Answer to Question #95949 in Statistics and Probability for Sana

Question

Prepare a frequency distribution from the following marks related to statistics exam and calculate mean, median and mode.
68,62,58,73,101,90,86,61,84,63,56,72,102,68,83,92,87,60,83,69,57,71,103,57,87,93,88,69,76,70,54,70,104,58,88,94,68,57,64,86,55,88,105,88,89,84,90,74,90,84,82,70,60,60,90,81,91,77,60,83,78,80,75,68,96,82,93,102,99,91,75,90,70,68,94,93,83,101,70,94,77,100,75,69,93,84,95,96,72,100,75,101,80,67,92,85,90,90

Take class intervals of 10.

Answer 1

$54,55,56,57,57,57,58,58,60,60,60,60,61,62,63,$

$64,67,68,68,68,68,68,69,69,69,70,70,70,70,70,$

$71,72,72,73,74,75,75,75,75,76,77,77,78,80,80,$

$81,82,82,83,83,83,83,84,84,84,84,85,86,86,87,$

$87,88,88,88,88,89,90,90,90,90,90,90,90,91,91,$

$92,92,93,93,93,93,94,94,94,95,96,96,99,100,$

$100,101,101,101,102,102,103,104,105$

Take class intervals of 10.

$$Frequency\ Distribution\ Table$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Class & Count & Percentage \\ \hline 50-59 & 8 & 8.16\\ \hdashline 60-69 & 17 & 17.35\\ \hdashline 70-79 & 18 & 18.37\\ \hdashline 80-89 & 23 & 23.47\\ \hdashline 90-99 & 22 & 22.45\\ \hdashline 100-109 & 10 & 10.20\\ \hline Total & 98 & 100 \end{array}$$

$$mean={1 \over n}\displaystyle\sum_{i=1}^nx_i$$

$$mean={1 \over n}(54+55+56+3\cdot57+2\cdot58+4\cdot60+$$ $$+61+62+63+64+67+5\cdot68+3\cdot69+5\cdot70+$$ $$+71+2\cdot72+73+74+4\cdot75+76+2\cdot77+78+$$ $$+2\cdot80+81+2\cdot82+4\cdot83+4\cdot84+85+2\cdot86+$$ $$+2\cdot87+4\cdot88+89+7\cdot90+2\cdot91+2\cdot92+4\cdot93+$$ $$+3\cdot94+95+2\cdot96+99+2\cdot100+3\cdot101+2\cdot102+$$ $$+103+104+105)$$

$$mean\approx80.367347$$

Median: The value that lies in the middle of the data when the data set is ordered. If the data set has an odd number of entries, then the median is the middle data entry. If the data has an even number of entries, then the median is obtained by adding the two numbers in the middle and dividing result by two

To find the median, we place all the numbers in order. The data has an even number of entries

$${98 \over 2}=49$$

We consider 49th variable with value of 83 and 50th variable with value of 83

$$median={83+83 \over 2}=83$$

Mode: The data entry that occurs with the greatest frequency. A data set may have one mode, more than one mode, or no mode. If no entry is repeated the data set has no mode.

We see that the variable with value of 90 occurs most often (7 times)

$$mode=90,\ unimodal$$