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7       The mean and standard deviation of a group of   observations were found to be and , respectively. Later on it was found that three observations were incorrect, which were recorded as   and  . Find the mean and standard deviation if the incorrect observations are omitted.

Given,  Initial Number of observations, n = 100    Thus, incorrect sum = 2000 Hence, New sum of observations = 2000 - 21-21-18 = 1940 New number of observation, n' = 100-3 =97 Therefore, New Mean = New Sum)/100   = 20 Now, Standard Deviation,  ,which is the incorrect sum. Thus,  New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18) = 40900 - 441 - 441 - 324 = 39694 Hence, Correct...

6.   The mean and standard deviation of marks obtained by   students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

 Subject Mathematics Physics Chemistry Mean Standard deviation

Which of the three subjects shows highest variability in marks and which shows the lowest?

Given, Standard deviation of physics = 15 Standard deviation of chemistry = 20 Standard deviation of mathematics = 12 We know , C.V. The subject with greater C.V will be more variable than others. Hence, Mathematics has lowest variability and Chemistry has highest variability.

5.   The mean and standard deviation of observations are found to be   and , respectively. On rechecking, it was found that an observation was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(ii) If it is replaced by

Given,  Number of observations, n = 20 Also, Incorrect mean = 10 And, Incorrect standard deviation = 2    Thus, incorrect sum = 200 Hence, correct sum of observations = 200 – 8 + 12 = 204 Therefore, Correct Mean = (Correct Sum)/20   = 10.2 Now, Standard Deviation,  ,which is the incorrect sum. Thus,  New sum = Old sum - (8x8) + (12x12) = 2080 – 64 + 144 = 2160 Hence, Correct Standard...

5. The mean and standard deviation of    observations are found to be   and , respectively. On rechecking, it was found that an observation was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.

Given,  Number of observations, n = 20 Also, Incorrect mean = 10 And, Incorrect standard deviation = 2    Thus, incorrect sum = 200 Hence, correct sum of observations = 200 – 8 = 192 Therefore, Correct Mean = (Correct Sum)/19   = 10.1 Now, Standard Deviation,  ,which is the incorrect sum. Thus,  New sum = Old sum - (8x8) = 2080 – 64 = 2016 Hence, Correct Standard Deviation =  =2.02

4.  Given that is the mean and    is the variance of observations  .Prove that the mean and variance of the observations  , are   and   respectively, .

Given, Mean =  and variance =  Now, Let   be the the resulting observations if each observation is multiplied by a:   Hence the mean of the new observations   is  We know,     Now, Substituting the values of  : Hence the variance of the new observations   is  Hence proved.

3.   The mean and standard deviation of six observations are   and  , respectively. If each observation is multiplied by , find the new mean and new standard deviation of the resulting observations.

Given, Mean = 8 and Standard deviation = 4 Let the observations be Mean,  Now, Let   be the the resulting observations if each observation is multiplied by 3: New mean,  = 24 We know that, Standard Deviation =             -(i) Now, Substituting the values of  in (i): Hence, the variance of the new observations =  Therefore, Standard Deviation =  =  = 12

2.  The mean and variance of observations are and , respectively. If five of the observations are   . Find the remaining two observations.

Given, The mean and variance of 7 observations are 8 and 16, respectively Let the remaining two observations be x and y, Observations: 2, 4, 10, 12, 14, x, y ∴ Mean,  42 + x + y = 56 x + y = 14          -(i) Now, Variance               (Using (i))      -(ii) Squaring (i), we get       (iii) (iii) - (ii) : 2xy = 96  (iv) Now, (ii) - (iv):          -(v) Hence, From (i) and (v): x – y = 2  x =...

1.    The mean and variance of eight observations are    and  , respectively. If six of the observations are    and   , find the remaining two observations.

Given, The mean and variance of 8 observations are  9  and  9.25, respectively Let the remaining two observations be x and y, Observations: 6, 7, 10, 12, 12, 13, x, y. ∴ Mean,  60 + x + y = 72 x + y = 12          -(i) Now, Variance               (Using (i))      -(ii) Squaring (i), we get       (iii) (iii) - (ii) : 2xy = 64  (iv) Now, (ii) - (iv):          -(v) Hence, From (i) and (v): x – y...

5. The sum and sum of squares corresponding to length (in cm) and weight (in gm) of 50 plant products are given below:

Which is more varying, the length or weight?

For lenght x, Mean,  We know, Variance,  We know,  Standard Deviation =  C.V.(x) =  For weight y, Mean, Mean,  We know, Variance,  We know,  Standard Deviation =  C.V.(y) =  Since C.V.(y) > C.V.(x) Therefore, weight is more varying.

4.  The following is the record of goals scored by team A in a football session:

 No. of goals scored 0 1 2 3 4 No. of matches 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard deviation    goals. Find which team may be considered more consistent?

No. of goals scored  Frequency 0 1 0 0 0 1 9 1 9 9 2 7 4 14 28 3 5 9 15 45 4 3 16 12 48    =N = 25   = 50 =130 For Team A, Mean, We know, Variance,  We know,  Standard Deviation =  C.V.(A) =  For Team B, Mean = 2 Standard deviation,  = 1.25 C.V.(B) =  Since C.V. of firm B is more than C.V. of A. Therefore, Team A is more consistent.

3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

 Firm A Firm B No. of wage earners Mean of monthly wages Variance of the distribution of wages

(ii) Which firm, A or B, shows greater variability in individual wages?

Given, Variance of firm A = 100 Standard Deviation =  Again, Variance of firm B = 121 Standard Deviation =  Since , firm B has greater variability in individual wages.

3.      An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

 Firm A Firm B No. of wage earners Mean of monthly wages Variance of the distribution of wages

(i) Which firm A or B pays larger amount as monthly wages?

Given, Mean of monthly wages of firm A = 5253 Number of wage earners = 586 Total amount paid = 586 x 5253 = 30,78,258 Again, Mean of monthly wages of firm B = 5253 Number of wage earners = 648 Total amount paid = 648 x 5253 = 34,03,944 Hence firm B pays larger amount as monthly wages.

2   From the prices of shares X and Y below, find out which is more stable in value:

 X 35 54 52 53 56 58 52 50 51 49 Y 108 107 105 105 106 107 104 103 104 101

X() Y() 35 108 1225 11664 54 107 2916 11449 52 105 2704 11025 53 105 2809 11025 56 106 8136 11236 58 107 3364 11449 52 104 2704 10816 50 103 2500 10609 51 104 2601 10816 49 101 2401 10201 =510 = 1050 =26360 =110290 For X, Mean ,  Variance,  We know,  Standard Deviation =  C.V.(X) =   Similarly, For Y, Mean ,  Variance,  We know,  Standard Deviation...

1.From the data given below state which group is more variable, A or B?

 Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7

The group having a higher coefficient of variation will be more variable. Let the assumed mean, A = 45 and h = 10 For Group A Marks Group A Midpoint 10-20 9 15 -3 9 -27 81 20-30 17 25 -2 4 -34 68 30-40 32 35 -1 1 -32 32 40-50 33 45 0 0 0 0 50-60 40 55 1 1 40 40 60-70 10 65 2 4 20 40 70-80 9 75 3 9 27 81    =N = 150       = -6 =342 Mean, We know,...

10.  The diameters of circles (in mm) drawn in a design are given below:

 Diameters 33-36 37-40 41-44 45-48 49-52 No. of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.
[ Hint First make the data continuous by making the classes as   and then proceed.]

Let the assumed mean, A = 92.5 and h = 5 Diameters No. of circles  Midpoint 32.5-36.5 15 34.5 -2 4 -30 60 36.5-40.5 17 38.5 -1 1 -17 17 40.5-44.5 21 42.5 0 0 0 0 44.5-48.5 22 46.5 1 1 22 22 48.5-52.5 25 50.5 2 4 50 100    =N = 100       = 25 =199 Mean, We know, Variance,  We know,  Standard Deviation =  Hence, Mean = 43.5, Variance = 30.84 and Standard...

9. Find the mean, variance and standard deviation using short-cut method.

 Height in cms 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115 No. of students 3 4 7 7 15 9 6 6 3

Let the assumed mean, A = 92.5 and h = 5 Height in cms Frequency Midpoint 70-75 3 72.5 -4 16 -12 48 75-80 4 77.5 -3 9 -12 36 80-85 7 82.5 -2 4 -14 28 85-90 7 87.5 -1 1 -7 7 90-95 15 92.5 0 0 0 0 95-100 9 97.5 1 1 9 9 100-105 6 102.5 2 4 12 24 105-110 6 107.5 3 9 18 54 110-115 3 112.5 4 16 12 48    =N = 60       = 6 =254 Mean, We know,...

8.  Find the mean and variance for the following frequency distributions.

 Classes 0-10 10-20 20-30 30-40 40-50 Frequencies 5 8 15 16 6

Classes Frequency Mid-point  0-10 5 5 25 -22 484 2420 10-20 8 15 120 -12 144 1152 20-30 15 25 375 -2 4 60 30-40 16 35 560 8 64 1024 40-50 6 45 270 18 324 1944    = N = 50   = 1350     =6600   We know, Variance,  Hence, Mean = 27 and Variance = 132

7.Find the mean and variance for the following frequency distributions.

 Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2

Classes Frequency Mid point  0-30 2 15 30 -92 8464 16928 30-60 3 45 135 -62 3844 11532 60-90 5 75 375 -32 1024 5120 90-120 10 105 1050 2 4 40 120-150 3 135 405 28 784 2352 150-180 5 165 825 58 3364 16820 180-210 2 195 390 88 7744 15488    = N = 30   = 3210     =68280   We know, Variance,  Hence, Mean = 107 and Variance = 2276

6     Find the mean and standard deviation using short-cut method.

 60 61 62 63 64 65 66 67 68 2 1 12 29 25 12 10 4 5

Let the assumed mean, A = 64 and h = 1 60    2 -4 16 -8 32 61 1 -3 9 -3 9 62 12 -2 4 -24 48 63 29 -1 1 -29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80   =100     = 0 =286 Mean, We know, Variance,  We know,  Standard Deviation =  Hence, Mean = 64 and Standard Deviation = 1.691

5. Find the mean and variance for each of the data.

 92 93 97 98 102 104 109 3 2 3 2 6 3 3

92  3 276 -8 64 192 93 2 186 -7 49 98 97 3 291 -3 9 27 98 2 196 -2 4 8 102 6 612 2 4 24 104 3 312 4 16 48 109 3 327 9 81 243   = 22 = 2200     =640 We know, Variance,  Hence, Mean = 100 and Variance = 29.09
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