Question

Marks

10-19

20-29

30-39

40-49

50-59

60-69

Number of students

4

8

a

22

48

b

The mean score of the students was 46.5

         (i) Show that the values of a and b are 12 and 6 respectively.                       (5marks)

         (ii) Estimate the median for the above sample.                                               (3marks)

        (iii) Determine the variance and the standard deviation                                  (4 marks)     

        (iv) Given that 32% of the students failed the examination, what was the cut off points.

(3 marks)

(v)Estimate the mode for this data.                                                                (3marks)

Answer 1

Marks frequency$(f)$ class mark$(x)$ $fx$

10-19 4 14.5 58

20-29 8 24.5 196

30-39 $a$ 34.5 34.5a

40-49 22 44.5 979

50-59 48 54.5 2616

60-69 $b$ 64.5 64.5b

Total frequency is given as, $\sum(f)=100$.

Therefore,

$\sum(f)=4+8+a+22+48+b=100$

$82+a+b=100$

$a+b=18.........(i)$

Given that the mean $\bar{x}=46.5$ and its formula is given by the formula'

$\bar{x}=\sum(fx)/\sum (f)$

Since we already have the quantity $\sum(f)$, let us find $\sum(fx)$ given by,

$\sum(fx)=(58+196+34.5a+979+2616+64.5b)=(3849+34.5a+64.5b)$

Now,

$\bar{x}=(3849+34.5a+64.5b)/100=46.5$

This implies that, $3849+34.5a+64.5b=4650$

Hence,

$34.5a+64.5b=801...............(ii)$

To find the values of unknowns $a$ and $b,$ we solve for equations $(i)$ and $(ii)$ using substitution method.

In Equation $(i)$ the value of $a$ can be given as, $a=18-b$. Substituting this value of $a$ in equation $(ii)$ gives,

$34.5(18-b)+64.5b=801$

$621-34.5b+64.5b=801$

$30b=180$

$b=6$

The value of $a$ is found using the fact that $a=18-b$ .

Therefore,

$a=18-6=12$

Thus, the values for $a$ and $b$ are 12 and 6 respectively as required.

b.

$Table 1$

Marks frequency$(f)$ class mark$(x)$ $fx$ class boundary C.F

10-19 4 14.5 58 9.5-19.5 4

20-29 8 24.5 196 19.5-29.5 12

30-39 12 34.5 414 29.5-39.5 24

40-49 22 44.5 979 39.5-49.5 46

50-59 48 54.5 2616 49.5-59.5 94

60-69 6 64.5 387 59.5-69.5 100

C.F is the cumulative frequency.

To find the median, let us first determine the median class.

The median value is,

Median value=$\sum(f)/2=100/2=50$

The median value is at the $50^{th}$ position therefore the median class is, 50-59.

To find the median, we apply the formula below,

$median=k+((\sum(f)/2)-C.F)*c)/f$ where,

$k$ is the lower class boundary of the median class=49.5

$C.F$ is the cumulative frequency of the class preceding the median class=46

$c$ is the class width of the median class=10

$f$ is the frequency of the median class.

Hence,

$median=49.5+((50-46)*10/48)=49.5+(40/48)=49.5+0.8333=50.33(2\space decimal \space places)$

c.

$Table 2$

Marks $f$ $x$ $fx$ C.F $fx^2$

10-19 4 14.5 58 4 841

20-29 8 24.5 196 12 4802

30-39 12 34.5 414 24 14283

40-49 22 44.5 979 46 43565.5

50-59 48 54.5 2616 94 142572

60-69 6 64.5 387 100 24961.5

C.F is the cumulative frequency.

The variance is given by,

$variance=(1/(\sum(f)-1))*(\sum(fx^2)-(\sum(fx))^2/(\sum(f)))$

$variance=(1/99)*(231025-(21622500/100))$

$variance=(1/99)*(14800)=149.495(3\space decimal\space places)$

Standard deviation(SD) is given as,

$SD=\sqrt{variance}=\sqrt{149.495}=12.23(2\space decimal \space places)$

d.

To find the cut off points, we have to determine the $32^{nd}$ percentile given as,

$P_{32}=k+(((32*\sum(f)/100))-C.F)*c/f$ where,

$k$ is the lower class boundary of the class containing $P_{32}$

$f$ is the frequency of the class containing $P_{32}$

$c$ is the width of the class containing $P_{32}$

$C.F$ tis the cumulative frequency of the class preceding the class containing $P_{32}$

Let us find the class containing $P_{32}$ given as,

$32*100/100=32^{nd}$ position. This position is in the class 40-49 as shown in $Table1$.

We shall use $Table1$ to tackle this question.

Now,

$P_{32}=39.5+(32-24)*10/22=39.5+3.6364=43.1364\approx43$.

Therefore the cut off point is 43.

e.

To find the mode we shall use data in $Table1$

Mode is defined by the formula,

$mode=k+(f_m-f_1*c)/(2*f_m-f_1-f_2)$

where,

$k$ 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

The modal class is the class with the highest frequency=48 thus, modal class is, 50-59

Therefore, $mode=49.5+((48-22)*10/(2*48-6-22)=49.5+(260/68)=49.5+3.8235=53.324(3\space d\space p).$