Answer to Question #280276 in Statistics and Probability for Kelvin

Question

A sample of 80 students was taken, and these adults were asked about the number hours they spend on academic studies per weeks. The following table gives the frequency distribution.

Studying time.  Number of students

0 - 3.           18

4 - 7.            26

8-11.            22

12 - 15.          11

16 - 19.           3

(a) find the class boundaries midpoints of the third class?

(b) Do all classes have the same width? Of so what is the width?

(c)mean

(d) mode

(e) median

(F) standard deviation

(G) prepare the frequency columns

(H) what percentage of these students that study for 8 hour?

(I) construct the Histogram for this data set and describe the shape of the distribution

(J) construct a relative frequency polygon

Answer 1

$a)$

The third class is 8-11,

The class boundaries are,

Upper class boundary=11+0.5=11.5

Lower class boundary=8-0.5=7.5

The midpoint is,${(8+11)\over2}={19\over2}=9.5$

$b)$

All the classes have the same class boundaries which can be determined as follows.

For the first class, the upper limit is, 3+0.5=3.5

For the second class, the upper limit is, 7+0.5=7.5.

Since the class width is the difference between the upper limits of two consecutive classes, taking the first and the second class, we have that,

$class\space width=(7.5-3.5)=4$

c)

Studying time.  Number of students(f). Midpoint(x). fx

0 - 3.            18. 1.5. 27

4 - 7.             26. 5.5. 143

8-11.             22. 9.5. 209

12 - 15.          11. 13.5. 148.5

16 - 19.            3 . 17.5. 52.5

The mean is given by,

$\bar{x}={\sum fx\over\sum f}={580\over 80}=7.25$

$d)$

Table 1.

Studying time.   f. class boundaries. cf

0 - 3.            18. 0.5-3.5 18

4 - 7.             26. 3.5-7.5 44

8-11.             22. 7.5-11.5 66

12 - 15.           11. 11.5-15.5 77

16 - 19.            3. 15.5-19.5 80

To find the mode, we need to determine the modal class.

The modal class is the class with the highest frequency.

Therefore, the modal class is 4-7

The mode is can therefore determined using the formula below.

$mode=l+{(f_m-f_1)\times c\over2f_m-f_1-f_2}$ where,

$l$ is the lower class boundary of the modal class

$c$ is the class width of the modal class

$f_m$ is the frequency of the modal class

$f_1$ is the frequency of the class preceding the modal class

$f_2$ is the frequency of the class succeeding the modal class.

$mode=3.5+{(26-18)\times4\over(2\times 26)-18-22}=3.5+{32\over12}=6.1667(4dp)$

$e)$

To find the median, we shall use the table in part (d) above.

First, we determine the median class,

$n=80$

The median value is the ${n\over2}^{th}={80\over2}=40^{th}$ position.

The median class is 4-7.

We use the following formula to find the median,

$median=l+{({n\over2}-cf)\times c\over f}$ where,

$l$ is the lower class boundary of the median class

$cf$ is the cumulative frequency of the class preceding the median class.

$f$ is the frequency of the median class

$c$ is the width of the median class.

Therefore,

$median=3.5+{(40-18)\times 4\over26}=3.5+3.3846=6.8846(4dp)$

$f)$

Studying time.  (f). (x). fx x² fx²

0 - 3.            18 1.5. 27 2.25 40.5

4 - 7.             26 5.5. 143 30.25 786.5

8-11.             22. 9.5. 209 90.25 1985.5

12 - 15.          11. 13.5. 148.5 182.25 2004.75

16 - 19.            3 . 17.5. 52. 306.25 918.75

The variance is given as,

$variance={1\over n-1}\times (\sum fx^2-{(\sum fx)^2\over n})= {1\over79}\times(5736-{579.5^2\over 80})={1538.25\over79}=19.4715$ Thus, the standard deviation is,

$standard \space deviation=\sqrt{variance}=\sqrt{19.4715}=4.4126$

$g)$

Number of students is the frequency and the study time is the class interval.

Class interval.  frequency.

0 - 3.            18

4 - 7.            26

8-11.             22

12 - 15.          11

16 - 19.            3

$h)$

To find the  percentage of these students that study for 8 hours, we need to determine the $i^{th}$ percentile whose value is $8$.That is,

$P_i=l+({i\times n\over 100}-cf)\times{c\over f}=8$ where,

$l$ is the lower class boundary of the class containing $P_i$

$f$ is the frequency of the class containing $P_i$

$c$ is the width of the class containing $P_i$

$cf$ is the cumulative frequency of the class preceding the class containing $P_i$

We shall consider the class 8-11.

Now, we need to find the value for $i$ as follows.

$7.5+({i\times 80\over100}-44)\times{4\over22}=8$

$(0.8i-44)\times{4\over 22}=0.5\implies{3.2i\over22}=8.5\implies i=58.4375$

$i=59$ is the cumulative frequency. The frequency for students who study for 8 hours is,

$f=58.4375-44=14.4375$

The percentage of students who study for 8 hours is therefore, ${14.4375\over 80}\times 100=18.05\%(2dp)$

$i)$

Below is the Histogram for the given frequency table.

The scale used is,

x-axis: 1cm=2 hours

y-axis: 1cm=5 students

The Histogram above is Right-skewed since it has a peak that is left of the center and a gradual tapering to the right side of the graph.

 

$j)$

To construct the relative frequency polygon, we need to find the midpoint and the relative frequency for each class as follows.

class midpoint relative frequency

0-3 1.5 (18/80)*100=22.5

4-7 5.5 (26/80)*100=32.5

8-11 9.5 (22/80)*100=27.5

12-15 13.5 (11/80)*100=13.75

16-19 17.5 (3/80)*100=3.75

The relative frequency polygon of Relative frequency against the Midpoint is shown below.