Mathematics Part II Solutions Solutions for Class 10 Math Chapter 5 Statistics are provided here with simple step-by-step explanations. These solutions for Statistics are extremely popular among Class 10 students for Math Statistics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 10 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

The table below classifies the number of days of a month according to the amount of rainfall received in a certain locality.

Find the mean daily rainfall during this month.

 Rainfall (mm) Number of days 54 56 58 55 50 47 44 41 3 5 6 3 2 4 5 2

We can make the complete table as follows:

 Rainfall  (mm) Number of Days Total Rainfall (mm) 54 3 162 56 5 280 58 6 348 55 3 165 50 2 100 47 4 188 44 5 220 41 2 82 Total 30 1545

Mean daily rainfall =

#### Question 1:

The table below classifies the members of a committee according to their ages.

 Age Number of Members 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60 6 14 16 22 5 4 3

Calculate the mean age of the members of this committee.

We can make the complete table as follows:

 Age Number of Member Class Average Total Age 25 − 30 6 27.5 165 30 − 35 14 32.5 455 35 − 40 16 37.5 600 40 − 45 22 42.5 935 45 − 50 5 47.5 237.5 50 − 55 4 52.5 210 55 − 60 3 57.5 172.5 Total 70 2775

Mean age of the members of the committee =

#### Question 2:

The table below shows the number of students in class 10 of a school, classified according to their heights:

 Height (cm) Number of Students 120 − 125 125 − 130 130 − 135 135 − 140 140 − 145 145 − 150 150 − 155 155 − 160 19 36 23 23 43 21 23 12

Calculate the mean height

We can make the complete table as follows:

 Height  (cm) Number of Student Class Average Total Height 120 − 125 19 122.5 2327.5 125 − 130 36 127.5 4590.0 130 − 135 23 132.5 3047.5 135 − 140 23 137.5 3162.5 140 − 145 43 142.5 6127.5 145 − 150 21 147.5 3097.5 150 − 155 23 152.5 3507.5 155 − 160 12 157.5 1890.0 Total 200 27750

Mean height =

#### Question 1:

The table below classifies according to weight, the infants born during a week in a hospital:

 Weight (kg) Number of Infants 2,500 2,600 2,750 2,800 3,000 3,150 3,250 3,300 3,500 4 6 8 10 12 10 8 7 5

Find the median weight.

The given data can be shown as:

 Weight  (kg) Number of Infant Up to 2.500 4 Up to 2.600 10 Up to 2.750 18 Up to 2.800 28 Up to 3.000 40 Up to 3.150 50 Up to 3.250 58 Up to 3.300 65 Up to 3.500 70

Here, we need to find the weight of infant.

From the table it can be inferred that the weight of the 29th infant to the 40th infant is 3.000 kg.

So, the weight of the 35th infant is 3.000 kg.

Thus, the median weight of the infants born during the week in the hospital is 3.000 kg.

#### Question 1:

The table below shows the number of employees of an office, classified according to the income-tax paid by them.

 Income Tax (Rupees) Number of Employees 1000 − 2000 2000 − 3000 3000 − 4000 4000 − 5000 5000 − 6000 6000 − 7000 7000 − 8000 8000 − 9000 8 10 15 18 22 8 6 3

Compute the median income-tax.

The given data can be shown as:

 Income Tax (Rupees) Number of Employee Below 2000 8 Below 3000 18 Below 4000 33 Below 5000 51 Below 6000 73 Below 7000 81 Below 8000 87 Below 9000 90

Let us tabulate the numbers 2000, 3000, … in the first row and 8, 18, … in the second row.

 x 2000 3000 4000 5000 6000 7000 8000 9000 y 8 18 33 51 73 81 87 90

We know that the change in y is proportional to the corresponding change in x.

We have to find that value of x for which y =

The number 45 is between the numbers y = 33 and y = 51.

The corresponding numbers are x = 4000 and x = 5000.

Thus, we must have:

Thus, the median income tax of the employees of the office is Rs.4666.67.

#### Question 2:

The table below classifies the candidates who took and examination, according to the marks scored by them:

 Marks Number of Candidates 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100 44 40 35 20 12 10 8 6 4 1

Find the median mark.

Given data can be shown as:

 Marks Number of candidate Below 10 44 Below 20 84 Below 30 119 Below 40 139 Below 50 151 Below 60 161 Below 70 169 Below 80 175 Below 90 179 Below 100 180

Let us tabulate the numbers 10, 20, … in the first row and 44, 84, … in the second row.

 x 10 20 30 40 50 60 70 80 90 100 y 44 84 119 139 151 161 169 175 179 180

We know that the change in y is proportional to the corresponding change in x.

We have to find that value of x for which y =

The number 90 is between the numbers y = 84 and y = 119.

The corresponding numbers are x = 20 and x = 30.

Thus, we must have:

Thus, the median marks of the candidates who took an examination is 21.71.

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