Answer to Question #255260 in Statistics and Probability for Tonnie

Question

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

a.

To answer this question, let us summarize our data first by defining the class marks(x), frequency(f), class boundaries and product(fx), of frequency(f) and class mark(x) for each class in order to find the values of $a$ and $b$.

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

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

We can find the values of $a$ and $b$ by forming two equations, the mean$(\bar{x})$ and the total frequency as shown below.

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

$82+a+b=100$

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

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,

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

The two equations can be solved by substitution method to get values for $a$ and $b$.

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 given.

The parts b and d and e of the question will use the table below in order for us to determine the required measure.

$Table 1$

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.

b.

For us to determine the median, we first find the median value then we can use it to get the median class as shown below.

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.

We apply the formula below to find the median.

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

$l$ 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.

To find variance and standard deviation, we shall use $Table2$ and apply the formula required.

$Table 2$

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

We apply the formula below to find the variance.

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

Therefore, variance is,

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

variance=(1/99)∗(14800)=149.495(3 decimal 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,

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

$l$ 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,

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

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

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

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