Exercise 9.2

1. Weights of parcels in a transport office are given below.

Find the mean weight of the parcels.

Solution:

Given:

Weights of parcels and their frequencies.

To Find Mean weight of the parcels:

Mean (x̄) = ΣfxΣf / ΣfΣfx = / = kg

Therefore the mean weight of the parcels is 85 kg.

2. Number of families in a village in correspondence with the number of children are given below.

Find the mean number of children per family.

Solution:

Given:

Number of children and their corresponding number of families.

To Find Mean number of children per family:

Mean (x̄) = ΣfxΣf = / = 1.711.61 (approx.)

Therefore the mean number of children per family is approximately 1.71.

3. If the mean of the following frequency distribution is 7.2 find value of 'K'.

Solution:

Given:

Frequency distribution and mean = .

To Find value of K:

Mean (x̄) = ΣfxΣf

7.2 = (260 + 10K250 + 10K) / (40 + K40)

7.2(40 + K) = 260 + 10K

+ = 260 + 10K

288 - = 10K -

= K

K = 28 / 2.8 =

Therefore the value of K is 10.

4. Number of villages with respect to their population as per India census 2011 are given below.

Find the average population in each village.

Solution:

Given:

Population (in thousands) and number of villages.

To Find Average population in each village:

{.reveal(when="blank-7")Mean (x̄) = ΣfxΣf = / = 17.7316.72 (approx.)

Therefore the average population in each village is approximately 17.73 thousand.

5. AFLATOUN social and financial educational program initiated savings program among the high school children in Hyderabad district. Mandal wise savings in a month are given in the following table.

Find arithmetic mean of school wise savings in each mandal. Also find the arithmetic mean of saving of all schools.

Solution:

Given:

Mandal wise savings.

To Find Arithmetic mean of school wise savings in each mandal and arithmetic mean of savings of all schools:

School wise savings in each mandal:

• Amberpet: 21546 =

* Thirumalgiri: 24786 =

* Saidabad: 9755 =

* Khairathabad: 9124 =

* Secundrabad: 6003 =

* Bahadurpura: 75339 =

Total number of schools: 6 + 6 + 5 + 4 + 3 + 9 =

Total amount saved: 2154 + 2478 + 975 + 912 + 600 + 7533 =

Arithmetic mean of savings of all schools: 14652 / 33 = 444455 (approx.)

Therefore:

Arithmetic mean of school wise savings in each mandal: 359, 413, 195, 228, 200, 837.

Arithmetic mean of savings of all schools: 444 (approx.)

6. The heights of boys and girls of IX class of a school are given below.

Compare the heights of the boys and girls. [Hint: Find median heights of boys and girls]

Solution:

Given:

Heights of boys and girls.

To Find:

Compare the heights of boys and girls using median.

Solution:

Boys:

Total number of boys: 2 + 5 + 12 + 10 + 7 + 1 = .

Median position: (37 + 1) / 2 = 19th20th position.

Cumulative frequencies: 2, 7, 19, 29, 36, 37.

Median height of boys: cm.

Girls:

Total number of girls: 1 + 2 + 10 + 5 + 6 + 5 = .

Median position: (29 + 1) / 2 = 15th30th position.

Cumulative frequencies: 1, 3, 13, 18, 24, 29.

Median height of girls: cm.

Therefore the median height of boys and girls is the same, 147 cm.

7. Centuries scored and number of cricketers in the world are given below.

Find the mean, median and mode of the given data.

Solution:

Given:

Centuries scored and number of cricketers.

To Find Mean, median, and mode:

Solution:

Mean: ΣfxΣf = / = 11.1912.20 (approx.)

Median: Median position = ( + 1) / 2 = 7080th position.

Median = centuries.

Mode: Mode = centuries (highest frequency is 56).

Therefore:

Mean: 11.19 (approx.)

Median: 10 centuries

Mode: 5 centuries

8. On the occasion of New year's day a sweet stall prepared sweet packets. Number of sweet packets and cost of each packet are given as follows.

Find the mean, median, and mode of the data.

Solution:

Given:

Cost of sweet packets and number of packets.

To Find Mean, median, and mode:

Solution:

Mean: ΣfxΣf = / =

Median: Median position = ( + 1) / 2 = 75.5th76.5th position.

Median = 7574

Mode: Mode = 5075 (highest frequency is 36).

Therefore:

Mean: 80

Median: 75

Mode: 50

9. The mean (average) weight of three students is 40 kg. One of the students Ranga weighs 46 kg. The other two students, Rahim and Reshma have the same weight. Find Rahim's weight.

Solution:

Given:

Mean weight of three students is 40 kg. Ranga weighs 46 kg. Rahim and Reshma have the same weight.

To Find Rahim's weight:

Solution:

Total weight of three students: 40 × 3 = kg.

Weight of Ranga: kg.

Combined weight of Rahim and Reshma: - 4640 = kg.

Since Rahim and Reshma have the same weight, Rahim's weight = / 2 = kg.

Therfore Rahim's weight is 37 kg.

10. The donations given to an orphanage home by the students of different classes of a secondary school are given below.

Find the mean, median, and mode of the data.

Solution:

Given:

Donations by students of different classes.

To Find Mean, Median, and Mode:

Mean: ΣfxΣf = / = 11.2512.25

Median:

Median position = ( + 1) / 2 = 40.541.5th position.

Median = 1015

Mode:

Mode = 1015 (highest frequency is 20).

Therefore:

Mean: 11.25

Median: 10

Mode: 10

11. There are four unknown numbers. The mean of the first two numbers is 4 and the mean of the first three is 9. The mean of all four numbers is 15. If one of the four numbers is 2, find the other numbers.

Solution:

Given:

Four unknown numbers.

Mean of first two numbers is .

Mean of first three numbers is .

Mean of all four numbers is .

One of the four numbers is .

To Find the other three numbers:

Solution:

Let the four numbers be a, b, c, and d.

Given that one of the numbers is 2. Let a = 23.

Mean of first two numbers is:

(a + ba - b) / 2 = 4

( + b) / 2 = 4

2 + b =

b = 610

Mean of first three numbers is:

(a + b + ca - b + c) / 3 = 9

( + + c) / 3 = 9

+ c =

c =

Mean of all four numbers is:

(a + b + c + d) / 4 = 15

(2 + 6 + + d) / 4 = 15

+ d =

d =

Therefore the other three numbers are 6, 19, and 33.