class 10 Maths ncert book Question No 1 ,2 Exercise 14.4
Answers
Answer:1 The following distribution gives the daily income of 50 workers if a factory.
48
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Sol. Converting the given distribution to a less than type cumulative frequency distribution, we get
49
Let us now plot the points corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and join them by a free hand smooth curve.
213 The curve thus obtained is less than type ogive.
Q.2 During the medical check-up of 35 students of a class, their weights were recorded as follows :
50
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Sol. Here the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals. To represent the data in the table graphically, we mark the upper limits on the class intervals on x-axis and
their corresponding cumulative frequencies on the y-axis, choosing a convenient scale. Let us now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them by a free hand smooth curve.
215
The curve thus obtained is less than type ogive To locate n2=352= 17.5 on the y-axis. From this point draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis gives the median of the data. In this case it is 46.95. Let us make the following table in order to find mode by using the formula
51
The class 46 – 48 has the maximum frequency, therefore, this is modal class
Here, ℓ = 46, h = 2, f1 = 14, f0 = 5 and f2 = 4
Now, Mode = ℓ+(f1−f02f1−f0−f2)×h
= 46+14−528−5−4×2 = 46+919×2
= 46 + 0.95 = 46.95
Thus, mode is verified.
Step-by-step explanation: