Class 10 Mathematics

Statistics

Exercise 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year.

Age (in years)5-1515-2525-3535-4545-5555-65
Number of patients6112123145

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution: Solution: Modal class `= 35 – 45`, `l = 35`, `h = 10`, `f_1 = 23`, `f_0 = 21` and `f_2 = 14`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=35+(23-21)/(2xx23-21-14)xx10`

`=35+2/11xx10=36.8`


Calculations for Mean:
Class Intervalfixifixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
Σ fi = 80Σ fixi = 2830

`x=(Σf_i\x_i)/(Σf_i)=(2830)/(80)=35.37`

The mode of the data shows that maximum number of patients is in the age group of 26.8, while average age of all the patients is 35.37.

2. The following data gives the information on the observed lifetime (in hours) of 225 electrical components:

Lifetime (in hours)0-2020-4040-6060-8080-100100-120
Frequency103552613829

Determine the modal lifetimes of the components.

Solution: Modal class `= 60-80`, `l = 60`, `f_1 = 61`, `f_0 = 52`, `f_2 = 38` and `h = 20`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=60+(61-52)/(2xx61-52-38)xx20`

`=60+9/32xx20=65.62`


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.

ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007

Solution: Modal class `= 1500-2000`, `l = 1500`, `f_1 = 40`, `f_0 = 24`, `f_2 = 33` and `h = 500`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=1500+(40-24)/(2xx40-24-33)xx500`

`=1500+16/23xx500=1847.82`

Calculations for mean:
Class Intervalfixidi = xi - aui = di/hfiui
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
Σ fi = 200Σ fiui = -35

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=2750-(35)/(200)xx500`

`=2750-87.5=2662.50`


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.

Number of students per teacherNumber of states/UT
15-203
20-258
25-309
30-3510
35-403
40-450
45-500
50-552

Solution: Modal class `= 30-35`, `l = 30`, `f_1 = 10`, `f_0 = 9`, `f_2 = 3` and `h = 5`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=30+(10-9)/(2xx10-9-3)xx5`

`=30+1/8xx5=30.625`

Calculation for mean:
Class Intervalfixidi = xi - aui = di/hfiui
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
Σ fi = 35Σ fiui = -23

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=32.5-23/35xx5`

`=32.5-22/7=29.22`

The mode shows that maximum number of states has 30-35 students per teacher. The mean shows that average ratio of students per teacher is 29.22

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scoredNumber of batsmen
3000-40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-10001
10000-110001

Find the mode of the data.

Solution: Modal class `= 4000-5000`, `l = 4000`, `f_1 = 18`, `f_0 = 4`, `f_2 = 9` and `h = 1000`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=4000+(18-4)/(2xx18-4-9)xx1000`

`=4000+14/23xx1000=4608.70`

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data.

Number of cars0-1010-2020-3030-4040-5050-6060-7070-80
Frequency71413122011158

Solution: Modal class `= 40 – 50`, `l = 40`, `f_1 = 20`, `f_0 = 12`, `f_2 = 11` and `h = 10`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=40+(20-12)/(2xx20-12-11)xx10`

`=40+8/17xx10=44.70`


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Exercise 1

Exercise 3