Ans. (B) 29. The points scored by basket – ball team in a series of matches are as follows: 15, 3, 8, 10, 22, 5, 27, 11, 12, 19, 18, 21, 13, 14. Its median is (A) 13 (B) 13.4 (C) 13.5 (D) 14.5 Sol. Arranging in ascending order to we get 3, 5, 8, 10, 11, 12, 13, 14, 15, 18, 19, 21, 22, 27 Median = mean of 7th and 8th term. Ans. (C) 30. One of the methods of determining mode is (A) Mode = 2 median – 3 mean (B) Mode = 2 median + 3 mean (C) Mode = 3 median –2 mean (D) Mode = 3 median + 2 mean Sol. Ans. (C) 31. The best average of dealing a qualitative data is (A) Mean (B) Median (C) Mode (D) Harmonic mean Sol. Ans. (B) 32. If a variable takes discrete values x + 4, x – (7/2), x – (5/2), x –3, x - 2, x + (1/2), x +5, x – (1/2), (x  0), then median is (A) x+(5/4) (B) x-(5/4) (C) x+(3/2) (D) x-(1/2) Sol. Arranging in ascending order to we get x – (7/2), x –3, x – (5/2), x - 2, x – (1/2), x + (1/2), x + 4, x +5, (x  0). Hence (D) is correct answer. 33. If mode of a data is 27 and mean is 24, then median is (A) 24 (B) 25 (C) 26 (d) none of these Sol. Mode + 2(Mean) = 3 (Median) ⇒ Median = = 25 34. Median of 16, 10, 14, 11, 9, 8, 12, 6, 5 is (A) 10 (B) 12 (C) 11 (D) 14 Sol. Arranging in ascending order we get 5, 6, 8, 9, 10, 11, 12, 14, 16 Clearly median is 10. Ans. (A) 35. Mode of the data 5, 7, 2, 7, 7, 4, 7, 7, 6, 7 is (A) 5 (B) 4 (C) 6 (D) 7 Sol. ‘7’ occurs maximum number of times, so frequency is maximum for ‘7’. So 7 is Mode. 36. Geometric mean of 1, 2, 22, 23, ... 2n is (A) 2n/2 (B) 2n/2 (C) 2(n–1)/2 (D) 2(n + 1)/2 Sol. G.M. = (1. 2. 22. ……… 2n)1/(n+1) = [2n(n+1)/2]1/(n+1) = 2n/2. 37. If the first quartile is 104 and quartile deviation is 8, then the third quartile is (A) 130 (B) 140 (C) 136 (D) 146 Sol. Q1 = 104 and = 18 ⇒ Q3 = 36 + Q1 = 36 + 104 = 140 38. The variance of firs n natural number is (A) (B) (C) (D) none of these Sol. 2 = = (n + 1) 39. Om Shankar Dwivedi’s monthly earning for a year in Rupees are 139, 150, 151, 151, 157, 158, 160, 161, 162, 162, 173, 175. The coefficient of range for his earnings is (A) 0.115 (B) 0.215 (C) 0.315 (D) 1.15 Sol. Coefficient of Range = = 0.115 40. You are provided with the following raw data of two variable x and y. x = 235, y = 250, x2 = 6750, y2 = 6840. Ten pairs of values are included in the survey. The standard deviation are (A) 11.08, 7.68 (B) 11.02, 7.58 (C) 11.48, 7.48 (D) none of these Sol. ∑ x = 235, n = 10, ∑ x2 = 6750 ⇒ 2 = . 6750 − = 675 − (23.5)2 = 675 − 552.25 = 122.75 ⇒  = 11.08 similarly for ∑y = 250, n = 10, ∑y2 = 6840 ⇒  = 7.68 41. If an observation in a series is 0 , then its GM will be (A) Indeterminate (B) Positive (C) 0 (D) none of these Sol. If an observation is zero, then product of the observations will be zero. Therefore G.M. will be zero 42. The coefficient of variation of a series is 50. Its S.D. is 21.2. Its arithmetic means is (A) 36.6 (B) 22.6 (C) 26.6 (D) 36.1 Sol. Coefficient of variations = ⇒ 50 = Mean = 42.4 43. Variance is independent of (A) origin only (B) scale only (C) origin and change of scale (D) none of these Sol. Variance is always independent of origin only 44. The G.M. of 4, 5, 20, 25 is (A) 10 (B) 20 (C) 100 (D) none of these Sol. G.M. = (4. 5. 20. 25)1/4 = (42. 54)1/4 = 2  5 = 10 45. The A.M. of nCo, nC1, nC2, ......., nCn, is : (A) (B) (C) (D) none of these Sol. Hence (B) is correct answer. 46. The mean wage of 1000 workers in a factory running in two shifts of 700 and 300 workers is Rs. 500. the mean wage of 700 workers working in day shift is Rs. 450. The mean wage of workers working in the night shift is : (A) Rs. 570 (B) Rs. 616.67 (C) Rs. 543.67 (D) none of these Sol. n1 = 700, n2 = 300, , It mean wage of workers in the shift is , then Hence (B) is correct answer. 47. The average weight of 25 boys was calculated to be 78.4 kg. If was later discovered that one weight was misread as 69 kg instead of 96 kg. The correct average is : (A) 79 kg (B) 79.48 kg (C) 81.32 kg (D) none of these Sol. Incorrect sum of all items = 78.4 × 25 = 1960 kg. ∴ Correct sum = 1960 – 69 + 96 = 1987 kg ∴ Correct average Hence (B) is correct answer. 48. A car owner buys petrol at Rs. 7.50, Rs. 8.00 and Rs. 8.50 per litre for the 3 successive years. If he spends Rs. 4000 each year, then the average cost per litre of petrol is : (A) Rs. 8 (B) Rs. 8.25 (C) Rs. 7.98 (D) none of these Sol. Petrol bought in first year litres Petrol bought in second year litres Petrol bought in third year litres Total money = Rs. 12000. ∴ Average cost per litre = Rs. 7.98 Hence (C) is correct answer. 49. The mean of following frequency table is 50. Class Frequency 0-20 20-40 40-60 60-80 80-100 17 f1 32 f2 19 Total 120 The missing frequencies are : (A) 28, 24 (B) 24, 36 (C) 36, 28 (D) none of these Sol. Mid point (x) Frequency (f) fx 10 30 50 70 90 17 f1 32 f2 19 170 30f1 1600 70f2 1710 120 30f1 + 70f2 + 3480 ⇒ 600 = 3f1 + 7f2 + 348 ⇒ 3f1 + 7f2 = 252 .....(1) Also, f1 + f2 + 68 = 120 ⇒ f1 + f2 = 52 .....(2) Solving (1) and (2), we get f2 = 24, f1 = 28. Hence (A) is correct answer. 50. Geometric mean of 1, 2, 22, 23, ......, 2n is : (A) (B) (C) (D) Sol. Hence (B) is correct answer. 51. If is the mean of a distribution, then = (A) 0 (B) M.D. (C) S.D. (D) none of these Sol. Algebraic sum of deviations from A.M. is always zero Hence (A) is correct answer. 52. The mean square deviation of n observations x1, x2, .... xn about –2 and 2 are 18 and 10 respectively. Then S.D. of the given set is : (A) 1 (B) 2 (C) 3 (D) 4 Sol. and ∴ Adding and subtracting, we get and and Hence (C) is correct answer. 53. Mean of n times is . If these x items are successively increased by 2, 22, 23, ....., 2n, then the new mean is : (A) (B) (C) (D) none of these Sol. New mean = . Hence (B) is correct answer. 54. If and are means to two distributions such that and is the mean of the combined distribution, then : (A) (B) (C) (D) Sol. If n1 and n2 are the number of items in the two distributions having means and . Similarly, Hence (D) is correct answer. 55. The quartile deviation of daily wages (in Rs.) of 7 persons is given below : 12, 7, 15, 10, 17, 17, 25 is : (A) 14.5 (B) 5 (C) 9 (D) 4.5 Sol. n = 7 + 10 + 12 + 15 + 17 + 17 + 25 = 103 Q1 = size of item = size of 26th item = 12 Q3 = size of 3 item = size of 78th item = 17 ∴ Q3 – Q1 = 17 – 12 = 5. Hence (B) is correct answer. 56. Mean deviation of numbers 3, 4, 5, 6, 7 is : (A) 0 (B) 1.2 (C) 5 (D) 25 Sol. , n = 5 x 3 4 5 6 7 2 1 0 1 2 6 Hence (B) is correct answer. 57. The mean of a set of observations is . If each observation is divided by ,  ≠ 0 and then is increased by 10, then the mean of the new set is (A) (B) (C) (D) Sol: (C) 58. The median of the data 13, 14, 16, 18, 20, 22 is (A) 17 (B) 16 (C) 18 (D) none of these Sol: (A) 59. Quartile deviation is nearly equal to (A)  (B) 3/2 (C) 2/3 (D) none of these Sol: (C) 60. Which of the following is not a measure of dispersion? (A) mean (B) variance (C) mean deviation (D) range Sol: (A) 61. A boy goes to school from his home at a speed of x km/ hr and comes back at a speed of y km/ h then the average speed of the boy is given by (A) km/hr (B) km/hr (C) km/hr (D) none of these Sol: (C)