Board | CBSE |
Class | Class 11 |
Subject | Statistics for Economics |
Chapter | Chapter 11 |
Chapter Name | Measures of Dispersion |
Number of questions solved | 22 |
Category | T.R. Jain and V.K. Ohri |
T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 11 – Measures of Dispersion
Question 1:
5 students obtained following marks in statistics:
20, 35, 25, 30, 15
Find out range and coefficient of range.
Answer:
Here,
Highest value (H)= 35
Lowest value (L) = 15
Range= Highest value − Lowest value
i.e. R= H− L
Substituting the given values in the formula
R= 35 − 15= 20
Coefficient of Range is as follows:
CR =H-LH+Lor, CR =35-1535+15=2050⇒ CR =0.4Hence, the range (R) of the above data is 20 and coefficient of Range (CR) is 0.4
Question 2:
Prices of shares of a company were note as under from Monday through Saturday. Find out range and the coefficient of range.
Day | Mon. | Tues. | Wed. | Thu. | Fri. | Sat. |
Price (₹) | 200 | 210 | 208 | 160 | 220 | 250 |
Answer:
Here,
Highest value among the prices of shares= 250
Lowest Value among the prices of shares= 160
Range (R) = Highest value (H)− Lowest Value (L)
or, R = 250 − 160
⇒ R = 90
Coefficient of Range (CR)=H-LH+L or, CR=250-160250+160=90410⇒ CR =0.219 or 0.22 ( approx.)Hence, the Range (R) of the above data is 90 and Coefficient of Range (CR) is 0.22
Question 3:
You know share market is going bullish during the last several months. Collect weekly data on the share price of any two important industries during the past six months. Calculate the range of share prices. Comment on how volatile are the share prices.
Answer:
Month | Price of shares Tata Motors |
Price of shares Reliance |
Oct. Nov. Dec. Jan. Feb. Mar. |
325 397 405 415 420 388 |
913.35 900.25 750.90 780.70 799.25 850.35 |
For Tata Motors
Highest Value=420
Lowest Value=325
Range (R) = Highest Value (H)− Lowest Value (L)
or, R₁ = 420 − 325
⇒ R₁ = 95
Coefficient of Range (CR1)=H-LH+L=420-325420+325=95745=0.127For Reliance
Highest Value= 913.35
Lowest Value= 750.90
Range (R)= Highest value (H)− Lowest value (L)
or, R₂ = 913.35 − 750.90
⇒ R₂ = 162.45
Coefficient of Range (CR2)=H-LH+L=913.35-750.90913.35+750.90=162.451664.25=0.097From the above results we can observe that the prices of the Tata Motors shares are less volatile as compared to the prices of Reliance shares.
Question 4:
Calculate range and the coefficient of range of the following series:
Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
Number of Students | 15 | 18 | 25 | 30 | 16 | 10 | 9 |
Answer:
Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
No. of Students | 15 | 18 | 25 | 30 | 16 | 10 | 9 |
Here,
Highest value=70
Lowest value=10
Range (R) = Highest value (H) − Lowest Value (L)
= 70 − 10
= 60
Coefficient of Range (CR)=70-1070+10=6080=0.75Hence, the Range (R) of the above series is 60 and Coefficient of Range (CR) is 0.75
Question 5:
Find out the range and the coefficient of range from the following data:
Daily Wage (₹) | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 15 |
Number of Workers | 10 | 15 | 12 | 18 | 25 | 20 | 10 | 4 |
Answer:
Daily Wage | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 15 |
No. of workers | 10 | 15 | 12 | 18 | 25 | 20 | 10 | 4 |
Here,
Highest value =15
Lowest value = 6
Range (R) = Highest value (H) − Lowest value (L)
= 15 − 6
= 9
Coefficient of Range (CR)=H-LH+L=15-615+6=921=0.429Hence, the Range (R) of the above series is 9 and Coefficient of Range (CR) is 0.429
Question 1:
Marks obtained by 100 students of a class are given below. Find out range and coefficient of range of the marks.
Marks | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 | 90−100 |
Number of Students | 4 | 10 | 16 | 22 | 20 | 18 | 8 | 2 | 5 |
Answer:
Marks | No. of Student |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100 |
4 10 16 22 20 18 8 2 5 |
Range (R)= Upper limit of last class interval − Lower limit of the first class interval
(R)= 100 − 10
(R) = 90
Coefficient of Range (CR)=H-LH+Lor, CR=100-10100+10or, CR=90110⇒ CR=0.81Hence, the Range (R) of the marks is 90 and Coefficient of Range (CR) is 0.8
Question 2:
In an examination, 25 students obtained the following marks. Find out coefficient of range of the marks.
Marks | 5−9 | 10−14 | 15−19 | 20−24 | 25−29 | 30−34 | 35−39 |
Number of Students | 1 | 3 | 8 | 5 | 4 | 2 | 2 |
Answer:
Converting the inclusive series into exclusive series,
Marks | No. of Students |
4.5 − 9.5 9.5 − 14.5 14.5 −19.5 19.5 −24.5 24.5 −29.5 29.5 −34.5 34.5 −39.5 |
1 3 8 5 4 2 2 |
Here,
Highest Marks value (H)=39.5
Lowest Marks value (L)=4.5
Coefficient of Range (CR)=H-LH+Lor, CR=39.5-4.539.5+4.5or, CR=3544⇒ CR=0.79Hence, the Coefficient of range of marks in the above series is 0.79
Question 1:
Estimate quartile deviation and the coefficient of quartile deviation of the following data:
8, 9, 11, 12, 13, 17, 20, 21, 23, 25, 27
Show that QD is the average of the difference between two quartiles.
Answer:
Sr. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
Data | 8 | 9 | 11 | 12 | 13 | 17 | 20 | 21 | 23 | 25 | 27 |
In order to find the quartile deviation in case of individual series, find out the values of first quartile and third quartile using the following equations:
Q1 = Size of
N+14thitem
or, Q1 = Size of
11+14thitem
or, Q1 = Size of 3rd item
⇒ Q1 = 11
Q3 = Size of
3N+14thitem
or, Q3 = Size of
311+14thitem
or, Q3 = Size of 9th item
⇒ Q3 = 23
Calculating Quartile Deviation and Coefficient of Quartile Deviation
Quartile deviation (Q.D.)=Q3-Q12or, Q.D.=23-112or, Q.D.=122⇒ Q.D.=6Coefficient of quartile deviation=Q3-Q1Q3+Q1or, Coefficient of Q.D.=23-1123+11=1234=0.353We know that the difference between the third quartile (Q3) and first Quartile (Q1) of a series is called the inter Quartile range i.e.
Inter Quartile Range= Q3
-Q1
Where as Quartile deviation is half of the Inter Quartile Range. i.e.
Quartile Deviation (Q.D.)=
Q3-Q12
Hence, from the above formula it is clear that quartile deviation is the average of the difference between the two quartiles.
Question 2:
Find out quartile deviation and coefficient of quartile deviation of the following series:
28, 18, 20, 24, 30, 15, 47, 27
Answer:
Arranging the data in ascending order.
Sr. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Observation | 15 | 18 | 20 | 24 | 27 | 28 | 30 | 47 |
In order to find the quartile deviation in case of individual series, we need to find out the values of third quartile and first quartile using the following equations:
Q1 = Size of
N+14thitem
or, Q1= Size of
8+14thitem
or, Q1= Size of 2.25th item
or, Q1= Size of 2nd item +
14Size of 3rd item-Size of 2nd item
or, Q1=18+1420-18
⇒Q1 = 18.5
Q3 = Size of
3N+14thor, Q3 = Size of 6.75th item
or, Q3 = Size of 6th item +
34Size of 7th item-Size of 6th item
or, Q3=28+3430-28or, Q3=27+342or, Q3=28+1.5⇒ Q3=29.5Quartile deviation (Q.D.)=Q3-Q12or, Q.D.=29.5-18.52or, Q.D.=112⇒ Q.D.=5.5Coefficient of Quartile deviation=Q3-Q1Q3+Q1 or, Coefficient of Q.D. =29.5-18.529.5+18.5 or, Coefficient of Q.D. =1148⇒Coefficient of Q.D. =0.229=0.23Hence the Q.D. of the series is 5.5 and Coefficient of the Q.D. is 0.23
Question 1:
Find out quartile deviation and the coefficient of quartile deviation of the following data:
Age | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
Members | 3 | 61 | 132 | 153 | 140 | 51 | 3 |
Answer:
Age | Numbers (f) |
Cumulative Frequency (c.f.) |
20 30 (Q₁)40 50 (Q₃)60 70 80 |
3 61 132 153 140 51 3 |
3 3 + 61 = 64 64 + 132 = 196 196 + 153 = 349 349 + 140 = 489 489 + 51 = 540 540 + 3 = 543 |
Σf=N=543 |
Q1 = Size of
N + 14thitem
= Size of
543+14thitem
= Size of 136th item
136th item lies in 196th cumulative frequency of the series. Age corresponding to 196th (c.f.) is 40.
Hence, Q1 = 40
Q3 = Size of
3N+14thitem
= Size of
3543+14thitem
= Size of 408th item
408th item lies in the 489th c.f. of the series. Age corresponding to 489th c.f. is 60.
Hence, Q3 = 60.
Calculating Quartile Deviation and Coefficient of Quartile Deviation:
Quartile deviation (Q.D.)=Q3-Q12 or, Q.D. =60-402 or, Q.D. =202 ⇒ Q.D. =10Coefficient of quartile deviation=Q3-Q1Q3+Q1or, Coefficient of Q.D.=60-4060+40or, Coefficient of Q.D.=20100⇒Coefficient of Q.D. =0.2Hence, Q.D. of the series is 10 and Coefficient of Q.D. is 0.2
Question 2:
Estimate quartile deviation and the coefficient of quartile deviation of the following series:
Height (inches) | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 |
Number of Students | 15 | 20 | 32 | 35 | 33 | 22 | 20 | 10 | 8 |
Answer:
Height (inches) |
No. of student (f) | Cumulative frequency (c.f.) |
58 59 60 61 62 63 64 65 66 |
15 20 32 35 33 22 20 10 8 |
15 15 + 20 = 35 35 + 32 = 67 67 + 35 = 102 102 + 33 = 135 135 + 22 = 157 157 + 20 = 177 177 + 10 = 187 187 + 8 = 195 |
Σf=N=195 |
Q1 = Size of
N+14thitem
= Size of
195+14thitem
= Size of 49th item
49th item lies in 67th c.f. of the series. Height corresponding to 67th (c.f.) is 60.
Hence, Q1 = 60 inches
Q3 = Size of
3N+14thitem
= Size of
3195+14thitem
= Size of 147th item
147th item lies in 157th c.f. of the series. Height corresponding to 157th c.f. is 63.
Hence, Q3 = 63 inches
Calculating Quartile Deviation and Coefficient of Quartile Deviation
Quartile deviation (Q.D.)=Q3-Q12or, Q.D.=63-602or, Q.D.=32⇒ Q.D.=1.5Coefficient of quartile deviation=Q3-Q1Q3+Q1or, Coefficient of Q.D.=63-6063+60or, Coefficient of Q.D.=3123⇒ Coefficient of Q.D.=0.024Hence, Q.D. of the series is 1.5 and Coefficient of Q.D. is 0.024
Question 1:
Given the following data, find out quartile deviation and the coefficient of quartile deviation:
Wages (₹) | 0−5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 |
Number of Workers | 4 | 6 | 3 | 8 | 12 | 7 |
Answer:
Wage | No. of worker (f) | Cumulative Frequency (c.f.) |
0 − 5 5 − 10 10 − 15 15 − 20 20 −25
25 − 30 |
4 6 3 8 12 7 |
4 4 + 6 = 10 10 + 3 = 13 13 + 8 = 21 21 + 12 = 33 33 + 7 = 40 |
Σf=N=40 |
Q1 = Size of
N4thitem
= Size of
404thitem
= Size of 10th item
10th item lies in group 5 − 10 and falls within 10th c.f. of the series.
Q1=l1+N4-c.f.f×iHere,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval
Thus,Q1=5+10-46×5Q1 =5+66×5⇒Q1 =10Like wise,
Q3 = Size of
3N4thitem
= Size of 3
404th item = Size of 30th item
30th item lies in the group 20-25 within the 33rd c.f. of the series.
Thus, Q3=l1+3N4-c.ff×ior, Q3 =20+30-2112×5or, Q3 =20+9×512or, Q3 =20+3.75⇒ Q3 =23.75Quartile deviation (Q.D.)=Q3-Q12=23.75-102=13.752=6.87Coefficient of quartile deviation=Q3-Q1Q3+Q1=23.75-1023.75+10=13.7533.75=0.4
Question 2:
Find out quartile deviation and coefficient of quartile deviation from the following data:
Class Interval | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
Frequency | 4 | 8 | 5 | 4 | 9 | 10 |
Answer:
Class interval | Frequency (f) |
Cumulative Frequency (c.f.) |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 |
4 8 5 4 9 10 |
4 4 + 8 = 12 12 + 5 = 17 17 + 4 = 21 21 + 9 = 30 30 + 10 = 40 |
Σf=N = 40 |
Q1 = Size of
N4thitem
= Size of
404thitem
= Size of 10th item
10th item lies in group 10 − 20 and fall within 12th c.f. of the series.
Q1=l1+N4-c.f.f×iHere,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval
Thus,Q1 =10+10-48×10or, Q1 =10+608or, Q1 =10+7.5⇒ Q1 =17.5Likewise,
Q3 = Size of
3N4thitem
= Size of
3404thitem
= Size of 30th item
30th item lies in group 40-50 and fall within the 30th c.f. of the series
Thus, Q3=l1+3N4-c.f.f×i or, Q3 =40+30-219×10 or, Q3 =40+99×10 or, Q3=40+10⇒ Q3=50Quartile deviation (QD)=Q3-Q12=50-17.52=32.52=16.25Coefficient of Q.D=Q3-Q1Q3+Q1=50-17.550+17.5=32.567.5=0.48
Question 1:
Find out mean deviation of the monthly income of the five families given below, using arithmetic mean of the data:
852, 635, 792, 836, 750
Answer:
Sr. No. | Monthly Income (X) |
Deviation from Arithmetic mean dX=X-X X¯=773 |
1 2 3 4 5 |
852 635 792 836 750 |
852-773=79 138 19 63 23 |
N=5 | ΣX = 3865 | ΣdX¯=322 |
Mean X=ΣXN=38655=773Mean deviation MDx¯=ΣdXN=3225=64.4Hence, mean deviation of the monthly income is 64.4.
Question 2:
Weight of nine students of a class is given below. Calculate mean deviation, using median and arithmetic mean of the series. Also calculate coefficient of mean deviation:
Weight (kg) : 47, 50, 58, 45, 53, 59, 47, 60, 49
Answer:
Sr. No | Weight (X) |
Deviation from Median dm=X-M(M = 50) |
1 2 3 4 5 6 7 8 9 |
45 47 47 49 50 53 58 59 60 |
5 3 3 1 3 8 9 10 |
N=9 | Σdm=42 |
Sr. No | Weight (X) |
Deviation from Mean dX=X-X¯X=52 |
1 2 3 4 5 6 7 8 9 |
45 47 47 49 50 53 58 59 60 |
7 5 5 3 2 1 6 7 8 |
N=9 | ∑X=468 | Σdx¯=44 |
From Median
(a) M = Size of
N+12thitem
= Size of
9+12thitem
= Size of 5th item
= 50 kgs
(b)
MDm=ΣdmN=429=4.666=4.67 kgs approx.(c)
Coefficient of MDm=MDmM=4.6750=0.0934From Mean
(a)
X=ΣXN=4689=52 kgs(b)
MDX=ΣdXN=449=4.89 kgs(c)
Coefficient of MDX=MDXX=4.8952=0.094
Question 1:
Find out mean deviation and its coefficient of the following data:
Items | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
Frequency | 8 | 16 | 18 | 22 | 14 | 9 | 6 | 7 |
Answer:
Items (X) | Frequency (f) | fX |
Deviation from Mean X-X= dX |
fdX |
5 10 15 20 25 30 35 40 |
8 16 18 22 14 9 6 7 |
40 160 270 440 350 270 210 280 |
15.2 10.2 5.2 0.2 4.8 9.8 14.8 19.8 |
121.6 163.2 93.6 4.4 67.2 88.2 88.8 138.6 |
Σf = 100 | Σfx = 2020 | ΣfdX=765.6 |
(a)
Mean (X¯)=ΣfxΣf=2020100=20.2(b) Mean deviation from Arithmetic Mean
MDX=ΣfdXΣf=765.6100=7.656(c) Coefficient of
MDX¯=MDXX=7.65620.2=0.379=0.38 approx.
Question 2:
Calculate mean deviation from the following data, using mean and median, respectively.
Size | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
Frequency | 2 | 4 | 5 | 3 | 2 | 1 | 4 |
Answer:
Calculation of mean deviation from median
Size (X) | Frequency (f) | Cumulative frequency (c.f.) | Deviation from median |dM| = |X − M| M = 8 |
f|dM| |
4 6 8 10 12 14 16 |
2 4 5 3 2 1 4 |
2 2 + 4 = 6 6 + 5 = 11 11 + 3 = 14 14 + 2 = 16 16 + 1 = 17 17 = 4 = 21 |
4 2 2 4 6 8 |
8 8 6 8 6 32 |
Σf=N= 21 | Σf|dM| = 68 |
(a) Median or (M) = Size of
N+12thitem
= Size of
21+12thitem
= Size of 11th item
= 8
(b) Mean Deviation from median
MDM=ΣfdMN=6821=3.24Calculation of mean deviation from mean
Size (X) | Frequency (f) | fX |
From mean deviation from mean dX=X-XX=9.71 |
fdX¯ |
4 6 8 10 12 14 16 |
2 4 5 3 2 1 4 |
8 24 40 30 24 14 64 |
5.71 3.71 1.71 0.29 2.29 4.29 6.29 |
11.42 14.84 8.55 0.87 4.58 4.29 25.16 |
Σf = 21 | ΣfX = 204 | ΣfdX¯= 69.71 |
(a)
Mean (X¯)=ΣfXΣf=20421=9.71(b) Mean deviation from A.M.
MDX=ΣfdXΣf=69.7121=3.32
Question 1:
The following table gives distribution of marks for 50 students of a class. Calculate mean deviation from the mean and median respectively from the data:
Marks Obtained | 140−150 | 150−160 | 160−170 | 170−180 | 180−190 | 190−200 |
Frequency | 4 | 6 | 10 | 18 | 9 | 3 |
Answer:
Calculation of M.D from Median
Marks | Mid value m | Frequency (f) |
Cumulative frequency | |dM| = |m − M| M = 172.78 |
f|dM| |
140 − 150 150 − 160 160 − 170 170 − 180 180 − 190 190 − 200 |
145 155 165 175 185 195 |
4 6 10 18(f) 9 3 |
4 4 + 6 = 10 10 + 10 = 20(c.f) 20 + 18 = 38 38 + 9 = 47 47 + 3 = 50 |
27.78 17.78 7.78 2.22 12.22 22.22 |
111.12 106.68 77.8 39.96 109.98 66.66 |
Σf = N = 50 | Σf|dM| = 512.2 |
Median = Size of
N2thitem
= Size of
502thitem
= Size of 25th item
25th item lies under the 38th cumulative frequency therefore 170 − 180 is the median class interval.
Thus, median value will be as follows:
M=l1+N2-c.ff×ior, M =170+25-2018×10or, M =170+5018or, M =170+2.78⇒ M =172.78Hence, Mean deviation MDM=ΣfdMΣf=512.250=10.24Calculation of M.D from Mean
Marks | Mid value m |
Frequency (f) |
fm | dX=m-XX=171.2 |
fdX |
140 − 150 150 − 160 160 − 170 170 − 180 180 − 190 190 − 200 |
145 155 165 175 185 195 |
4 6 10 18(f) 9 3 |
580
930 1650 3150 1665 585
|
26.2 16.2 6.2 3.8 13.8 23.8 |
104.8 97.2 62 68.4 124.2 71.4 |
Σf = 50 | Σfm = 8560 | ΣfdX=528 |
Mean (X)=ΣfmΣf=856050=171.2Hence, Mean deviation from the arithmetic mean is:
MDX=ΣfdXΣf=52850=10.56
Question 2:
Estimate the coefficient of mean deviation from the median from the following data:
Age Group | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
Number of Workers | 8 | 12 | 20 | 16 | 4 |
Answer:
Calculation of MD. from Median
Age | Mid value (m) |
No. of workers (f) | Cumulative Frequency |
|dm| = |m − M| M = 45 |
f|dm| |
20 − 30 30 − 40 40 − 50 50 − 60 60 −70 |
25 35 45 55 65 |
8 12 20 16 4 |
8 8 + 12 = 20 20 + 20 =40 40 + 16 =56 56 + 4 = 60 |
20 10 10 20 |
160 120 160 80 |
N=Σf = 60 | Σf|dm| = 520 |
Median = Size of
N2thitem
= Size of
602thitem
= Size of 39th item
39th item lies under the 40th cumulative frequency of the series. So, 40 − 50 will be the median class interval.
Hence, the median value of the series is:
M=l1+N2-c.f.f×ior, M =40+30-2020×10or, M =40+10×1020⇒ M =45Thus,Mean Deviation from median MDM=ΣfdmΣf=52060=8.67Coefficient of MDM=MDMMedian=8.6745=0.19
Question 1:
The following table gives marks obtained by 7 students of a class. Find out standard deviation of the marks.
40, 42, 38, 44, 46, 48, 50
Answer:
S. No. | Marks (X) |
Deviation x=X-XX=44 |
Square of the deviation x2=X-X2 |
1 2 3 4 5 6 7 |
40 42 38 44 46 48 50 |
−4 −2 −6 2 4 6 |
16 4 36 4 16 36 |
N = 7 | ΣX = 308 | Σx2 = 112 |
Mean (X)=ΣXN=3087=44Standard deviation (σ )=x2N=1127=16=4Hence, standard deviation of the marks is 4.
Question 2:
Weight of some students is given below in kilograms. Find out standard deviation.
41, 44, 45, 49, 50, 53, 55, 55, 58, 60
Answer:
S. No. | Weight (X) |
Deviation x=X-XX=51 |
Square of deviation x2=X-X2 |
1 2 3 4 5 6 7 8 9 10 |
41 44 45 49 50 53 55 55 58 60 |
−10 −7 −6 −2 −1 2 4 4 7 9 |
100 49 36 4 1 4 16 16 49 81 |
N = 10 | ΣX = 510 | Σx2 = 356 |
Mean (X)=ΣXN=51010=51Standard deviation (σ )=x2N=35610=35.6=5.97Hence, the Standard deviation is 5.97.
Question 3:
Using step-deviation method, calculate standard deviation of the following series:
Marks | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
Number of Students | 5 | 10 | 20 | 40 | 30 | 20 | 10 | 4 |
Answer:
Marks (X) |
Mid value (m) |
Frequency (f) |
fm |
Deviation from mean value x=X-X X¯=39.39 |
x2 | fx2 |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
5 15 25 35 45 55 65 75 |
5 10 20 40 30 20 10 4 |
25 150 500 1400 1350 1100 650 300 |
−34.39 −24.39 −14.39 −4.39 5.61 15.61 25.61 35.61 |
1182.67 594.87 207.07 19.27 31.47 243.67 655.87 1268.07 |
5913.35 5948.7 4141.4 770.8 944.1 4873.4 6558.7 5072.28 |
Σf = 139 | Σfm = 5475 | Σfx2 = 34222.73 |
Mean (X)=ΣfmN=5475139=39.39Standard deviation (σ )=Σfx2N=34222.73139=246.21=15.7Hence, the standard deviation of the above series is 15.7
Question 1:
Find out standard deviation of the savings of the following 10 persons:
Persons | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Savings (₹) | 114 | 108 | 100 | 98 | 101 | 109 | 117 | 119 | 121 | 126 |
Answer:
Persons | Savings (X) |
Deviation x=X-X X¯=111.3 |
Square of the deviation x2 |
1 2 3 4 5 6 7 8 9 10 |
114 108 100 98 101 109 117 119 121 126 |
2.7 −3.3 −11.3 −13.3 −10.3 −2.3 5.7 7.7 9.7 14.7 |
7.29 10.89 127.69 176.89 106.09 5.29 32.49 59.29 94.09 216.09 |
N = 10 | ΣX = 1113 | Σx2 = 836.1 |
Mean (X)=ΣXN=111310=111.3Standard deviation (σ )=Σx2N=836.110=83.61=9.14Hence, standard deviation of the savings is 9.14
Question 2:
Find out standard deviation and its coefficient of the following series:
Size | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
Frequency | 6 | 8 | 16 | 15 | 33 | 11 | 12 |
Answer:
Size (X) |
Frequency (f) |
fX |
Deviation from mean x=X-X |
Square of deviation x2 | fx2 |
10 20 30 40 50 60 70 |
6 8 16 15 33 11 12 |
60 160 480 600 1650 660 840 |
−34.06 −24.06 −14.06 −4.06 5.94 15.94 25.94 |
1160.08 578.88 197.68 16.48 35.28 254.08 672.88 |
6960.48 4631.04 3162.88 247.2 1164.24 2794.88 8074.60 |
Σf = 101 | Σfx = 4450 | Σfx2 = 27035.32 |
Mean (X)=ΣfXΣf=4450101=44.06Standard deviation (σ )=Σfx2Σf=27035.32101=267.68=16.36Coefficient of S.D.=σX=16.3644.06=0.37Hence, the S.D. is 16.36 and Coefficient of S.D. is 0.37
Question 1:
Calculate standard deviation of the following series:
Size | 0−5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 |
Frequency | 2 | 5 | 7 | 13 | 21 | 16 | 8 | 3 |
Answer:
Size | Mid Value m |
Frequency | Deviation from assumed Average (dx = m − A) |
dx’=dxC C=5 |
fdx‘ | fdx‘2 |
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40 |
2.5 7.5 12.5 17.5(A) 22.5 27.5 32.5 37.5 |
2 5 7 13 21 16 8 3 |
−15 −10 −5 5 10 15 20 |
−3 −2 −1 1 2 3 4 |
−6 −10 −7 21 32 24 12 |
18 20 7 21 64 72 48 |
N = 75 | Σfdx‘ = 66 | Σfdx’2 = 250 |
Calculating Standard deviation through Step deviation method:
Standard deviation (σ )=Σfdx’2N-Σfdx’N2×Cor, σ=Σfdx’2N-Σfdx’N2×5or, σ =25075-66752×5or, σ =3.33-0.7744 ×5or, σ =2.5556×5or, σ =1.598×5⇒ σ =7.99Hence, standard deviation of the above series is 7.99.
Question 2:
Calculate standard deviation of the following data, using step-deviation method.
Age | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 |
Frequency | 3 | 61 | 132 | 153 | 140 | 51 | 2 |
Answer:
Age | Mid value m |
Frequency (f) |
Deviation assumed from average (dx = m − A) |
dx’=dxc=dx10 |
fdx’=f×dx’ |
fdx’2=f×dx’2 |
20−30 30 −40 40 −50 50 −60 60 −70 70 −80 80 −90 |
25 35 45 55=A 65 75 85 |
3 61 132 153 140 51 2 |
−30 −20 −10 10 20 30 |
−3 −2 −1 1 2 3 |
−9 −122 −132 140 102 6 |
27 244 132 140 204 18 |
N = 542 | Σfdx‘ = -15 | Σfdx‘2 = 765 |
Standard deviation (σ)=Σfdx’2N-Σfdx’N2×cor, σ =765542–155422×10or, σ =1.41–0.032×10or, σ =1.41-0.0009×10or, σ =1.19×10⇒ σ =11.9Hence, standard deviation of the above data is 11.9
Question 3:
Find out standard deviation of the distribution of population in 104 villages of a Tehsil, as given below by step-deviation method.
Distribution of Population |
Population | No. of Villages |
0−200 200−400 400−1,000 1,000−2,000 2,000−5,000 |
10 28 42 18 6 |
Answer:
Population |
Mid Value (m) |
No. of villages (f) |
Deviation assumed from average
dx=m-A
|
dx’=dxi=dx40 |
fdx’=f×dx’ |
fdx’2=f×dx’2 |
0-200 200-400 400-1000 1000-2000 2000-5000 |
100 300 700=A 1500 3500 |
10 28 42 18 6 |
-600 -400 800 2800 |
-15 -10 20 70 |
-150 -280 360 420 |
2250 2800 7200 29400 |
N=104 | ∑fdx’= -350 |
∑fdx’2=41650 |
Calculating Standard deviation through Step deviation method:
Standard Deviation (σ)=Σfdx’2N-Σfdx’N2×ior, σ =41650104–3501042×40or, σ =400.480-11.296×40or, σ =389.184×40or, σ =19.72775×40⇒ σ =789.11Hence, standard deviation of the above series is 789.11
Question 4:
Calculate mean and standard deviation of the following data by short-cut method:
Class Interval | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
Frequency | 5 | 10 | 15 | 20 | 25 | 18 | 7 |
Answer:
C.I. | Mid Value (m) |
Frequency (f) |
Deviation from Assumed mean dx = m − A (A = 45) |
fdx | f×dx2=fdx2 |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
15 25 35 45=A55 65 75 |
5 10 15 20 25 18 7 |
−30 −20 −10 10 20 30 |
−150 −200 −150 250 360 210 |
4500 4000 1500 2500 7200 6300 |
Σf = N = 100 | Σfdx = 320 | Σfdx2 = 26000 |
Calculating Mean and standard deviation through Short-cut method:Mean (X)=A+ΣfdxΣf or, X=45+320100 or, X=45+3.2⇒ X=48.2Standard deviation (σ)=Σfdx2N-ΣfdxN2or, σ =26000100-3201002or, σ =260-3.22or, σ =260-10.24or, σ =249.76⇒ σ =15.80
Question 1:
Following are the marks obtained by 20 students in statistics. Find out coefficient of variation of the marks.
62 | 85 | 73 | 81 | 74 | 58 | 66 | 72 | 54 | 84 |
65 | 50 | 83 | 62 | 85 | 52 | 80 | 86 | 71 | 75 |
Answer:
S. No. | Marks (X) |
Deviation from mean x=X-X |
Square deviation x2 |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 |
62 85 73 81 74 58 66 72 54 84 65 50 83 62 85 52 80 86 71 75 |
−8.9 +14.1 +2.1 10.1 3.1 −12.9 −4.9 1.1 −16.9 13.1 −5.9 −20.9 12.1 −8.9 14.1 −18.9 9.1 15.1 .1 4.1 |
79.21 198.81 4.41 102.01 9.61 166.41 24.01 1.21 285.61 171.61 34.81 436.81 146.81 79.21 198.81 357.21 82.81 228.01 .01 16.81 |
ΣX = 1418 | Σx2 = 2623.8 |
Mean (X)=ΣXN=141820=70.9Standard deviation (σ)=x2Nor, σ =2623.820or, σ =131.19⇒ σ = 11.45Coefficient of variation (CV)=σX×100or, CV=11.4570.9×100⇒ CV=16.14Hence, coefficient of variation (CV) of the marks is 16.14
Question 2:
Calculate coefficient of variation of the following data:
S. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Value | 25 | 42 | 33 | 48 | 45 | 29 | 43 | 39 |
Answer:
S. No. | Value (X) |
Deviation from mean x=X-X |
x2 |
1 2 3 4 5 6 7 8 |
25 42 33 48 45 29 43 39 |
−13 4 −5 10 7 −9 5 1 |
169 16 25 100 49 81 25 1 |
N = 8 | ΣX = 304 | Σx2 = 466 |
Mean (X)=ΣXN=3048=38Standard deviation (σ )=Σx2N=4668=58.25=7.63Coefficient of variation (CV)=σX×100=7.6338×100=76338=20.08Hence, coefficient of variation is 20.08
Question 3:
Given the following data, calculate coefficient of variation:
Age | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 |
Number of Students | 3 | 61 | 132 | 153 | 140 | 51 | 2 |
Answer:
Age | Mid Value (m) |
Frequency (f) |
Deviation from Assumed mean (dx = X − A) (A = 55) |
dx’=dxi=dx10 |
f×dx’=fdx’ |
f×dx’2=fdx’2 |
20−30 30 −40 40 −50 50 −60 60 −70 70 −80 80 −90 |
25 35 45 55=A 65 75 85 |
3 61 132 153 140 51 2 |
−30 −20 −10 10 20 30 |
−3 −2 −1 1 2 3 |
−9 −122 −132 140 102 6 |
27 244 132 140 204 18 |
N = 542 | Σfdx‘= -15 | Σfdx‘2 =765 |
Mean (X)=A+Σfdx’N×ior, X=55+-15542×10or, X =55-.28⇒ X =54.72Standard Deviation (σ)=Σfdx’2N-Σfdx’N2×ior, σ =765542–155422×10or, σ =1.41–0.032×10or, σ =1.41-0.0009×10or, σ =1.187×10⇒ #160;σ =11.87Coefficient of Variation (CV)=σX×100or, CV=11.8754.72×100or, CV=119054.72⇒CV =21.69Hence, Coefficient of variation is 21.69
Question 1:
The following table shows number of firms in two different areas ‘A’ and ‘B’ according to their annual profits. Present the data by way of Lorenz curve.
Profit (‘000’ ₹) | 6 | 25 | 60 | 84 | 105 | 150 | 170 | 400 |
Firms in Area A | 6 | 11 | 13 | 14 | 15 | 17 | 10 | 14 |
Firms in Area B | 2 | 38 | 52 | 28 | 38 | 26 | 12 | 4 |
Answer:
Firm in Area A | Firms in Area B | |||||||
Profit (x) |
Cumulative sum | Cumulative % |
No. of firm | c.f | Cumulative % |
No. of firms | c.f. | Cumulative % |
6 25 60 84 105 150 170 400 |
6 31 91 175 280 430 600 1000 |
.6 3.1 9.1 17.5 28 43 60 100 |
6 11 13 14 15 17 10 14 |
6 17 30 44 59 76 86 100 |
6 17 30 44 59 76 86 100 |
2 38 52 28 38 26 12 4 |
20 40 92 120 158 184 196 200 |
10 20 46 60 79 82 98 100 |
Σx = 1000 |
Question 2:
Make a Lorenz curve of the following data:
Income | 500 | 1,000 | 2,000 | 3,000 | 3,500 |
Number in Class ‘A’ (000) | 4 | 6 | 8 | 12 | 10 |
Number in Class ‘B’ (000) | 8 | 7 | 5 | 3 | 2 |
Answer:
CLASS A | CLASS B | |||||||
Income (x) |
Cumulative sum | Cumulative % |
No. of class | c.f. | Cumulative % |
No. of class | c.f. | cumulative |
500 1000 2000 3000 3500 |
500 1500 3500 6500 10,000 |
5 15 35 65 100 |
4 6 8 12 10 |
4 10 18 30 40 |
10 25 45 75 100 |
8 7 5 3 2 |
8 15 20 23 25 |
32 60 80 92 100 |
Question 1:
Calculate range and coefficient of range from the following data:
4, 7, 8, 46, 53, 77, 8, 1, 5, 13
Answer:
Given:
Highest value (H) = 77
Lowest value (L) = 1
Range = Highest value − Lowest value
i.e. R = H − L
Substituting the given values in the formula.
R = 77 − 1 = 76
Coefficient of Range=H-LH+L=77-177+1=7678=0.97Thus, range is 76 and coefficient of range is 0.97
Question 2:
Given the following data-set, calculate range and the coefficient of range:
Size | 4.5 | 5.5 | 6.5 | 7.5 | 8.5 | 9.5 | 10.5 | 11.5 |
Frequency | 4 | 5 | 6 | 3 | 2 | 1 | 3 | 5 |
Answer:
Highest value (H) = 11.5
Lowest value (L) = 4.5
Range = Highest value − Lowest value
i.e. R = H − L
or, R = 11.5 − 4.5 = 7
Coefficient of Range=H-LH+L=11.5-4.511.5+4.5=716=0.437Note– As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.
Question 3:
Find out quartile deviation and the coefficient of range, given the following data-set:
Class Interval | 1−5 | 6−10 | 11−15 | 16−20 | 21−25 | 26−30 | 31−35 |
Frequency | 2 | 8 | 15 | 35 | 20 | 10 | 14 |
Answer:
In order to calculate range and its coefficient, we must first convert the given inclusive class intervals in exclusive class intervals.
Class Interval | Exclusive Class Interval | Mid Value= L1+L22 |
Frequency |
1 − 5 6 − 10 11 − 15 16 − 20 21 − 25 26 − 30 31 − 35 |
0.5 − 5.5 5.5 − 10.5 10.5 − 15.5 15.5 − 20.5 20.5 − 25.5. 25.5. − 30.5 30.5 − 35.5 |
3 8 13 18 23 28 33 |
2 8 15 35 20 10 14 |
Range = Upper Limit of the Highest Class Interval – Lower Limit of the Lowest Class Intervalor, Range = 35.5 – 0.5 = 35Coefficient of Range=H-LH+L=35.5-0.535.5+0.5=3536=0.972
Question 4:
Find out quartile deviation and the coefficient of quartile deviation of the following series.
Wages of 9 workers in Rupees:
170, 82, 110, 100, 150, 150, 200, 116, 250
Answer:
Arranging the data in the ascending order as presented below.
82, 100, 110, 116, 150, 150, 170, 200, 250
Here, N = 9
Q1=size of 9+14th=size of 2.5th item =size of 2nd item+0.5 size of 3rd item-size of 2nd item =100+0.5110-100=105Q3=size of 39+14thitem=size of 7.5th item =size of 7th item+0.5 size of 8th item-size of 7th item =170+0.5200-170=185Q.D.=Q3-Q12=185-1052=40 Coefficient of Q.D.=Q3 – Q1Q3 + Q1=185-105185+105=0.276
Question 5:
Given the following data, estimate the coefficient of QD:
15, 20, 23, 23, 25, 25, 27, 40
Answer:
Given data series:
15, 20, 23, 23, 25, 25, 27, 40
Here, N = 8
Q1=size of 8+14th=size of 2.25th item =size of 2nd item+0.25 size of 3rd item-size of 2nd item =20+0.2523-20=20.75Q3=size of 38+14thitem=size of 6.75th item =size of 6th item+0.75 size of 7th item-size of 6th item =25+0.7527-25=26.5
Coefficient of Q.D.=Q3 – Q1Q3 + Q1=26.5-20.7526.5+20.75=0.121
Question 6:
Find out mean deviation of the following series from mean and median:
Size | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
Frequency | 2 | 4 | 5 | 31 | 2 | 1 | 4 |
Answer:
x | f | fx | dX=X-X |
fdX |
4 6 8 10 12 14 16 |
2 4 5 31 2 1 4 |
8 24 40 310 24 14 64 |
5.87 3.87 1.87 0.13 2.13 4.13 6.13 |
11.74 15.48 9.35 4.03 4.26 4.13 24.52 |
∑f = 49 | ∑fx = 484 | ∑fdX= 73.51 |
Mean X=∑fx∑f=48449=9.87Thus calculate the deviation of the values from 9.87.Mean deviation from mean MDX=∑fdX∑f=73.5149=1.50
x | f | c.f. | dM = |x − M| | f|dM| |
4 6 8 10 12 14 16 |
2 4 5 31 2 1 4 |
2 6 11 42 44 45 49 |
6 4 2 2 4 6 |
12 16 10 4 4 24 |
∑f = 49 | ∑f|dM| = 70 |
N = 49
Median=size of N+12thor, Median=size of 49+12thitem= size of 25th itemNow, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25th is 42 (in the c.f. column), which is corresponding to 10.
Hence, median is 10.
Thus, we calculate the deviation of the values from 10.
Mean Deviation from median=MDM=∑fdM∑f=7049=1.428 (or 1.43 approximately)
Question 7:
Calculate mean deviation and coefficient of mean deviation with the help of median:
Class Interval | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
Frequency | 15 | 19 | 14 | 20 | 18 | 14 |
Answer:
x | Mid Value (m) |
f | c.f. | |m−M| = |dM| | f|dM| |
0 − 10 10 − 20 20 − 30 |
5 15 25 |
15 19 14 |
15 34 48 |
26 16 6 |
390 304 80 |
30-40 | 35 | 20 | 68 | 4 | 252 |
40 − 50 50 − 60 |
45 55 |
18 14 |
86 100 |
4 14 |
252 336 |
∑fx = 100 | ∑f|dM| = 1446 |
Here, N or ∑fx = 100
Median class is given by the size of
N2thitem, i.e.
1002thitem, which is 50th item.
This corresponds to the class interval of 30 − 40, so this is the median class.
Median=l1+N2-c.f.f×iso, Median=30+1002-4820×10or, Median=30+220×10or, Median=30+1Therefore, Median=31Thus, we calculate the deviation of values from 31.
Mean Deviation from median=MDM=∑fdM∑f=1446100=14.46Coefficient of Mean Deviation = Mean DeviationMedian=14.4631= 0.466
Question 8:
Calculate mean deviation from mean of the following series:
Size of Items | 3−4 | 4−5 | 5−6 | 6−7 | 7−8 | 8−9 | 9−10 |
Frequency | 3 | 7 | 22 | 60 | 85 | 32 | 9 |
Answer:
Class Interval |
Mid Value (m) |
f | fm | m-X=dX¯ |
fdX |
3 − 4 4 − 5 5 − 6 6 − 7 7 − 8 8 − 9 9 − 10 |
3.5 4.5 5.5 6.5 7.5 8.5 9.5 |
3 7 22 60 85 32 9 |
10.5 31.5 121 390 637.5 272 85.5 |
3.6 2.6 1.6 0.6 0.4 1.4 2.4 |
10.8 18.2 35.2 36 34 44.8 21.6 |
∑f = 218 | ∑fm = 1548 | ∑fdX = 200.6 |
Mean X=∑fm∑f=1548218=7.10Thus, we calcualte the value of deviations from 7.10Mean Deviation from Mean MDX=∑fdX∑f=200.6218=0.920
Question 9:
Given below are the marks obtained by the students of a class. Calculate mean deviation, and its coefficient, using median of data.
Marks | 17 | 35 | 38 | 16 | 42 | 27 | 19 | 11 | 40 | 25 |
Answer:
The given data is an individual series, therefore, to calculate median, we first need to arrange the data in ascending order. This is done as below.
11, 16, 17, 19, 25, 27, 35, 38, 40, 42Here, N =10 (even)Median=size of N2thitem+size of N2+1th item2or, Median=size of 5thitem+size of 6th item2=25+272=26so, Median is 26.Thus, we calculate the deviation of values from 26
X | |x – M| = |dM| |
11 16 17 19 25 27 35 38 40 42 |
15 10 9 7 1 1 9 12 14 16 |
∑|dM| = 94 |
Mean Deviation from Median MDM=∑dMn=9410=9.4Coefficient of MDM=MDMMedian=9.426=0.36
Question 10:
Nine students of a class obtained following marks. Calculate mean deviation from median.
S. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Marks | 68 | 49 | 32 | 21 | 54 | 38 | 59 | 66 | 41 |
Answer:
The given data series is discrete in nature, therefore, to calculate median, we first have to arrange the data in ascending order. This is done as below.
21, 32, 38, 41, 49, 54, 59, 66, 68Here, N =9 (odd)Median=size of N+12=9+12=size of 5th item
Thus, we calculate the deviation of values from 49.
Marks | Ascending Order | |X– M| = |dM| |
68 49 32 21 54 38 59 66 41 |
21 32 38 41 49 54 59 66 68 |
28 17 11 8 5 10 17 19 |
∑|dM| = 115 |
Mean Deviation from Median MDM=∑dMN=1159=12.77
Question 11:
Following data relate to the age-difference of husbands and wives of a particular community. Find out mean deviation from mean.
Age-difference | 0−5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 |
Frequency | 449 | 705 | 507 | 281 | 109 | 52 | 16 | 4 |
Answer:
X | Mid Value m |
f | fm | m-X=dX |
fdX |
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40 |
2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5 |
449 705 507 281 109 52 16 4 |
1122.5 5287.5 6337.5 4917.5 2452.5 1430 520 150 |
7.96 2.96 2.04 7.04 12.02 17.04 22.04 27.04 |
3574.04 2086.8 1034.28 1978.24 1312.36 886.08 352.64 108.16 |
∑f = 2123 | ∑fm = 22217.5 | ∑ fdX= 11332.6 |
Mean X=∑fm∑f=22217.52123=10.46Mean Deviation from Mean MDX=∑fdXX=11332.62123=5.34
Question 12:
Find out the mean deviation and its coefficient using median of the following data:
S. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Number of Victims of Accidents | 16 | 21 | 10 | 17 | 8 | 4 | 2 | 1 | 2 | 2 | 2 | 2 |
Answer:
The given data series is individual in nature, therefore, to calculate median, we first have to arrange the data in ascending order. This is done as below.
1, 2, 2, 2, 2, 2, 4, 8, 10, 16, 17, 21Here, N =12 (even)Median=size of N2thitem+size of N2+1th item2or, Median=size of 6thitem+size of 7th item2=2+42=3so, Median is 3.
No. of Victims | Ascending Order | |x – M| = |dM| |
16 21 10 17 8 4 2 1 2 2 2 2 |
1 2 2 2 2 2 4 8 10 16 17 21 |
2 1 1 1 1 1 1 5 7 13 14 18 |
∑|dM| = 65 |
Mean Deviation from Median MDM=∑dMn=6512=5.42Coefficient of MDM=MDMMedian=5.423=1.81
Question 13:
Calculate standard deviation, given the following data:
10, 12, 14, 16, 18, 22, 24, 26, 28
Answer:
X | x=X-X | x2 |
10 12 14 16 18 22 24 26 28 |
−8.88 −6.88 −4.88 −2.88 −.88 3.12 5.12 7.12 9.12 |
78.85 47.33 23.81 8.29 0.77 9.73 26.21 50.69 83.17 |
∑X = 170 | ∑x2 = 328.85 |
Mean X=∑XN=1709=18.88Standard Deviation σ=∑x2N=328.859=6.045
Question 14:
Calculate standard deviation and the coefficient of standard deviation, given the following data:
Income (₹) | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
Number of Workers | 26 | 29 | 40 | 35 | 26 | 18 | 14 | 12 |
Answer:
Income (X) |
No. of workers (f) |
fX | x=X-X | x2 | fx2 |
5 10 15 20 25 30 35 40 |
26 29 40 35 26 18 14 12 |
130 290 600 700 650 540 490 280 |
−14.4 −9.4 −4.4 0.6 5.6 10.6 15.6 20.6 |
207.36 88.36 19.36 0.36 31.36 112.36 243.36 424.36 |
5391.36 2562.44 774.4 12.6 815.36 2022.48 3407.04 5092.32 |
∑f = 200 | ∑fX = 3880 | ∑fx2 = 20078 |
X=∑fX∑f=3880200=19.4Standard Deviation σ=∑fx2∑f=20078200=10.02Coefficient of Standard Deviation=σX=10.0219.4=0.516
Question 15:
Of the two sets of income distribution of five and seven persons respectively, as given below calculate standard deviation:
(i) Income (₹) | 4,000 | 4,200 | 4,400 | 4,600 | 4,800 | ||
(ii) Income (₹) | 3,000 | 4,000 | 4,200 | 4,400 | 4,600 | 4,800 | 5,800 |
Answer:
For Income Group I
X1 | x1=X1-X1 | xI2 |
4000 4200 4400 4600 4800 |
−400 −200 200 400 |
160000 40000 40000 160000 |
∑X1 = 22,000 | ∑ xI2= 4,00,000 |
X1=∑X1n1=220005=4400Standard Deviation σ1=∑x12n1=4,00,0005=282.84Thus, Standard Deviation for the first group of income is Rs 282.84For Income Group II
X2 | x2=X2-X2 | x22 |
3000 4000 4200 4400 4600 4800 5800 |
−1400 −400 −200 200 400 1400 |
1960000 160000 40000 40000 160000 1960000 |
30,800 | ∑ x22= 4320000 |
X2=∑X2n2=30,8007=4,400Standard Deviation σ2=∑x22n2=4,32,0007=785.58Thus, Standard Deviation for the second group of income is Rs 785.58
Question 16:
Find out the standard deviation of the marks secured by 10 students:
S. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Marks | 43 | 48 | 65 | 57 | 31 | 60 | 37 | 48 | 78 | 59 |
Answer:
S.No. | Marks (X) |
x=X-X | x2 |
1 2 3 4 5 6 7 8 9 10 |
43 48 65 57 31 60 37 48 78 59 |
−9.6 −4.6 12.4 4.4 −21.6 7.4 −15.6 −4.6 25.4 6.4 |
92.16 21.16 153.76 19.36 466.56 54.76 243.36 21.16 645.16 40.96 |
∑X = 526 | ∑x2 = 1758.4 |
Mean X=∑XN=52610=52.6Standard Deviation σ=∑x2N=1758.410=13.26
Question 17:
Data of daily sale proceeds of a shop are as below. Calculate mean deviation and standard deviation.
Daily Sales | 102 | 100 | 110 | 114 | 118 | 122 | 126 |
Days | 3 | 9 | 25 | 35 | 17 | 10 | 1 |
Answer:
Daily Sales (X) |
Days (f) |
fX | X-X=dX |
fdXor fX-X |
fX-X2 or fdX2 |
102 100 110 114 118 122 126 |
3 9 25 35 17 10 1 |
306 900 2750 3990 2006 1220 126 |
10.98 12.98 2.98 1.02 5.02 9.02 13.02 |
32.94 116.82 745 35.7 85.34 90.2 13.02 |
361.68 1516.32 222 36. 428.4 813.6 169.52 |
∑f = 100 | ∑fX = 11298 | ∑ fdX= 1119.02 |
∑ = 3547.92 |
Mean X=∑fX∑f=11298100=112.98Mean Deviation from mean MDX=∑fdX∑f=1119.02100=11.19Standard Deviation (σ) =∑fX-X2∑f=3547.92100= 5.96
Question 18:
Calculate range, standard deviation and coefficient of variation of marks secured by students.
50 | 55 | 57 | 49 | 54 | 61 | 64 | 59 | 58 | 56 |
Answer:
X | x=X-X |
x2 |
50 55 57 49 54 61 64 59 58 56 |
−6.3 −1.3 .7 −7.3 −2.3 4.7 7.7 2.7 1.7 −.3 |
39.69 1.69 .49 53.29 5.29 22.09 59.29 7.29 2.89 .09 |
∑X=563 |
∑x2=192.1 |
Here,
N = 10
Highest value (H) = 64
Lowest value (Z) = 49
Range (R) = H − L = 64 − 49 = 15
So, range is 15.
Mean X=∑XN=56310=56.3Standard Deviation σ=∑x2N=192.110=4.38Coefficient of Variation=σX¯×100=4.3856.3×100=7.78
Question 19:
Following data show the number of runs made by Sachin and Sourabh in different innings. Find out who is a good scorer and who is a consistent player?
Sachin | 92 | 17 | 83 | 56 | 72 | 76 | 64 | 45 | 40 | 32 |
Sourabh | 28 | 70 | 31 | 00 | 59 | 108 | 82 | 14 | 3 | 95 |
Answer:
Sachin | Saurav | ||||
X1 | x1 | x12 | X2 | x2 | x22 |
92 17 83 56 72 76 64 45 40 30 |
34.3 −40.7 25.3 −1.7 14.3 18.3 6.3 −12.7 −17.7 −25.7 |
1176.49 1656.49 640.09 2.89 204.49 334.89 39.69 161.29 313.29 660.49 |
28 70 31 00 59 108 82 14 3 95 |
−27 25 −24 −55 4 53 27 −41 −52 40 |
729 625 576 3025 16 2809 729 1681 2704 1600 |
∑X1= 577 | ∑x12 = 5190.1 | ∑X2 = 550 | ∑x22 = 14494 |
XSachin=∑X1n1=57710=57.7σSachin=∑x12n1=5190.110=22.78Coefficient of Variation (Sachin)=σSachinXSachin×100=22.7857.7×100=39.48XSaurav=∑X2n2==55010=55σSaurav=∑x22n2=1449410=38.07Coefficient of Variation (Saurav)=σSauravXSaurav×100=38.0755×100=69.21Observation and Conclusion
- In order to regard anyone as a better batsman, we have to analyse the mean runs of both the players. Analysing the mean runs, we can say that Sachin is a better scorer, since his mean runs (average of 57.7) is higher than that of Saurav (average of 55).
- In order to regard one as more consistent, we have to take the help of coefficient of variation. The one who has lower coefficient of variation will be regarded as more consistent than the other. In this way, we can say that Sachin is more consistent than Saurav, as the coefficient of variation of Sachin (39.48) is lower than that of Saurav’s (69.21).
Thus, on the basis of the above two observations, we can conclude that Sachin is not only a better scorer, but also a more consistent player than Saurav.
Question 20:
Calculate standard deviation of marks secured by 100 examinees in the examination:
Marks | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 |
Number of Examinees | 19 | 3 | 2 | 49 | 24 | 2 | 0 | 1 |
Answer:
X | Mid Values m |
f | fX | x=X- X | x2 | fx2 |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 |
15 25 35 45 55 65 75 85 |
19 3 2 49 24 2 1 |
285 75 70 2205 1320 130 85 |
−26.70 −16.70 −6.7 3.3 13.3 23.3 33.3 43.3 |
712.89 278.89 44.89 10.89 176.89 542.89 1108.89 1874.89 |
13544.91 836.67 89.78 533.61 4245.36 1085.78 1874.89 |
∑f = 100 | ∑fX = 4170 | ∑fx2 = 22211 |
X=∑fX∑f=4170100=41.70Standard Deviation σ=∑fx2∑f=22211100=14.90
Question 21:
Calculate coefficient of variation from the following data:
Variables | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
Frequencies | 6 | 8 | 16 | 15 | 32 | 11 | 12 |
Answer:
X | f | fX | x=X- X |
x2 | fx2 |
10 20 30 40 50 60 70 |
6 8 16 15 32 11 12 |
60 160 480 600 1600 660 840 |
−34 −24 −14 −4 6 16 26 |
1156 576 196 16 36 256 676 |
6936 4608 3136 240 1152 2816 8112 |
∑f = 100 | ∑fX = 4400 | ∑f x2 = 27000 |
X=∑fX∑f=4400100=44Standard Deviation σ=∑fx2∑f=27000100=16.43Coefficient of Variation =σX×100=27000100×100 = 37.34
Question 22:
Estimate coefficient of variation of the following data:
Weight (kg) | 0−20 | 20−40 | 40−60 | 60−80 | 80−100 |
Number of Persons | 81 | 40 | 66 | 49 | 14 |
Answer:
Class Interval | Mid Value (m) |
f | fm | x=X- X |
x2 | fx2 |
0 − 20 20 − 40 40 − 60 60 − 80 80 − 100 |
10 30 50 70 90 |
81 40 66 49 14 |
810 1200 3300 3430 1260 |
−30 −10 10 30 50 |
900 100 100 900 2500 |
72900 4000 6600 44100 35000 |
∑f = 250 | ∑fm = 10000 | ∑fx2 = 162600 |
X=∑fm∑f=10000250=40Standard Deviation σ=∑fx2∑f=162600250=25.50Coefficient of Variation =σX×100=25.5040×100 = 63.75