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Mathematics

The monthly income of a group of 320 employees in a company is given below :

Monthly Income (in ₹)No. of Employees
6000 - 700020
7000 - 800045
8000 - 900065
9000 - 1000095
10000 - 1100060
11000 - 1200030
12000 - 130005

Draw an ogive of the given distribution on a graph sheet taking 2 cm = ₹1000 on one axis and 2 cm = 50 employees on the other axis. From the graph, determine :

(i) The median wage.

(ii) The number of employees whose income is below ₹8500.

(iii) If the salary of a senior employee is above ₹11500, find the number of senior employees in the company.

(iv) The upper quartile.

Measures of Central Tendency

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Answer

  1. The cumulative frequency table for the given continuous distribution is :
Monthly Income (in ₹)No. of EmployeesCumulative frequency
6000 - 70002020
7000 - 80004565
8000 - 900065130
9000 - 1000095225
10000 - 1100060285
11000 - 1200030315
12000 - 130005320
  1. Take 2 cm along x-axis = 1000 (₹)

  2. Take 1 cm along y-axis = 50 (Employees)

  3. Since, scale on x-axis starts at 6000, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 6000.

  4. Plot the points (7000, 20), (8000, 65), (9000, 130), (10000, 225), (11000, 285), (12000, 315) and (13000, 320) representing upper class limits and the respective cumulative frequencies.
    Also plot the point representing lower limit of the first class i.e. 6000 - 7000.

  5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

The monthly income of a group of 320 employees in a company is given below. Draw an ogive of the given distribution on a graph sheet taking 2 cm = ₹1000 on one axis and 2 cm = 50 employees on the other axis. From the graph, determine the median wage, the number of employees whose income is below ₹8500. If the salary of a senior employee is above ₹11500, find the number of senior employees in the company. The upper quartile. Measures of Central Tendency, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

(i) Here, n (no. of students) = 320.

To find the median :

Let A be the point on y-axis representing frequency = n2=3202\dfrac{n}{2} = \dfrac{320}{2} = 160.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 9300.

Hence, the required median income = ₹9300.

(ii) Let N be the point on x-axis representing income = ₹8500.

Through N, draw a vertical line to meet the ogive at Q. Through Q, draw a horizontal line to meet the y-axis at B. The ordinate of the point B represents 95.

Hence, no. of employees earning less than ₹8500 = 95.

(iii) Let O be the point on x-axis representing income = ₹11500.

Through O, draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 300.

No. of employees earning less than ₹11500 = 300

So, no. of employees earning more than ₹11500 = Total employees - employees earning less than ₹11500 = 320 - 300 = 20.

Hence, there are 20 senior employees in the company.

(iv) To find upper quartile :

Let D be the point on y-axis representing frequency = 3n4=9604\dfrac{3n}{4} = \dfrac{960}{4} = 240.

Through D, draw a horizontal line to meet the ogive at S. Through S, draw a vertical line to meet the x-axis at T. The abscissa of the point T represents ₹10250.

Hence, upper quartile = ₹10250.

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