Sunday, 26 January 2014

CBSE Class 10 - Maths - CH 14 - Statistics (Ex 14.1)

Statistics

(NCERT Ex 14.1)
CBSE Class 10 - Maths - CH 14 - Statistics (Ex 14.1)

Q1: A survey was conducted by a group of students as a part of their environment awareness programme, 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.


Answer:

We use the following relation to find the class mark (xi) for each interval:

Class mark (xi) =

Calculating xi and fixi as follows:

Number of plantsNumber of houses(fi)xifixi
0 -2 1 1 1 ☓ 1 = 1
2 -4 2 3 2 ☓ 3 = 6
4 -6 1 5 1 ☓ 5 = 5
6 -8 5 7 5 ☓ 7 = 35
8 - 10 6 9 6 ☓ 9 = 54
10 -12 2 11 2 ☓ 11 = 22
12 - 14 3 13 3 ☓ 13 = 39
Total 20 162

From the table, it is observed that

Σ fi = 20

Σ fixi = 162

Mean , x =

= = 8.1

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (xi) and fi are small.


Q2: Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages(in Rs)100 - 120120 - 140140 - 160160 - 180180 - 200
Number of workers12148610

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer:



To find the class mark (xi) for each interval, using the following relation:

Class mark (xi) =

Class size (h) = 20

Taking 150 as assumed mean (a), di, ui, and fiui can be calculated as follows:

Daily wages(in Rs)Number of workers(fi)xidi = xi - 150ui = fiui
100 - 12012110-40-2-24
120 - 14014130-20-1-14
140 - 1608150000
160 - 18061702016
180 - 2001019040220
Total50-12

Σ fi = 50

Σ fiui = -12

Mean , x = a +()h

             = 150 + ()20

             = 150 -

             = 150 - 4.8

             = 145.2

∴ the mean daily wage of the workers of the factory = Rs 145.20.

Q3: 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.

Daily pocket allownace(in Rs)11 - 1313 - 1515 - 1717 - 1919 - 2121 - 2323 - 25
Number of workers76913f54

Answer:

To find the class mark (xi) for each interval, using the following relation:

xi =

Given that, mean pocket allowance, x = Rs 18

Taking 18 as assumed mean (a), di and fidi are calculated as:

Daily pocket allowance(in Rs)Number of children(fi)Class mark(xi)di = xi - 18fidi
11 - 13712-6-42
13 - 15614-4-24
15 - 17916-2-18
17 - 19131800
19 - 21f2022f
21 - 23522420
23 - 25424624
TotalΣ fi = 44 + f2f - 40

From the table:

Σ fi = 44 + f

Σ fidi = 2f - 40

x = a +

18 = 18 + ()

0 =

2f - 40 = 0

2f = 40

f = 20

∴ the missing frequency, f, is 20.

Q4: 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.

Number of heart beats per minute65 - 6868 - 7171 - 7474 - 7777 - 8080 - 8383 - 86
Number of woman2438742

Answer:

To find the class mark (xi) for each interval, using the following relation:

xi =

Class size, h, of this data = 3

Considering 75.5 as assumed mean (a), di, ui, fiui are calculated as follows.

Number of heart beats per minuteNumber of woman(fi)xidi = xi - 75.5ui = fiui
65 - 68266.5-9-3-6
68 - 71469.5-6-2-8
71 - 74372.5-3-1-3
74 - 77875.5000
77 - 80778.5317
80 - 83481.5628
83 - 86284.5936
Total304

From the table:

Σ fi = 30

Σ fiui = 4

Mean , x = a +()☓ h

              = 75.5 + 4/30 ☓ 3

              = 75.5 + 0.4 = 75.9

∴ mean heart beats per minute for these women are 75.9 beats per minute.

Q5: 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.

Number of mangoes50 - 5253 - 5556 - 5859 - 6162 - 64
Number of boxes
1511013511525

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer:

Number of mangoesNumber of boxes fi
50 - 5215
53 - 55110
56 - 58135
59 - 61115
62 - 6425

Since the class intervals are not continuous, there is a gap of 1 between two class intervals. Therefore adding    to the upper class limit and subtract   from the lower class limit of each interval.

Obtaining Class mark (xi) by using the following relation.

xi =

Class size (h) of this data = 3

Taking 57 as assumed mean (a), di, ui, fiuare calculated as:

Class intervalfixidi = xi - 57ui = fiui
49.5 - 52.51551-6-2-30
52.5 - 55.511054-3-1-110
55.5 - 58.513557000
58.5 - 61.51156031115
61.5 - 64.525636250
Total40025

From the table:

Σ fi = 400

Σ fiui = 25

Mean , x = a +()☓ h

              = 57 + ()☓ 3

              = 57 + 3/16= 57 + 0.1875

              = 57.1875

              ≈ 57.19

Mean number of mangoes kept in a packing box = 57.19.

Using step deviation method as the values of fi, di are big and also, there is a common multiple between all di.

Q6: The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure(in Rs)100 - 150150 - 200200 - 250250 - 300300 - 350
Number of households451222

Find the mean daily expenditure on food by a suitable method.

Answer:

To find the class mark (xi) for each interval, using the following relation:

xi =

Class size = 50

Taking 225 as assumed mean (a), di, ui, fiui are calculated as:

Daily expenditure(in Rs)fixidi = xi - 225ui = fiui
100 - 1504125-100-2-8
150 - 2005175-50-1-5
200 - 25012225000
250 - 30022755012
300 - 350232510024
Total25-7

From the table:

Σ fi = 25

Σ fiui = -7

Mean , x = a +()☓ h

              = 225 + ()☓ 50

              = 225 - 14

              = 211

Thus the mean daily expenditure on food is Rs 211.

Q7: To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

concentration of SO2 (in ppm)Frequency
0.00 - 0.044
0.04 - 0.089
0.08 - 0.129
0.12 - 0.162
0.16 - 0.204
0.20 - 0.242

Answer:

To find the class mark (xi) for each interval, using the following relation:

xi =

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), calculating di, ui, fiui  as:

concentration of SO2 (in ppm)Frequency(fi)Class mark(xi)di = xi - 0.14ui = fiui
0.00 - 0.0440.02-0.12-3-12
0.04 - 0.0890.06-0.08-2-18
0.08 - 0.1290.10-0.04-1-9
0.12 - 0.1620.14000
0.16 - 0.2040.180.0414
0.20 - 0.2420.220.0824

From the table:

Σ fi = 30

Σ fiui = -31

Mean , x = a +()☓ h

              = 0.14 + ()☓ 0.04

              = 0.14 - 0.04133

              = 0.09867

              ≈ 0.099 ppm

Therefore, mean concentration of SO2 in the air = 0.099 ppm.

Q8: A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days0 - 66 - 1010 - 1414 - 2020 - 2828 - 3838 - 40
Number of students111074431

Answer:

To find the class mark (xi) for each interval, using the following relation:

xi =

Taking 17 as assumed mean (a), calculating di and fidi as:

Number of daysNumber of students(fi)xidi = xi - 17fidi
0 - 6 11 3 - 14 - 154
6 - 10 10 8 - 9 - 90
10 - 14 7 12 - 5 - 35
14 - 20 4 17 0 0
20 - 28 4 24 7 28
28 - 38 3 33 16 48
38 - 40 1 39 22 22
Total 40 - 181


From the table:

Σ fi = 40

Σ fidi = -181

Mean, x = a +

            = 17 + ()

            = 17 - 4.525

            = 12.475

            ≈12.48

Therefore, the mean number of days = 12.48 days for which a student was absent.

Q9: The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate(in %)45 - 5555 - 6565 - 7575 - 8585 - 95
Number of cities3101183

Answer:

To find the class mark (xi) for each interval, using the following relation:

xi =

Class size(h) for this data = 10

Taking 70 as assumed mean (a), calculating di, ui, fiui :

Literacy rate(in %)Number of cities(fi)xidi = xi - 70ui = fiui
45 - 55350-20-2-6
55 - 651060-10-1-10
65 - 751170000
75 - 858801018
85 - 953902026
Total35-2

From the table:

Σ fi = 35

Σ fiui = -2

Mean , x = a +()☓ h

              = 70 + ()☓ 10

              = 70 -

              = 70 -

              = 70 - 0.57

              = 69.43

Therefore, mean literacy rate is 69.43%.

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