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TOPICS
Exercise - 14.2

Question-1 :-  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.

Solution :-
```  From given table, to find the class mark (xi) for each inerval, the following relation is used.
Class mark (xi) = (upper class limit + lower class limit)/2
Class size (h) = 10
Taking assumed mean (a) = 30

Age (in years)Number of patients (fi)xidi = xi - 30fidi

5 - 15610-20-120

15 - 251120-10-110

25 - 35213000

35 - 45234010230

45 - 55145020280

55 - 6556030150

TotalΣfi = 80Σfidi = 430

Now, x = a + (Σfidi/Σfi) = 30 + (430/80)
= 30 + 43/8
= 30 + 5.375
= 35.375
= 35.38 (approx.)

Therefore, the mean of this data is 35.38 years.
It represents that on an average, the age of patient admitted to hospital was 35.38 years.
```
```  Here the maximum class frequency is 23, and the class corresponding to this frequency is 35 – 45.
So, the modal class is 35 – 45.
Now, Modal class = 35 - 45,
lower limit(l) of modal class = 23,
class size(h) = 10
frequency(f1) of modal class = 23,
frequency(f0) of modal class = 21,
frequency(f2) of modal class = 14,

Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 35 + [(23-21)/{(2x23)-21-14}] x 10
= 35 + [2/(46-35)] x 10
= 35 + (2/11 x 10)
= 35 + 20/11
= 35 + 1.81
= 36.81

Therefore, Mode is 36.8.
It represents that the age of maximum number of patients admitted in hospital was 36.81 years.
```

Question-2 :-  The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Determine the modal lifetimes of the components.

Solution :-
```  Here the maximum class frequency is 61, and the class corresponding to this frequency is 60 – 80.
So, the modal class is 60 – 80.
Now, Modal class = 60 - 80,
lower limit(l) of modal class = 60,
class size(h) = 20
frequency(f1) of modal class = 61,
frequency(f0) of modal class = 52,
frequency(f2) of modal class = 38,

Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 60 + [(61-52)/{(2x61)-52-38}] x 20
= 60 + [9/(122-90)] x 20
= 60 + (9/32 x 20)
= 60 + 45/8
= 60 + 5.625
= 65.625
= 65.63 (approx.)

Therefore, modal lifetime of electrical coponents is 65.63 hours.
```

Question-3 :-  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 : Solution :-
```  Here the maximum class frequency is 40, and the class corresponding to this frequency is 1500 – 2000.
So, the modal class is 1500 – 2000.
Now, Modal class = 1500 - 2000,
lower limit(l) of modal class = 1500,
class size(h) = 500
frequency(f1) of modal class = 40,
frequency(f0) of modal class = 24,
frequency(f2) of modal class = 33,

Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 1500 + [(40-24)/{(2x40)-24-33}] x 500
= 1500 + [16/(80-57)] x 500
= 1500 + (16/23 x 500)
= 1500 + 8000/23
= 1500 + 347.826
= 1847.826
= 1847.83 (approx.)

Therefore, modal monthly expenditure was Rs. 1847.83.
```
```  From given table, to find the class mark (xi) for each inerval, the following relation is used.
Class mark (xi) = (upper class limit + lower class limit)/2
Class size (h) = 500
Taking assumed mean (a) = 2750

Expenditure (in Rs.)Number of families (fi)xidi = xi - 2750ui = di/500fiui

1000 - 1500241250-1500-3-72

1500 - 2000401750-1000-2-80

2000 - 2500332250-500-1-33

2500 - 3000282750000

3000 - 3500303250500130

3500 - 40002237501000244

4000 - 45001642501500348

4500 - 5000747502000428

TotalΣfi = 200Σfiui = -35

Now, x = a + (Σfiui/Σfi) x h = 2750 + (-35/200) x 500
= 2750 - 175/2
= 2750 - 87.5
= 2662.5

Therefore, the mean monthly expenditure was Rs. 2662.5.
```

Question-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. Solution :-
```  Here the maximum class frequency is 10, and the class corresponding to this frequency is 30 – 35.
So, the modal class is 30 – 35.
Now, Modal class = 30 - 35,
lower limit(l) of modal class = 30,
class size(h) = 5
frequency(f1) of modal class = 10,
frequency(f0) of modal class = 9,
frequency(f2) of modal class = 3,

Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 30 + [(10-9)/{(2x10)-9-3}] x 5
= 30 + [1/(20-12)] x 5
= 30 + (1/8 x 5)
= 30 + 5/8
= 30 + 0.625
= 30.625
= 30.63 (approx.)

Therefore, the mode of this data is 30.63.
It represents that most of the states/U.T have a teacher-student ratio as 30.63.
```
```  From given table, to find the class mark (xi) for each inerval, the following relation is used.
Class mark (xi) = (upper class limit + lower class limit)/2
Class size (h) = 5
Taking assumed mean (a) = 32.5

Expenditure (in Rs.)Number of families (fi)xidi = xi - 32.5ui = di/5fiui

15 - 20317.5-15-3-9

20 - 25822.5-10-2-16

25 - 30927.5-5-1-9

30 - 351032.5000

35 - 40337.5513

40 - 45042.51020

45 - 50047.51530

50 - 55252.52048

TotalΣfi = 35Σfiui = -23

Now, x = a + (Σfiui/Σfi) x h = 32.5 + (-23/35) x 5
= 32.5 - 23/7
= 32.5 - 3.28
= 29.22

Therefore, the mean of data is 29.22.
It represents that on an average, teacher-student ratio was 29.22.
```

Question-5 :-  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.

Solution :-
```  Here the maximum class frequency is 18, and the class corresponding to this frequency is 4000 – 5000.
So, the modal class is 4000 – 5000.
Now, Modal class = 4000 - 5000,
lower limit(l) of modal class = 4000,
class size(h) = 1000
frequency(f1) of modal class = 18,
frequency(f0) of modal class = 4,
frequency(f2) of modal class = 9,

Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 4000 + [(18-4)/{(2x18)-4-9}] x 1000
= 4000 + [14/(36-13)] x 1000
= 4000 + (14/23 x 1000)
= 4000 + 14000/23
= 4000 + 608.695
= 4608.695
= 4608.7 (approx.)

Therefore, the mode of this data is 4608.7 runs.
```

Question-6 :- 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 : Solution :-
```  Here the maximum class frequency is 20, and the class corresponding to this frequency is 40 – 50.
So, the modal class is 40 – 50.
Now, Modal class = 40 - 50,
lower limit(l) of modal class = 40,
class size(h) = 10
frequency(f1) of modal class = 20,
frequency(f0) of modal class = 12,
frequency(f2) of modal class = 11,

Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 40 + [(20-12)/{(2x20)-12-11}] x 10
= 40 + [8/(40-23)] x 10
= 40 + (8/17 x 10)
= 40 + 80/17
= 40 + 4.7
= 44.7

Therefore, the mode of this data is 44.7 cars.
```
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