Chapter 15 Statistics Ex 15.3 Solutions
Question - 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 |
Answer - 1 : -
each series. Theseries having greater C.V. is said to be more variable than the other. Theseries having lesser C.V. is said to be more consistent than the other.
Co-efficient ofvariation (C.V.) = (σ/ x̅) × 100
Where, σ = standarddeviation, x̅ = mean
For Group A
| Marks | Group A fi | Mid-point Xi | Yi = (xi – A)/h | (Yi)2 | fiyi | fi(yi)2 |
| 10 – 20 | 9 | 15 | ((15 – 45)/10) = -3 | (-3)2 = 9 | – 27 | 81 |
| 20 – 30 | 17 | 25 | ((25 – 45)/10) = -2 | (-2)2 = 4 | – 34 | 68 |
| 30 – 40 | 32 | 35 | ((35 – 45)/10) = – 1 | (-1)2 = 1 | – 32 | 32 |
| 40 – 50 | 33 | 45 | ((45 – 45)/10) = 0 | 02 | 0 | 0 |
| 50 – 60 | 40 | 55 | ((55 – 45)/10) = 1 | 12 = 1 | 40 | 40 |
| 60 – 70 | 10 | 65 | ((65 – 45)/10) = 2 | 22 = 4 | 20 | 40 |
| 70 – 80 | 9 | 75 | ((75 – 45)/10) = 3 | 32 = 9 | 27 | 81 |
| Total | 150 | | | | -6 | 342 |
Where A = 45,
and yi =(xi – A)/h
Here h = class size =20 – 10
h = 10
So, x̅ = 45 +((-6/150) × 10)
= 45 – 0.4
= 44.6
σ2 =(102/1502) [150(342) – (-6)2]
= (100/22500) [51,300– 36]
= (100/22500) × 51264
= 227.84
Hence, standarddeviation = σ = √227.84
= 15.09
∴ C.V for group A= (σ/ x̅) × 100
= (15.09/44.6) × 100
= 33.83
Now, for group B.
| Marks | Group B fi | Mid-point Xi | Yi = (xi – A)/h | (Yi)2 | fiyi | fi(yi)2 |
| 10 – 20 | 10 | 15 | ((15 – 45)/10) = -3 | (-3)2 = 9 | – 30 | 90 |
| 20 – 30 | 20 | 25 | ((25 – 45)/10) = -2 | (-2)2 = 4 | – 40 | 80 |
| 30 – 40 | 30 | 35 | ((35 – 45)/10) = – 1 | (-1)2 = 1 | – 30 | 30 |
| 40 – 50 | 25 | 45 | ((45 – 45)/10) = 0 | 02 | 0 | 0 |
| 50 – 60 | 43 | 55 | ((55 – 45)/10) = 1 | 12 = 1 | 43 | 43 |
| 60 – 70 | 15 | 65 | ((65 – 45)/10) = 2 | 22 = 4 | 30 | 60 |
| 70 – 80 | 7 | 75 | ((75 – 45)/10) = 3 | 32 = 9 | 21 | 63 |
| Total | 150 | | | | -6 | 366 |
Where A = 45,
h = 10
So, x̅ = 45 +((-6/150) × 10)
= 45 – 0.4
= 44.6
σ2 =(102/1502) [150(366) – (-6)2]
= (100/22500) [54,900– 36]
= (100/22500) × 54,864
= 243.84
Hence, standarddeviation = σ = √243.84
= 15.61
∴ C.V for group B= (σ/ x̅) × 100
= (15.61/44.6) × 100
= 35
By comparing C.V. ofgroup A and group B.
C.V of Group B >C.V. of Group A
So, Group B is morevariable.
| X | 35 | 54 | 52 | 53 | 56 | 58 | 52 | 50 | 51 | 49 |
| Y | 108 | 107 | 105 | 105 | 106 | 107 | 104 | 103 | 104 | 101 |
Answer - 2 : -
From the given data,
Let us make the tableof the given data and append other columns after calculations.
| X (xi) | Y (yi) | Xi2 | Yi2 |
| 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 |
| Total = 510 | 1050 | 26360 | 110290 |
We have to calculateMean for x,
Mean x̅ = ∑xi/n
Where, n = number ofterms
= 510/10
= 51
Then, Variance for x =
= (1/102)[(10× 26360) – 5102]
= (1/100) (263600 –260100)
= 3500/100
= 35
WKT Standard deviation= √variance
= √35
= 5.91
So, co-efficient ofvariation = (σ/ x̅) × 100
= (5.91/51) × 100
= 11.58
Now, we have tocalculate Mean for y,
Mean ȳ = ∑yi/n
Where, n = number ofterms
= 1050/10
= 105
Then, Variance for y =
= (1/102)[(10× 110290) – 10502]
= (1/100) (1102900 –1102500)
= 400/100
= 4
WKT Standard deviation= √variance
= √4
= 2
So, co-efficient ofvariation = (σ/ x̅) × 100
= (2/105) × 100
= 1.904
By comparing C.V. of Xand Y.
C.V of X > C.V. ofY
So, Y is more stablethan X.
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 wages earners | 586 | 648 |
| Mean of monthly wages | Rs 5253 | Rs 5253 |
| Variance of the distribution of wages | 100 | 121 |
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
Answer - 3 : -
(i) From the giventable,
Mean monthly wages offirm A = Rs 5253
and Number of wageearners = 586
Then,
Total amount paid =586 × 5253
= Rs 3078258
Mean monthly wages offirm B = Rs 5253
Number of wage earners= 648
Then,
Total amount paid =648 × 5253
= Rs 34,03,944
So, firm B pays largeramount as monthly wages.
(ii) Variance of firmA = 100
We know that, standarddeviation (σ)= √100
=10
Variance of firm B =121
Then,
Standard deviation(σ)=√(121 )
=11
Hence the standarddeviation is more in case of Firm B that means in firm B there is greatervariability in individual wages.
Question - 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 1.25 goals. Find which team may be considered more consistent?
Answer - 4 : -
From the given data,
Let us make the tableof the given data and append other columns after calculations.
| Number of goals scored xi | Number of matches fi | fixi | Xi2 | fixi2 |
| 0 | 1 | 0 | 0 | 0 |
| 1 | 9 | 9 | 1 | 9 |
| 2 | 7 | 14 | 4 | 28 |
| 3 | 5 | 15 | 9 | 45 |
| 4 | 3 | 12 | 16 | 48 |
| Total | 25 | 50 | | 130 |
Since C.V. of firm Bis greater
∴ Team A is moreconsistent.
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Answer - 5 : -
First we have tocalculate Mean for Length x,
