GSEB Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.4

 

Question 1. The following distribution gives the daily income of 50 workers of a factory. Convert the distribution above to a less than type cumulative frequency distribution, and draw it so give.

Answer: The less than type cumulative frequency distribution is shown in the table below. To draw the ogive, plot the upper class limits on the x-axis and the cumulative frequencies on the y-axis. Connect these plotted points with a smooth curve.In simple words: We create a new table that shows how many workers earn "less than" a certain amount each day. Then, we use these numbers to draw a smooth curve graph called an ogive. The graph goes up as the income limits increase, showing the total count of workers below each limit.

Exam Tip: Always remember that for a "less than" ogive, you plot the cumulative frequencies against the upper limits of the class intervals.

 

Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows: 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.

Answer: First, we construct the cumulative frequency distribution table from the given data. This table helps us determine the class intervals and their corresponding frequencies.Here, \( n = 35 \). So, \( \frac {n}{2} = \frac {35}{2} = 17.5 \). To obtain the median weight from the graph, locate 17.5 on the y-axis. From this point, you draw a line parallel to the x-axis, which cuts the ogive curve at a certain point. From this intersection point, you then draw a perpendicular line down to the x-axis. The point where this perpendicular intersects the x-axis gives you the median of the data, which is 46.5 kg. We can also verify this result using the formula for grouped frequency distribution. The value \( \frac {n}{2} = 17.5 \) lies in the class 46 - 48. So, 46-48 is the median class. Therefore, \( l = 46 \), \( h = 2 \), \( f = 14 \), \( c.f = 14 \). Median (by using formula) \( = l + [ \frac { \frac{n}{2} - c.f }{ f } ] \times h \) \( = 46 + [ \frac { 17.5 - 14 }{ 14 } ] \times 2 \) \( = 46 + [ \frac { 3.5 }{ 14 } ] \times 2 \) \( = 46 + [ \frac { 1 }{ 4 } ] \times 2 \) \( = 46 + \frac { 2 }{ 4 } \) \( = 46 + \frac { 1 }{ 2 } \) \( = 46 + 0.5 \) \( = 46.5 \) kg. Verification: We find that the median weight obtained from the graph is the same as the median weight obtained by using the formula.In simple words: We first make a table of total students up to each weight (cumulative frequency). Then, we draw a graph from this table. To find the middle weight (median) from the graph, we locate half the total students on the vertical axis, draw across to the curve, and then down to the horizontal axis. This gives 46.5 kg. We confirm this by using a math formula, which also gives the same result.

Exam Tip: When finding the median from an ogive, always ensure you correctly identify \( \frac{n}{2} \) on the cumulative frequency axis before drawing the lines.

 

Question 3. The following table gives production yield per hectare of wheat of 100 farms of a village. Change the distribution to a more than type distribution, and draw its ogive.

Answer: The "more than type" cumulative frequency distribution is presented in the table below. To draw the ogive, we plot the lower class limits on the x-axis and their corresponding cumulative frequencies on the y-axis. The points are then joined with a smooth curve.In simple words: We change the given table into one that shows how many farms produce "more than" a specific amount of wheat. Then, we use these new numbers to create a graph called an ogive. This graph will start high and go down as the production limits increase, showing the number of farms above each limit.

Exam Tip: For a "more than" ogive, always plot the cumulative frequencies against the lower limits of the class intervals.