statistics to find mean ,median mode

Question

for the frequency distribution given below $v=-\dfrac16,-\dfrac14 ,0,\dfrac12,\dfrac13$
$f= 12, 16, 21, 8, 27$

find mean ,median mode what percent of the population is non negative what percent of the population is negative valued

Answer 1

What we have are various values:

$v=-\dfrac16,-\dfrac14 ,0,\dfrac12,\dfrac13$

and their corresponding frequencies (how many times each of those values $v$ occurs):

$f= 12, 16, 21, 8, 27$

We can write this relationship as ordered pairs $$(v_i, f_i): \left(-\frac 16, 12\right), \left(-\frac 14,16\right), \,(0, 21), \left(\frac 12, 8\right), \left(\frac 13, 27\right)$$

That means there are $12 + 16 + 21 + 8 + 27 = 84$ individuals/scores in the population.

So, to compute the mean, you need to multiply each value $v_i$ by its corresponding frequency $f_i$, and then sum all the $v_i\cdot f_i$, then divide by $84$.

To find the median, you simply need to use the frequencies, $f$ to determine where half of the population/scores is at/below, and the other half at/above, and then find the corresponding value $v$. For example, in this case, the median value is given by $v =0$.

The mode is the most commonly occurring value: here, there are $27$ occurrences of the value $v = \frac 13$, more occurrences of that value than any other value. So mode is given by $v = \frac 13$.

Now for the last two questions: how many individuals in the population of $84$ scored $0, \frac 12,$ or $\frac 13$? Sum the corresponding frequencies, divide by $84$ and multiply by $\%100$.

How many individuals scored $-\frac 16$ or $-\frac 14$? Sum the corresponding frequencies, and divide the sum by $84$, then multiply by $\%100$ to obtain the percentage.

Your last two percentages should add to $\%100$.