Example Construction of Frequency Distribution
Construct a frequency distribution with the suitable class interval size of marks obtained by 50 students of a class, which are given below:
23, 50, 38, 42, 63, 75, 12, 33, 26, 39, 35, 47, 43, 52, 56, 59, 64, 77, 15, 21, 51, 54, 72, 68, 36, 65, 52, 60, 27, 34, 47, 48, 55, 58, 59, 62, 51, 48, 50, 41, 57, 65, 54, 43, 56, 44, 30, 46, 67, 53
Solution:
Arrange the marks in ascending order as
12, 15, 21, 23, 26, 27, 30, 33, 34, 35, 36, 38, 39, 41, 42, 43, 43, 44, 46, 47, 47, 48, 48, 50, 50, 51, 51, 52, 52, 53, 54, 54, 55, 56, 56, 57, 58, 59, 59, 60, 62, 63, 64, 65, 65, 67, 68, 72, 75, 77
Minimum Value = $$12$$ Maximum = $$77$$
Range = Maximum Value – Minimum Value = $$77 – 12$$ = $$65$$
Number of Classes = $$1 + 3.322\log N$$
Number of Classes = $$1 + 3.322\log 50$$
Number of Classes = $$1 + 3.322(1.69897)$$
Number of Classes = $$1 + 5.64$$ = $$6.64$$ or $$7$$, approximately.
Class Interval Size ($$h$$) = $$\frac{{Range}}{{No.{\text{ }}of{\text{ }}Classes}}$$ = $$\frac{{65}}{7}$$ = $$9.3$$ or $$10$$
|
Marks
Class Limits C.L |
Number of
Students $$f$$ |
Class
Boundary C.B |
Class
Marks $$x$$ |
|
$$10 – 19$$
|
$$2$$
|
$$9.5 – 19.5$$
|
$$\frac{{10 + 19}}{2} = 14.5$$
|
|
$$20 – 29$$
|
$$4$$
|
$$19.5 – 29.5$$
|
$$\frac{{20 + 29}}{2} = 24.5$$
|
|
$$30 – 39$$
|
$$7$$
|
$$29.5 – 39.5$$
|
$$\frac{{30 + 39}}{2} = 34.5$$
|
|
$$40 – 49$$
|
$$10$$
|
$$39.5 – 49.5$$
|
$$\frac{{40 + 49}}{2} = 44.5$$
|
|
$$50 – 59$$
|
$$16$$
|
$$49.5 – 59.5$$
|
$$\frac{{50 + 59}}{2} = 54.5$$
|
|
$$60 – 69$$
|
$$8$$
|
$$59.5 – 69.5$$
|
$$\frac{{60 + 69}}{2} = 64.5$$
|
|
$$70 – 79$$
|
$$3$$
|
$$69.5 – 79.5$$
|
$$\frac{{70 + 79}}{2} = 74.5$$
|
|
$$50$$
|
Note: To find the class boundaries, we take half of the difference between the lower class limit of the 2nd class and the upper class limit of the 1st class$$\frac{{20 – 19}}{2} = \frac{1}{2} = 0.5$$. This value is subtracted from the lower class limit and is added to the upper class limit to get the required class boundaries.
Frequency Distribution by Exclusive Method
|
Class Boundray C.B
|
Frequency
$$f$$ |
|
$$10 – 19$$
|
$$2$$
|
|
$$20 – 29$$
|
$$4$$
|
|
$$30 – 39$$
|
$$7$$
|
|
$$40 – 49$$
|
$$10$$
|
|
$$50 – 59$$
|
$$16$$
|
|
$$60 – 69$$
|
$$8$$
|
|
$$70 – 79$$
|
$$3$$
|
|
|
$$50$$
|
Collins Angwenyi
April 1 @ 1:03 pm
1)The following consists of marks obtained by 50 students of a class.
23, 50, 38, 42, 63, 75, 12, 33, 26, 39, 35, 47, 43, 52, 56, 59, 64, 77, 15,21, 51, 54,72, 68, 36,
65, 52, 60, 27, 34, 47, 48, 55, 58, 59, 62, 51, 48, 50, 41, 57, 65, 54, 43, 56, 44, 30, 46, 67, 53
i.Construct a frequency distribution with the suitable class interval size of marks.
ii.Compute:
a)The arithmetic mean.
b)The mode.
c)The median.
d)Construct a histogram and use it to estimate the mode.
The interquartile range (IQR) and the 6th decile.
2)A committee of 6 people is to be formed from a group of 20 people. The committee has to have the number of women double that of the men. In how many ways can this committee be formed if there are 12 men?
3)For each of the last five years the number of tourists, x thousands, visiting Sackton, and the average weekly sales, £ y thousands, in Sackton Stores were noted. The table shows the results.
Year 2007 2008 2009 2010 2011
x 250 270 264 290 292
y 4.2 3.7 3.2 3.5 3.0
i.Calculate the product moment correlation coefficient r between x and y.
ii.It is required to estimate the average weekly sales at Sackton Stores in a year when the number of tourists is 280 000. Calculate the equation of an appropriate regression line, and use it to find this estimate.
iii.Over a longer period, the value of r is -0.8. The mayor says “This shows that having more tourists causes sales at Sackton Stores to decrease.” Give a reason why this statement is not correct.