TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
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.

Number of plants0 – 22 – 44-66 – 88 – 1010 – 1212-14
Number of houses1215623

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 1
∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{162}{20}\) = 8.1
!! Since fi and xi are of small values we use direct method.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory. (AS4)

Daily wages in Rupees200 – 250250 – 300300 – 350350 – 400400- 450
Number of workers12148610

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 2
Here, the xi are of large numerical values in 250 – 300. So, a = 275.
So we use Assumed Mean method then,
\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
Here, the Assumed mean is taken as 275.
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = 275 + \(\frac{1900}{50}\) = 275 + 38 = 313

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f. (AS4)

Daily pocket allowance(in Rupees)11 – 1313 – 1515 – 1717 – 1919 – 2121 – 2323 – 25
Number of children76913f54

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 3
\(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
\(\overline{\mathrm{x}}\) = 18 (given)
⇒ 18 = \(\frac{752+20 f}{(44+f)}\)
⇒ 18(44 + f) = 752 + 20f ⇒ 20f – 18f = 792 – 752
⇒ 2f = 40
∴ f = \(\frac{40}{2}\) = 20

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 4.
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method. (AS4)

Number of heart beats/minute65-6868-7171-7474-7777-8080-8383-86
Number of women2438742

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 4
\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 75.5 + \(\frac{12}{30}\) = 75.5 + 0.4 = 75.9

Question 5.
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges. (AS4)

Number of oranges10-1415-1920-2425-2930-34
Number of baskets1511013511525

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose ?
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 5
Here, we use step division method where a = 22, h = 5
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
⇒ 22 + \(\frac{25}{400}\) × 5
⇒ 22 + 0.31 = 22.31

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rupees)100-150150-200200-250250-300300-350
Number of house holds451222

Find the mean daily expenditure on food by a suitable method. (AS4)
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 6
Here, a = 225, h = 50
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
= 225 + \(\frac{(-7)}{25}\) × 50 = 225 – 14 = 211
The average daily expenditure on food = ₹ 211

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collect for 30 localities in a certain city and is presented below.

Concentration of SO2 in ppm0.00-0.040.04-0.080.08-0.120.12-0.160.16-0.200.20-0.24
Frequency499242

Find the mean concentration of SO2 in the air. (AS4)
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 7
∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\) = 0.0986666 …….
≅ 0.099 ppm

Question 8.
A class teacher has the following attendence record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term. (AS4)

Number of days35-3838-4141-4444-4747-5050-5353-56
Number of students134471011

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 8
Here, a = 51.5
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 51.5 – \(\frac{99}{40}\)
⇒ 51.5 – 2.475 = 49.025 ≅ 49 days

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. (AS4)

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

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 9
Here a = 70, h = 10
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
⇒ \(\overline{\mathrm{x}}\) = 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\) = 70 – 0.5714
= 69.4285 ≅ 69.43%

Leave a Comment