Chapter 15 Statistics Ex 15.2 Solutions
Question - 1 : - Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Answer - 1 : -
So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
Let us make the table of the given data and append other columns after calculations.
| Xi | Deviations from mean (xi – x̅) | (xi – x̅)2 |
| 6 | 6 – 9 = -3 | 9 |
| 7 | 7 – 9 = -2 | 4 |
| 10 | 10 – 9 = 1 | 1 |
| 12 | 12 – 9 = 3 | 9 |
| 13 | 13 – 9 = 4 | 16 |
| 4 | 4 – 9 = – 5 | 25 |
| 8 | 8 – 9 = – 1 | 1 |
| 12 | 12 – 9 = 3 | 9 |
| | 74 |
We know that Variance,
σ2 =(1/8) × 74
= 9.2
∴Mean = 9 and Variance= 9.25
Answer - 2 : -
We know that Mean =Sum of all observations/Number of observations
∴Mean, x̅ = ((n(n +1))2)/n
= (n + 1)/2
and also WKT Variance,
By substitute thatvalue of x̅ we get,
WKT, (a + b)(a – b) =a2 – b2
σ2 =(n2 – 1)/12
∴Mean = (n + 1)/2 andVariance = (n2 – 1)/12
Question - 3 : - Find the mean and variance for the data First 10 multiples of 3
Answer - 3 : -
First we have to writethe first 10 multiples of 3,
3, 6, 9, 12, 15, 18,21, 24, 27, 30
So, x̅ = (3 + 6 + 9 +12 + 15 + 18 + 21 + 24 + 27 + 30)/10
= 165/10
= 16.5
Let us make the tableof the data and append other columns after calculations.
| Xi | Deviations from mean (xi – x̅) | (xi – x̅)2 |
| 3 | 3 – 16.5 = -13.5 | 182.25 |
| 6 | 6 – 16.5 = -10.5 | 110.25 |
| 9 | 9 – 16.5 = -7.5 | 56.25 |
| 12 | 12 – 16.5 = -4.5 | 20.25 |
| 15 | 15 – 16.5 = -1.5 | 2.25 |
| 18 | 18 – 16.5 = 1.5 | 2.25 |
| 21 | 21 – 16.5 = – 4.5 | 20.25 |
| 24 | 24 – 16.5 = 7.5 | 56.25 |
| 27 | 27 – 16.5 = 10.5 | 110.25 |
| 30 | 30 – 16.5 = 13.5 | 182.25 |
| | 742.5 |
Then, Variance
= (1/10) × 742.5
= 74.25
∴Mean = 16.5 andVariance = 74.25
Question - 4 : - Find the mean and variance for the data | xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
| fi | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Answer - 4 : -
Let us make the tableof the given data and append other columns after calculations.
| Xi | fi | fixi | Deviations from mean (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
| 6 | 2 | 12 | 6 – 19 = 13 | 169 | 338 |
| 10 | 4 | 40 | 10 – 19 = -9 | 81 | 324 |
| 14 | 7 | 98 | 14 – 19 = -5 | 25 | 175 |
| 18 | 12 | 216 | 18 – 19 = -1 | 1 | 12 |
| 24 | 8 | 192 | 24 – 19 = 5 | 25 | 200 |
| 28 | 4 | 112 | 28 – 19 = 9 | 81 | 324 |
| 30 | 3 | 90 | 30 – 19 = 11 | 121 | 363 |
| N = 40 | 760 | | | 1736 |
Question - 5 : - Find the mean and variance for the data | xi | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
| fi | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Answer - 5 : -
Let us make the tableof the given data and append other columns after calculations.
| Xi | fi | fixi | Deviations from mean (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
| 92 | 3 | 276 | 92 – 100 = -8 | 64 | 192 |
| 93 | 2 | 186 | 93 – 100 = -7 | 49 | 98 |
| 97 | 3 | 291 | 97 – 100 = -3 | 9 | 27 |
| 98 | 2 | 196 | 98 – 100 = -2 | 4 | 8 |
| 102 | 6 | 612 | 102 – 100 = 2 | 4 | 24 |
| 104 | 3 | 312 | 104 – 100 = 4 | 16 | 48 |
| 109 | 3 | 327 | 109 – 100 = 9 | 81 | 243 |
| N = 22 | 2200 | | | 640 |
Question - 6 : - Find the mean and standard deviation using short-cut method. | xi | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
| fi | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |
Answer - 6 : -
Let the assumed mean A= 64. Here h = 1
We obtain thefollowing table from the given data.
| Xi | Frequency fi | Yi = (xi – A)/h | Yi2 | fiyi | fiyi2 |
| 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 |
| | | | 0 | 286 |
Mean,
Where A = 64, h = 1
So, x̅ = 64 + ((0/100)× 1)
= 64 + 0
= 64
Then, variance,
σ2 =(12/1002) [100(286) – 02]
= (1/10000) [28600 –0]
= 28600/10000
= 2.86
Hence, standarddeviation = σ = √2.886
= 1.691
∴ Mean = 64 andStandard Deviation = 1.691
Question - 7 : - Find the mean and variance for the following frequency distributions in Exercises 7 and 8. | Classes | 0 – 30 | 30 – 60 | 60 – 90 | 90 – 120 | 120 – 150 | 150 – 180 | 180 – 210 |
| Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Answer - 7 : -
Let us make the tableof the given data and append other columns after calculations.
| Classes | Frequency fi | Mid – points xi | fixi | (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
| 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 |
Question - 8 : - Find the mean and variance for the following frequency distributions in Exercises 7 and 8. | Classes | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 –50 |
| Frequencies | 5 | 8 | 15 | 16 | 6 |
Answer - 8 : -
Let us make the tableof the given data and append other columns after calculations.
| Classes | Frequency fi | Mid – points xi | fixi | (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
| 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 |
Question - 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 |
| Frequencies | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
Answer - 9 : -
Let the assumed mean,A = 92.5 and h = 5
Let us make the tableof the given data and append other columns after calculations.
| Height (class) | Number of children Frequency fi | Midpoint Xi | Yi = (xi – A)/h | Yi2 | fiyi | fiyi2 |
| 70 – 75 | 2 | 72.5 | -4 | 16 | -12 | 48 |
| 75 – 80 | 1 | 77.5 | -3 | 9 | -12 | 36 |
| 80 – 85 | 12 | 82.5 | -2 | 4 | -14 | 28 |
| 85 – 90 | 29 | 87.5 | -1 | 1 | -7 | 7 |
| 90 – 95 | 25 | 92.5 | 0 | 0 | 0 | 0 |
| 95 – 100 | 12 | 97.5 | 1 | 1 | 9 | 9 |
| 100 – 105 | 10 | 102.5 | 2 | 4 | 12 | 24 |
| 105 – 110 | 4 | 107.5 | 3 | 9 | 18 | 54 |
| 110 – 115 | 5 | 112.5 | 4 | 16 | 12 | 48 |
| N = 60 | | | | 6 | 254 |
Mean,
Where, A = 92.5, h = 5
So, x̅ = 92.5 +((6/60) × 5)
= 92.5 + ½
= 92.5 + 0.5
= 93
Then, Variance,
σ2 =(52/602) [60(254) – 62]
= (1/144) [15240 – 36]
= 15204/144
= 1267/12
= 105.583
Hence, standarddeviation = σ = √105.583
= 10.275
∴ Mean = 93,variance = 105.583 and Standard Deviation = 10.275
Question - 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 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]
Answer - 10 : -
Let the assumed mean,A = 42.5 and h = 4
Let us make the tableof the given data and append other columns after calculations.
| Height (class) | Number of children (Frequency fi) | Midpoint Xi | Yi = (xi – A)/h | Yi2 | fiyi | fiyi2 |
| 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,
Where, A = 42.5, h = 4
So, x̅ = 42.5 +(25/100) × 4
= 42.5 + 1
= 43.5
Then, Variance,
σ2 =(42/1002)[100(199) – 252]
= (1/625) [19900 –625]
= 19275/625
= 771/25
= 30.84
Hence, standarddeviation = σ = √30.84
= 5.553
∴ Mean = 43.5,variance = 30.84 and Standard Deviation = 5.553.