T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 11 – Measures of Dispersion

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.

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:

For Tata Motors
Highest Value=420
Lowest Value=325
Range (R) = Highest Value (H)− Lowest Value (L)
or, R_₁   = 420 − 325
⇒  R_₁   = 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, R_₂   = 913.35 − 750.90
⇒ R_₂   = 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:

Answer:

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:

Answer:

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.

Answer:

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.

Answer:

Converting the inclusive series into exclusive series,

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:

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 3^rd item
⇒  Q₁     = 11
Q₃ = Size of
3N+14thitem
or,  Q₃  = Size of
311+14thitem
or,   Q₃ = Size of 9^th 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
-Q_{1
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.

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 2.25^th item
or, Q₁= Size of 2^nd 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 6.75^th item
or, Q₃ = Size of 6^th 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:

Answer:

Q₁ = Size of
N + 14thitem
= Size of
543+14thitem
= Size of 136^th item
136^th item lies in 196^th cumulative frequency of the series. Age corresponding to 196^th (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:

Answer:

Q₁ = Size of
N+14thitem
= Size of
195+14thitem
= Size of 49^th item
49^th item lies in 67^th c.f. of the series. Height corresponding to 67^th (c.f.) is 60.
Hence, Q₁ = 60 inches
Q₃ = Size of
3N+14thitem
= Size of
3195+14thitem
= Size of 147^th item
147^th item lies in 157^th c.f. of the series. Height corresponding to 157^th 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:

Answer:

Q₁ = Size of
N4thitem
= Size of
404thitem
= Size of 10^th item
10^th item lies in group 5 − 10 and falls within 10^th c.f. of the series.
Q1=l1+N4-c.f.f×iHere,
l_₁=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 30^th item
30^th item lies in the group 20-25 within the 33^rd 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:

Answer:

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:

 
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:

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:

Answer:

(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.

Answer:

Calculation of mean deviation from median

(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

(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:

Answer:

Calculation of M.D from Median

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

 
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:

Answer:

Calculation of MD. from Median

Median = Size of
N2thitem
= Size of
602thitem
= Size of 39^th item
39^th item lies under the 40^th 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:

 
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:

 
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:

Answer:

 
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:

Answer:

 
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:

Answer:

 
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:

Answer:

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.

Answer:

 
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.

Answer:

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:

Answer:

 
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.

Answer:

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:

Answer:

 
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:

Answer:

 
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.

Answer:

Question 2:

Make a Lorenz curve of the following data:

Answer:

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:

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:

Answer:

In order to calculate range and its coefficient, we must first convert the given inclusive class intervals in exclusive class intervals.

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:

Answer:

 
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

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 25^th 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:

Answer:

Here, N or ∑fx = 100
Median class is given by the size of
N2thitem, i.e.
1002thitem, which is 50^th 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:

Answer:

 
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.

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

 
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.

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 5^th item
Thus, we calculate the deviation of values from 49.

 
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.

Answer:

 
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:

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.

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:

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:

Answer:

 
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:

Answer:

For Income Group I

 
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=∑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:

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

 
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:

Answer:

 
X=∑fX∑f=4170100=41.70Standard Deviation σ=∑fx2∑f=22211100=14.90

Question 21:

Calculate coefficient of variation from the following data:

Answer:

 
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:

Answer:

 
X=∑fm∑f=10000250=40Standard Deviation σ=∑fx2∑f=162600250=25.50Coefficient of Variation =σX×100=25.5040×100 = 63.75