{"id":119400,"date":"2022-05-04T15:08:28","date_gmt":"2022-05-04T09:38:28","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119400"},"modified":"2022-05-04T15:08:28","modified_gmt":"2022-05-04T09:38:28","slug":"chapter-14-statistics-questions-and-answers-ncert-solutions-for-class-10-maths","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-14-statistics-questions-and-answers-ncert-solutions-for-class-10-maths","title":{"rendered":"Chapter 14 – Statistics Questions and Answers: NCERT Solutions for Class 10 Maths"},"content":{"rendered":"

Exercise 14.1<\/h2>\n

1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
\nNumber of Plants
\n0-2
\n2-4
\n4-6
\n6-8
\n8-10
\n10-12
\n12-14
\nNumber of Houses
\n1
\n2
\n1
\n5
\n6
\n2
\n3
\nWhich method did you use for finding the mean, and why?<\/h2>\n

Solution:
\nIn order to find the mean value, we will use direct method because the numerical value of fiand xiare small.
\nFind the midpoint of the given interval using the formula.
\nMidpoint (xi) = (upper limit + lower limit)\/2
\nNo. of plants
\n(Class interval)
\nNo. of houses
\nFrequency (fi)
\nMid-point (xi)
\nfixi
\n0-2
\n1
\n1
\n1
\n2-4
\n2
\n3
\n6
\n4-6
\n1
\n5
\n5
\n6-8
\n5
\n7
\n35
\n8-10
\n6
\n9
\n54
\n10-12
\n2
\n11
\n22
\n12-14
\n3
\n13
\n39<\/h3>\n

Sum fi= 20<\/h3>\n

Sum fixi= 162
\nThe formula to find the mean is:
\nMean = x\u0304 = \u2211fixi\/\u2211fi
\n= 162\/20
\n= 8.1<\/h3>\n

Therefore, the mean number of plants per house is 8.1<\/h3>\n

2. Consider the following distribution of daily wages of 50 workers of a factory.
\nDaily wages (in Rs.)
\n100-120
\n120-140
\n140-160
\n160-180
\n180-200
\nNumber of workers
\n12
\n14
\n8
\n6
\n10
\nFind the mean daily wages of the workers of the factory by using an appropriate method.<\/h2>\n

Solution:
\nFind the midpoint of the given interval using the formula.
\nMidpoint (xi) = (upper limit + lower limit)\/2
\nIn this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150 and class interval is h = 20.
\nSo, ui= (xi\u2013 A)\/h = ui = (xi\u2013 150)\/20
\nSubstitute and find the values as follows:
\nDaily wages
\n(Class interval)
\nNumber of workers
\nfrequency (fi)
\nMid-point (xi)
\nui= (xi\u2013 150)\/20
\nfiui
\n100-120
\n12
\n110
\n-2
\n-24
\n120-140
\n14
\n130
\n-1
\n-14
\n140-160
\n8
\n150
\n0
\n0
\n160-180
\n6
\n170
\n1
\n6
\n180-200
\n10
\n190
\n2
\n20
\nTotal
\nSum fi= 50<\/h3>\n

Sum fiui= -12
\nSo, the formula to find out the mean is:
\nMean = x\u0304 = A + h\u2211fiui\/\u2211fi=150 + (20 \u00d7 -12\/50) = 150 \u2013 4.8 = 145.20
\nThus, mean daily wage of the workers = Rs. 145.20<\/h3>\n

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
\nDaily Pocket Allowance(in c)
\n11-13
\n13-15
\n15-17
\n17-19
\n19-21
\n21-23
\n23-35
\nNumber of children
\n7
\n6
\n9
\n13
\nf
\n5
\n4<\/h2>\n

Solution:
\nTo find out the missing frequency, use the mean formula.
\nHere, the value of mid-point (xi) meanx\u0304= 18
\nClass interval
\nNumber of children (fi)
\nMid-point (xi)
\nfixi
\n11-13
\n7
\n12
\n84
\n13-15
\n6
\n14
\n84
\n15-17
\n9
\n16
\n144
\n17-19
\n13
\n18 = A
\n234
\n19-21
\nf
\n20
\n20f
\n21-23
\n5
\n22
\n110
\n23-25
\n4
\n24
\n96
\nTotal
\nfi= 44+f<\/h3>\n

Sum fixi= 752+20f
\nThe mean formula is
\nMean = x\u0304 = \u2211fixi\/\u2211fi= (752+20f)\/(44+f)
\nNow substitute the values and equate to find the missing frequency (f)
\n\u21d2 18 = (752+20f)\/(44+f)
\n\u21d2 18(44+f) = (752+20f)
\n\u21d2 792+18f = 752+20f
\n\u21d2 792+18f = 752+20f
\n\u21d2 792 \u2013 752 = 20f \u2013 18f
\n\u21d2 40 = 2f
\n\u21d2 f = 20
\nSo, the missing frequency, f = 20.<\/h3>\n

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
\nNumber of heart beats per minute
\n65-68
\n68-71
\n71-74
\n74-77
\n77-80
\n80-83
\n83-86
\nNumber of women
\n2
\n4
\n3
\n8
\n7
\n4
\n2<\/h2>\n

Solution:
\nFrom the given data, let us assume the mean as A = 75.5
\nxi= (Upper limit + Lower limit)\/2
\nClass size (h) = 3
\nNow, find the uiand fiuias follows:
\nClass Interval
\nNumber of women (fi)
\nMid-point (xi)
\nui= (xi\u2013 75.5)\/h
\nfiui
\n65-68
\n2
\n66.5
\n-3
\n-6
\n68-71
\n4
\n69.5
\n-2
\n-8
\n71-74
\n3
\n72.5
\n-1
\n-3
\n74-77
\n8
\n75.5
\n0
\n0
\n77-80
\n7
\n78.5
\n1
\n7
\n80-83
\n4
\n81.5
\n3
\n8
\n83-86
\n2
\n84.5
\n3
\n6<\/h3>\n

Sum fi= 30<\/h3>\n

Sum fiui= 4
\nMean = x\u0304 = A + h\u2211fiui\/\u2211fi
\n= 75.5 + 3\u00d7(4\/30)
\n= 75.5 + 4\/10
\n= 75.5 + 0.4
\n= 75.9
\nTherefore, the mean heart beats per minute for these women is 75.9<\/h3>\n

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
\nNumber of mangoes
\n50-52
\n53-55
\n56-58
\n59-61
\n62-64
\nNumber of boxes
\n15
\n110
\n135
\n115
\n25
\nFind the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?<\/h2>\n

Solution:
\nSince, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 1
\nHere, assumed mean (A) = 57
\nClass size (h) = 3
\nHere, the step deviation is used because the frequency values are big.
\nClass Interval
\nNumber of boxes (fi)
\nMid-point (xi)
\ndi= xi\u2013 A
\nfidi
\n49.5-52.5
\n15
\n51
\n-6
\n90
\n52.5-55.5
\n110
\n54
\n-3
\n-330
\n55.5-58.5
\n135
\n57 = A
\n0
\n0
\n58.5-61.5
\n115
\n60
\n3
\n345
\n61.5-64.5
\n25
\n63
\n6
\n150<\/h3>\n

Sum fi= 400<\/h3>\n

Sum fidi= 75
\nThe formula to find out the Mean is:
\nMean = x\u0304 = A +h \u2211fidi\/\u2211fi
\n= 57 + 3(75\/400)
\n= 57 + 0.1875
\n= 57.19
\nTherefore, the mean number of mangoes kept in a packing box is 57.19<\/h3>\n

6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
\nDaily expenditure(in c)
\n100-150
\n150-200
\n200-250
\n250-300
\n300-350
\nNumber of households
\n4
\n5
\n12
\n2
\n2<\/h2>\n

Solution:
\nFind the midpoint of the given interval using the formula.
\nMidpoint (xi) = (upper limit + lower limit)\/2
\nLet is assume the mean (A) = 225
\nClass size (h) = 50
\nClass Interval
\nNumber of households (fi)
\nMid-point (xi)
\ndi= xi\u2013 A
\nui= di\/50
\nfiui
\n100-150
\n4
\n125
\n-100
\n-2
\n-8
\n150-200
\n5
\n175
\n-50
\n-1
\n-5
\n200-250
\n12
\n225
\n0
\n0
\n0
\n250-300
\n2
\n275
\n50
\n1
\n2
\n300-350
\n2
\n325
\n100
\n2
\n4<\/h3>\n

Sum fi= 25<\/h3>\n

Sum fiui= -7
\nMean = x\u0304 = A +h\u2211fiui\/\u2211fi
\n= 225+50(-7\/25)
\n= 225-14
\n= 211
\nTherefore, the mean daily expenditure on food is 211<\/h3>\n

7. To find out the concentration of SO2in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
\nConcentration of SO2( in ppm)
\nFrequency
\n0.00 \u2013 0.04
\n4
\n0.04 \u2013 0.08
\n9
\n0.08 \u2013 0.12
\n9
\n0.12 \u2013 0.16
\n2
\n0.16 \u2013 0.20
\n4
\n0.20 \u2013 0.24
\n2
\nFind the mean concentration of SO2in the air.<\/h2>\n

Solution:
\nTo find out the mean, first find the midpoint of the given frequencies as follows:
\nConcentration of SO2(in ppm)
\nFrequency (fi)
\nMid-point (xi)
\nfixi
\n0.00-0.04
\n4
\n0.02
\n0.08
\n0.04-0.08
\n9
\n0.06
\n0.54
\n0.08-0.12
\n9
\n0.10
\n0.90
\n0.12-0.16
\n2
\n0.14
\n0.28
\n0.16-0.20
\n4
\n0.18
\n0.72
\n0.20-0.24
\n2
\n0.20
\n0.40
\nTotal
\nSum fi= 30<\/h3>\n

Sum (fixi) = 2.96
\nThe formula to find out the mean is
\nMean = x\u0304 = \u2211fixi\/\u2211fi
\n= 2.96\/30
\n= 0.099 ppm
\nTherefore, the mean concentration of SO2in air is 0.099 ppm.<\/h3>\n

8. A class teacher has the following absentee record of 40 students of a class for the whole
\nterm. Find the mean number of days a student was absent.
\nNumber of days
\n0-6
\n6-10
\n10-14
\n14-20
\n20-28
\n28-38
\n38-40
\nNumber of students
\n11
\n10
\n7
\n4
\n4
\n3
\n1<\/h2>\n

Solution:
\nFind the midpoint of the given interval using the formula.
\nMidpoint (xi) = (upper limit + lower limit)\/2
\nClass interval
\nFrequency (fi)
\nMid-point (xi)
\nfixi
\n0-6
\n11
\n3
\n33
\n6-10
\n10
\n8
\n80
\n10-14
\n7
\n12
\n84
\n14-20
\n4
\n17
\n68
\n20-28
\n4
\n24
\n96
\n28-38
\n3
\n33
\n99
\n38-40
\n1
\n39
\n39<\/h3>\n

Sumfi= 40<\/h3>\n

Sumfixi= 499
\nThe mean formula is,
\nMean = x\u0304 = \u2211fixi\/\u2211fi
\n= 499\/40
\n= 12.48 days
\nTherefore, the mean number of days a student was absent = 12.48.<\/h3>\n

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
\nliteracy rate.
\nLiteracy rate (in %)
\n45-55
\n55-65
\n65-75
\n75-85
\n85-98
\nNumber of cities
\n3
\n10
\n11
\n8
\n3<\/h2>\n

Solution:
\nFind the midpoint of the given interval using the formula.
\nMidpoint (xi) = (upper limit + lower limit)\/2
\nIn this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class interval is h = 10.
\nSo, ui= (xi-A)\/h = ui= (xi-70)\/10
\nSubstitute and find the values as follows:
\nClass Interval
\nFrequency (fi)
\n(xi)
\ndi=xi\u2013 a
\nui= di\/h
\nfiui
\n45-55
\n3
\n50
\n-20
\n-2
\n-6
\n55-65
\n10
\n60
\n-10
\n-1
\n-10
\n65-75
\n11
\n70
\n0
\n0
\n0
\n75-85
\n8
\n80
\n10
\n1
\n8
\n85-95
\n3
\n90
\n20
\n2
\n6<\/h3>\n

Sumfi= 35<\/h3>\n

Sumfiui = -2
\nSo, Mean = x\u0304 = A+(\u2211fiui\/\u2211fi)\u00d7h
\n= 70+(-2\/35)\u00d710
\n= 69.42
\nTherefore, the mean literacy part = 69.42<\/h3>\n

Exercise 14.2<\/h2>\n

1. The following table shows the ages of the patients admitted in a hospital during a year:
\nAge (in years)
\n5-15
\n15-25
\n25-35
\n35-45
\n45-55
\n55-65
\nNumber of patients
\n6
\n11
\n21
\n23
\n14
\n5
\nFind the mode and the mean of the data given above. Compare and interpret the two
\nmeasures of central tendency.<\/h2>\n

Solution:
\nTo find out the modal class, let us the consider the class interval with high frequency
\nHere, the greatest frequency = 23, so the modal class = 35 \u2013 45,
\nl = 35,
\nclass width (h) = 10,
\nfm= 23,
\nf1= 21 and f2= 14
\nThe formula to find the mode is
\nMode= l+ [(fm-f1)\/(2fm-f1-f2)]\u00d7h
\nSubstitute the values in the formula, we get
\nMode = 35+[(23-21)\/(46-21-14)]\u00d710
\nMode = 35+(20\/11) = 35+1.8
\nMode = 36.8 year
\nSo the mode of the given data = 36.8 year
\nCalculation of Mean:
\nFirst find the midpoint using the formula, xi= (upper limit +lower limit)\/2
\nClass Interval
\nFrequency (fi)
\nMid-point (xi)
\nfixi
\n5-15
\n6
\n10
\n60
\n15-25
\n11
\n20
\n220
\n25-35
\n21
\n30
\n630
\n35-45
\n23
\n40
\n920
\n45-55
\n14
\n50
\n700
\n55-65
\n5
\n60
\n300<\/h3>\n

Sumfi= 80<\/h3>\n

Sumfixi= 2830
\nThe mean formula is
\nMean = x\u0304 = \u2211fixi\/\u2211fi
\n= 2830\/80
\n= 35.37 years
\nTherefore, the mean of the given data = 35.37 years<\/h3>\n

2. The following data gives the information on the observed lifetimes (in hours) of 225
\nelectrical components:
\nLifetime (in hours)
\n0-20
\n20-40
\n40-60
\n60-80
\n80-100
\n100-120
\nFrequency
\n10
\n35
\n52
\n61
\n38
\n29
\nDetermine the modal lifetimes of the components.<\/h2>\n

Solution:
\nFrom the given data the modal class is 60\u201380.
\nl = 60,
\nThe frequencies are:
\nfm= 61, f1= 52, f2= 38 and h = 20
\nThe formula to find the mode is
\nMode= l+ [(fm-f1)\/(2fm-f1-f2)]\u00d7h
\nSubstitute the values in the formula, we get
\nMode =60+[(61-52)\/(122-52-38)]\u00d720
\nMode = 60+((9 x 20)\/32)
\nMode = 60+(45\/8) = 60+ 5.625
\nTherefore, modal lifetime of the components = 65.625 hours.<\/h3>\n

3. The following data gives the distribution of total monthly household expenditure of 200
\nfamilies of a village. Find the modal monthly expenditure of the families. Also, find the
\nmean monthly expenditure:
\nExpenditure
\nNumber of families
\n1000-1500
\n24
\n1500-2000
\n40
\n2000-2500
\n33
\n2500-3000
\n28
\n3000-3500
\n30
\n3500-4000
\n22
\n4000-4500
\n16
\n4500-5000
\n7<\/h2>\n

Solution:
\nGiven data:
\nModal class = 1500-2000,
\nl= 1500,
\nFrequencies:
\nfm= 40 f1= 24, f2= 33 and
\nh = 500
\nMode formula:
\nMode= l+ [(fm-f1)\/(2fm-f1-f2)]\u00d7h
\nSubstitute the values in the formula, we get
\nMode =1500+[(40-24)\/(80-24-33)]\u00d7500
\nMode = 1500+((16\u00d7500)\/23)
\nMode = 1500+(8000\/23) = 1500 + 347.83
\nTherefore, modal monthly expenditure of the families = Rupees 1847.83
\nCalculation for mean:
\nFirst find the midpoint using the formula, xi=(upper limit +lower limit)\/2
\nLet us assume a mean, A be 2750
\nClass Interval
\nfi
\nxi
\ndi = xi \u2013 a
\nui = di\/h
\nfiui
\n1000-1500
\n24
\n1250
\n-1500
\n-3
\n-72
\n1500-2000
\n40
\n1750
\n-1000
\n-2
\n-80
\n2000-2500
\n33
\n2250
\n-500
\n-1
\n-33
\n2500-3000
\n28
\n2750
\n0
\n0
\n0
\n3000-3500
\n30
\n3250
\n500
\n1
\n30
\n3500-4000
\n22
\n3750
\n1000
\n2
\n44
\n4000-4500
\n16
\n4250
\n1500
\n3
\n48
\n4500-5000
\n7
\n4750
\n2000
\n4
\n28<\/h3>\n

fi = 200<\/h3>\n

fiui = -35
\nThe formula to calculate the mean,
\nMean = x\u0304 = a +(\u2211fiui\/\u2211fi)\u00d7h
\nSubstitute the values in the given formula
\n= 2750+(-35\/200)\u00d7500
\n= 2750-87.50
\n= 2662.50
\nSo, the mean monthly expenditure of the families = Rupees 2662.50<\/h3>\n

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures
\nNo of Students per teacher
\nNumber of states \/ U.T
\n15-20
\n3
\n20-25
\n8
\n25-30
\n9
\n30-35
\n10
\n35-40
\n3
\n40-45
\n0
\n45-50
\n0
\n50-55
\n2<\/h2>\n

Solution:
\nGiven data:
\nModal class = 30 \u2013 35,
\nl= 30,
\nClass width (h) = 5,
\nfm= 10, f1= 9 and f2= 3
\nMode Formula:
\nMode= l+ [(fm-f1)\/(2fm-f1-f2)]\u00d7h
\nSubstitute the values in the given formula
\nMode = 30+((10-9)\/(20-9-3))\u00d75
\nMode = 30+(5\/8) = 30+0.625
\nMode = 30.625
\nTherefore, the mode of the given data = 30.625
\nCalculation of mean:
\nFind the midpoint using the formula, xi=(upper limit +lower limit)\/2
\nClass Interval
\nFrequency (fi)
\nMid-point (xi)
\nfixi
\n15-20
\n3
\n17.5
\n52.5
\n20-25
\n8
\n22.5
\n180.0
\n25-30
\n9
\n27.5
\n247.5
\n30-35
\n10
\n32.5
\n325.0
\n35-40
\n3
\n37.5
\n112.5
\n40-45
\n0
\n42.5
\n0
\n45-50
\n0
\n47.5
\n0
\n50-55
\n2
\n52.5
\n105.5<\/h3>\n

Sumfi= 35<\/h3>\n

Sumfixi= 1022.5
\nMean = x\u0304 = \u2211fixi\/\u2211fi
\n= 1022.5\/35
\n= 29.2
\nTherefore, mean = 29.2<\/h3>\n

5. The given distribution shows the number of runs scored by some top batsmenof the world in one- day international cricket matches.
\nRun Scored
\nNumber of Batsman
\n3000-4000
\n4
\n4000-5000
\n18
\n5000-6000
\n9
\n6000-7000
\n7
\n7000-8000
\n6
\n8000-9000
\n3
\n9000-10000
\n1
\n10000-11000
\n1
\nFind the mode of the data.<\/h2>\n

Solution:
\nGiven data:
\nModal class = 4000 \u2013 5000,
\nl = 4000,
\nclass width (h) = 1000,
\nfm= 18, f1= 4 and f2= 9
\nMode Formula:
\nMode= l+ [(fm-f1)\/(2fm-f1-f2)]\u00d7h
\nSubstitute the values
\nMode = 4000+((18-4)\/(36-4-9))\u00d71000
\nMode = 4000+(14000\/23) = 4000+608.695
\nMode = 4608.695
\nMode = 4608.7 (approximately)
\nThus, the mode of the given data is 4608.7 runs<\/h3>\n

6. A student noted the number of cars passing through a spot on a road for 100 periods each of3 minutes and summarized it in the table given below. Find the mode of the data:
\nNumber of cars
\nFrequency
\n0-10
\n7
\n10-20
\n14
\n20-30
\n13
\n30-40
\n12
\n40-50
\n20
\n50-60
\n11
\n60-70
\n15
\n70-80
\n8<\/h2>\n

Solution:
\nGiven Data:
\nModal class = 40 \u2013 50, l = 40,
\nClass width (h) = 10, fm= 20, f1= 12 and f2= 11
\nMode= l+ [(fm-f1)\/(2fm-f1-f2)]\u00d7h
\nSubstitute the values
\nMode = 40+((20-12)\/(40-12-11))\u00d710
\nMode = 40 + (80\/17) = 40 + 4.7 = 44.7
\nThus, the mode of the given data is 44.7 cars<\/h3>\n

Exercise 14.3<\/h2>\n

1. The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
\nMonthly consumption(in units)
\nNo. of customers
\n65-85
\n4
\n85-105
\n5
\n105-125
\n13
\n125-145
\n20
\n145-165
\n14
\n165-185
\n8
\n185-205
\n4<\/h2>\n

Solution:
\nFind the cumulative frequency of the given data as follows:
\nClass Interval
\nFrequency
\nCumulative frequency
\n65-85
\n4
\n4
\n85-105
\n5
\n9
\n105-125
\n13
\n22
\n125-145
\n20
\n42
\n145-165
\n14
\n56
\n165-185
\n8
\n64
\n185-205
\n4
\n68<\/h3>\n

N=68<\/h3>\n

From the table, it is observed that, n = 68 and hencen\/2=34
\nHence, the median class is 125-145 with cumulative frequency = 42
\nWhere,l= 125, n = 68, Cf= 22, f = 20, h = 20
\nMedian is calculated as follows:<\/h3>\n

=125+((34\u221222)\/20) \u00d7 20
\n=125+12 = 137
\nTherefore, median = 137
\nTo calculate the mode:
\nModal class = 125-145,
\nf1=20, f0=13, f2=14& h = 20
\nMode formula:
\nMode= l+ [(f1-f0)\/(2f1-f0-f2)]\u00d7h
\nMode = 125 + ((20-13)\/(40-13-14))\u00d720
\n=125+(140\/13)
\n=125+10.77
\n=135.77
\nTherefore, mode = 135.77
\nCalculate the Mean:
\nClass Interval
\nfi
\nxi
\ndi=xi-a
\nui=di\/h
\nfiui
\n65-85
\n4
\n75
\n-60
\n-3
\n-12
\n85-105
\n5
\n95
\n-40
\n-2
\n-10
\n105-125
\n13
\n115
\n-20
\n-1
\n-13
\n125-145
\n20
\n135
\n0
\n0
\n0
\n145-165
\n14
\n155
\n20
\n1
\n14
\n165-185
\n8
\n175
\n40
\n2
\n16
\n185-205
\n4
\n195
\n60
\n3
\n12<\/h3>\n

Sumfi= 68<\/h3>\n

Sumfiui= 7
\nx\u0304 =a+h \u2211fiui\/\u2211fi=135+20(7\/68)
\nMean=137.05
\nIn this case, mean, median and mode are more\/less equal in this distribution.<\/h3>\n

2. If the median of a distribution given below is 28.5 then, find the value of x & y.
\nClass Interval
\nFrequency
\n0-10
\n5
\n10-20
\nx
\n20-30
\n20
\n30-40
\n15
\n40-50
\ny
\n50-60
\n5
\nTotal
\n60<\/h2>\n

Solution:
\nGiven data, n = 60
\nMedian of the given data = 28.5
\nWhere, n\/2= 30
\nMedian class is 20 \u2013 30 with a cumulative frequency = 25+x
\nLower limit of median class,l= 20,
\nCf= 5+x,
\nf = 20 & h = 10<\/h3>\n

Substitute the values
\n28.5=20+((30\u22125\u2212x)\/20) \u00d7 10
\n8.5 = (25 \u2013 x)\/2
\n17 = 25-x
\nTherefore, x =8
\nNow, from cumulative frequency, we can identify the value of x + y as follows:
\nSince,
\n60=5+20+15+5+x+y
\nNow, substitute the value of x, to find y
\n60 = 5+20+15+5+8+y
\ny = 60-53
\ny = 7
\nTherefore, the value of x = 8 and y = 7.<\/h3>\n

3. The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to thepersons whose age is 18 years onwards but less than the 60 years.
\nAge (in years)
\nNumber of policy holder
\nBelow 20
\n2
\nBelow 25
\n6
\nBelow 30
\n24
\nBelow 35
\n45
\nBelow 40
\n78
\nBelow 45
\n89
\nBelow 50
\n92
\nBelow 55
\n98
\nBelow 60
\n100<\/h2>\n

Solution:
\nClass interval
\nFrequency
\nCumulative frequency
\n15-20
\n2
\n2
\n20-25
\n4
\n6
\n25-30
\n18
\n24
\n30-35
\n21
\n45
\n35-40
\n33
\n78
\n40-45
\n11
\n89
\n45-50
\n3
\n92
\n50-55
\n6
\n98
\n55-60
\n2
\n100
\nGiven data: n = 100 andn\/2= 50
\nMedian class = 35-45
\nThen,l= 35, cf= 45, f = 33 & h = 5<\/h3>\n

Median = 35+((50-45)\/33) \u00d7 5
\n= 35 + (5\/33)5
\n= 35.75
\nTherefore, the median age = 35.75 years.<\/h3>\n

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:
\nLength (in mm)
\nNumber of leaves
\n118-126
\n3
\n127-135
\n5
\n136-144
\n9
\n145-153
\n12
\n154-162
\n5
\n163-171
\n4
\n172-180
\n2
\nFind the median length of leaves.<\/h2>\n

Solution:
\nSince the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.
\nClass Interval
\nFrequency
\nCumulative frequency
\n117.5-126.5
\n3
\n3
\n126.5-135.5
\n5
\n8
\n135.5-144.5
\n9
\n17
\n144.5-153.5
\n12
\n29
\n153.5-162.5
\n5
\n34
\n162.5-171.5
\n4
\n38
\n171.5-180.5
\n2
\n40
\nSo, the data obtained are:
\nn = 40 andn\/2= 20
\nMedian class = 144.5-153.5
\nthen,l= 144.5,
\ncf = 17, f = 12 & h = 9<\/h3>\n

Median = 144.5+((20-17)\/12)\u00d79
\n= 144.5+(9\/4)
\n= 146.75mm
\nTherefore, the median length of the leaves = 146.75 mm.<\/h3>\n

5. The following table gives the distribution of a life time of 400 neon lamps.
\nLifetime (in hours)
\nNumber of lamps
\n1500-2000
\n14
\n2000-2500
\n56
\n2500-3000
\n60
\n3000-3500
\n86
\n3500-4000
\n74
\n4000-4500
\n62
\n4500-5000
\n48
\nFind the median lifetime of a lamp.<\/h2>\n

Solution:
\nClass Interval
\nFrequency
\nCumulative
\n1500-2000
\n14
\n14
\n2000-2500
\n56
\n70
\n2500-3000
\n60
\n130
\n3000-3500
\n86
\n216
\n3500-4000
\n74
\n290
\n4000-4500
\n62
\n352
\n4500-5000
\n48
\n400
\nData:
\nn = 400 &n\/2= 200
\nMedian class = 3000 \u2013 3500
\nTherefore,l= 3000, Cf= 130,
\nf = 86 & h = 500<\/h3>\n

Median = 3000 + ((200-130)\/86) \u00d7 500
\n= 3000 + (35000\/86)
\n= 3000 + 406.97
\n= 3406.97
\nTherefore, the median life time of the lamps = 3406.97 hours<\/h3>\n

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:
\nNumber of letters
\n1-4
\n4-7
\n7-10
\n10-13
\n13-16
\n16-19
\nNumber of surnames
\n6
\n30
\n40
\n16
\n4
\n4
\nDetermine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.
\nSolution:
\nTo calculate median:
\nClass Interval
\nFrequency
\nCumulative Frequency
\n1-4
\n6
\n6
\n4-7
\n30
\n36
\n7-10
\n40
\n76
\n10-13
\n16
\n92
\n13-16
\n4
\n96
\n16-19
\n4
\n100
\nGiven:
\nn = 100 &n\/2= 50
\nMedian class = 7-10
\nTherefore,l= 7, Cf= 36, f = 40 & h = 3<\/h2>\n

\"\"<\/h3>\n

Median = 7+((50-36)\/40) \u00d7 3
\nMedian = 7+42\/40
\nMedian=8.05
\nCalculate the Mode:
\nModal class = 7-10,
\nWhere,l= 7, f1= 40, f0= 30, f2= 16 & h = 3<\/h3>\n

\"\"<\/h3>\n

Mode = 7+((40-30)\/(2\u00d740-30-16)) \u00d7 3
\n= 7+(30\/34)
\n= 7.88
\nTherefore mode = 7.88
\nCalculate the Mean:
\nClass Interval
\nfi
\nxi
\nfixi
\n1-4
\n6
\n2.5
\n15
\n4-7
\n30
\n5.5
\n165
\n7-10
\n40
\n8.5
\n340
\n10-13
\n16
\n11.5
\n184
\n13-16
\n4
\n14.5
\n51
\n16-19
\n4
\n17.5
\n70<\/h3>\n

Sum fi= 100<\/h3>\n

Sum fixi= 825
\nMean = x\u0304 = \u2211fixi\/\u2211fi
\nMean = 825\/100 = 8.25
\nTherefore, mean = 8.25<\/h3>\n

7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.
\nWeight(in kg)
\n40-45
\n45-50
\n50-55
\n55-60
\n60-65
\n65-70
\n70-75
\nNumber of students
\n2
\n3
\n8
\n6
\n6
\n3
\n2<\/h2>\n

Solution:
\nClass Interval
\nFrequency
\nCumulative frequency
\n40-45
\n2
\n2
\n45-50
\n3
\n5
\n50-55
\n8
\n13
\n55-60
\n6
\n19
\n60-65
\n6
\n25
\n65-70
\n3
\n28
\n70-75
\n2
\n30
\nGiven: n = 30 andn\/2= 15
\nMedian class = 55-60
\nl = 55, Cf= 13, f = 6 & h = 5<\/h3>\n

Median = 55+((15-13)\/6)\u00d75
\nMedian=55 + (10\/6) = 55+1.666
\nMedian =56.67
\nTherefore, the median weight of the students = 56.67<\/h3>\n

Exercise 14.4<\/h2>\n

1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
\nDaily income in Rupees
\n100-120
\n120-140
\n140-160
\n160-180
\n180-200
\nNumber of workers
\n12
\n14
\n8
\n6
\n10<\/h2>\n

Solution
\nConvert the given distribution table to a less than type cumulative frequency distribution, and we get
\nDaily income
\nFrequency
\nCumulative Frequency
\nLess than 120
\n12
\n12
\nLess than 140
\n14
\n26
\nLess than 160
\n8
\n34
\nLess than 180
\n6
\n40
\nLess than 200
\n10
\n50
\nFrom the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50)on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve<\/h3>\n

\"\"<\/p>\n

2.During the medical check-up of 35 students of a class, their weights were recorded asfollows:
\nWeight in kg
\nNumber of students
\nLess than 38
\n0
\nLess than 40
\n3
\nLess than 42
\n5
\nLess than 44
\n9
\nLess than 46
\n14
\nLess than 48
\n28
\nLess than 50
\n32
\nLess than 52
\n35
\nDraw a less than type ogive for the given data. Hence obtain the median weight from the graph and verifythe result by using the formula.<\/h2>\n

Solution:
\nFrom the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35)on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.<\/h3>\n

\"\"<\/p>\n

Locate the point 17.5 on the y-axis and draw a line parallelto the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the mode by making a table.
\nClass interval
\nNumber of students(Frequency)
\nCumulative Frequency
\nLess than 38
\n0
\n0
\nLess than 40
\n3-0=3
\n3
\nLess than 42
\n5-3=2
\n5
\nLess than 44
\n9-5=4
\n9
\nLess than 46
\n14-9=5
\n14
\nLess than 48
\n28-14=14
\n28
\nLess than 50
\n32-28=4
\n32
\nLess than 52
\n35-22=3
\n35
\nThe class 46 \u2013 48 has the maximum frequency, therefore, this is modal class
\nHere,l= 46, h = 2,f1= 14,f0= 5 andf2= 4
\nThe mode formula is given as:
\nNow, Mode =<\/h3>\n

\"\"<\/h3>\n

= 46 + 0.95 = 46.95
\nThus, mode is verified.<\/h3>\n

3. The following tables gives production yield per hectare of wheat of 100 farms of a village.
\nProduction Yield
\n50-55
\n55-60
\n60-65
\n65-70
\n70-75
\n75-80
\nNumber of farms
\n2
\n8
\n12
\n24
\n38
\n16
\nChange the distribution to a more than type distribution and draw its ogive.<\/h2>\n

Solution:
\nConverting the given distribution to a more than type distribution, we get
\nProduction Yield (kg\/ha)
\nNumber of farms
\nMore than or equal to 50
\n100
\nMore than or equal to 55
\n100-2 = 98
\nMore than or equal to 60
\n98-8= 90
\nMore than or equal to 65
\n90-12=78
\nMore than or equal to 70
\n78-24=54
\nMore than or equal to 75
\n54-38 =16
\nFrom the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on
\nthis graph paper. The graph obtained is known as more than type ogive curve.<\/h3>\n

\"\"<\/p>\n","protected":false},"excerpt":{"rendered":"

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