10th Maths 14.2
Chapter 14
Statistics
NCERT Class 10th solution of Exercise 14.1
Exercise 14.2
Q1. The following table shows the ages of the patients admitted in a hospital during a year.
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Sol. :
For Mode
maximum class frequency is 23
modal class = 35—45
lower limit l = 35
class size h = 10
frequency of modal class f1=23
frequency just before modal class f0=21
frequency just after modal class f2=14
Mode Z =l+(f1-f02f1-f0-f2)× h
Mode Z =35+(23-212×23-21-14)×10
Mode Z =35+(246-35)×10
Mode Z =35+2011
Mode Z =35+1.82
Mode Z =36.82
For Mean
Mean ˉx=∑fixi∑fi
ˉx=259076
ˉx=34.08
Answer:
Mode = 36.82 years, Mean = 34.08 years.
Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Determine the modal lifetimes of the components.
Sol. :
For Mode
maximum class frequency is 61
modal class = 60—80
lower limit l = 60
class size h = 20
frequency of modal class f1=61
frequency just before modal class f0=52
frequency just after modal class f2=38
Mode Z =l+(f1-f02f1-f0-f2)× h
Mode Z =60+(61-522×61-52-38)×20
Mode Z =60+(9122-90)×20
Mode Z =60+18032
Mode Z =60+5.63
Mode Z =65.63
Answer:
Mode = 65.625 hours.
Q3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
Sol. :
For Mode
maximum class frequency is 40
modal class = 1500—2000
lower limit l = 1500
class size h = 500
frequency of modal class f1=40
frequency just before modal class f0=24
frequency just after modal class f2=33
Mode Z =l+(f1-f02f1-f0-f2)× h
Mode Z =1500+(40-242×40-24-33)×500
Mode Z =1500+(1680-57)×500
Mode Z =1500+800023
Mode Z =1500+347.83
Mode Z =
text{For Mean}
text{Assume mean a = 2750 and h = 3000 - 2500 = 500}
text{Mean }overlinex = a + ((sumf_iu_i)/(sumf_i)) timestext{h}
overline x = 2750 + (-35times500)/200
overline x = 2750 - (175)/2
overline x = 2750 -87.50
overline x = text{₹ 2662.50}
text{Answer:}
text{Modal monthly expenditure = ₹ 1847.83},
text{Mean monthly expenditure = ₹ 2662.5.}
Q4. 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.
Sol. :
text{For Mode}
text{maximum class frequency is 10}
text{modal class = 30—35}
text{lower limit l = 30}
text{class size h = 5}
text{frequency of modal class} f_1 = 10
text{frequency just before modal class} f_0 = 9
text{frequency just after modal class} f_2 = 3
text{Mode Z} = l + ((f_1 - f_0)/(2f_1-f_0-f_2))times text{ h}
text{Mode Z} = 30 + ((10 - 9)/(2 times 10 - 9 - 3)) times 5
text{Mode Z} = 30 + ((1)/(20-12)) times 5
text{Mode Z} = 30 + (5)/(8)
text{Mode Z} = 30 + 0.625
text{Mode Z} = 30.625
text{For Mean}
text{Assume mean a =32.5 and h = 35 - 30 = 5}
text{Mean }overlinex = a + ((sumf_iu_i)/(sumf_i)) timestext{h}
overline x = 32.5 + (-23times5)/35
overline x = 32.5 - 3.3
overline x = 29.2text ( approx.)
text{Answer:}
text{Mode = 30.6, Mean = 29.2.}
Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Find the mode of the data.
Sol. :
text{For Mode}
text{maximum class frequency is 18}
text{modal class = 4000—5000}
text{lower limit l = 4000}
text{class size h = 1000}
text{frequency of modal class} f_1 = 18
text{frequency just before modal class} f_0 = 4
text{frequency just after modal class} f_2 = 9
text{Mode Z} = l + ((f_1 - f_0)/(2f_1-f_0-f_2))times text{ h}
text{Mode Z} = 4000 + ((18 - 4)/(2 times 18 - 4 - 9)) times 1000
text{Mode Z} = 4000 + ((14)/(36-13)) times 1000
text{Mode Z} = 4000 + (14000)/(23)
text{Mode Z} = 4000 + 608.7
text{Mode Z} = 4608.7
text{Answer:}
text{Mode = 4608.7 runs.}
Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
Sol. :
text{For Mdoe}
text{maximum class frequency is 20}
text{modal class = 40—50}
text{lower limit l = 40}
text{class size h = 10}
text{frequency of modal class} f_1 = 20
text{frequency just before modal class} f_0 = 12
text{frequency just after modal class} f_2 = 11
text{Mode Z} = l + ((f_1 - f_0)/(2f_1-f_0-f_2))times text{ h}
text{Mode Z} = 40 + ((20 - 12)/(2 times 20 - 12 - 11)) times 10
text{Mode Z} = 40 + ((8)/(40-23)) times 10
text{Mode Z} = 40 + (80)/(17)
text{Mode Z} = 40 + 4.71
text{Mode Z} = 44.71
text{Answer:}
text{Mode = 44.7 car}
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