Answer to Question #206765 in Statistics and Probability for Boli

Question

65,98,55,62,79,59,51,90,72,56,70,62,66,80,94,79,63,73,71,85

Organize the number of oil changes into a frequency distribution

Answer 1

$51,55,56,59,62,62,63,65,66,70,$

$71,72,73,79,79,80,85,90,94,98$

Sample size: $n=20.$

Maximum value $Max= 98.$

Minimum value $Min= 51.$

Measurement $MU=1$

Amount of classes

$k=1+3.322\log_{10}(n)$

$=1+3.322\log_{10}(20)\approx5.322$

We must approach to the upper integer: $k=6.$

Class interval

$CI=\dfrac{98-51+1}{6}=8$

$\def\arraystretch{1.5} \begin{array}{c:c:c} Class & Class & Frequency \\ interval & boundary & f\\ \hline 51-58 & 50.5-58.5 & 3 \\ \hdashline 59-66 & 58.5-66.5 & 6 \\ \hdashline 67-74 & 66.5-74.5 & 4 \\ \hdashline 75-82 & 74.5-82.5 & 3 \\ \hdashline 83-90 & 82.5-90.5 & 2 \\ \hdashline 91-98 & 90.5-98.5 & 2 \\ \hline \end{array}$