3-10u To find the minimum value in a list [OPTN]-[LIST]-[Min]K1(LIST) 6( g) 1(Min)6( g) 6( g) 1(List)
)wExample To find the minimum value in List 1 (36, 16, 58, 46, 56)AK1(LIST)6( g) 1(Min)6( g) 6( g) 1(List) b)wu To find which of two lists contains the greatest value [OPTN]-[LIST]-[Max]K1(LIST) 6( g) 2(Max) 6( g) 6( g) 1(List)
,1(List)
)w• The two lists must contain the same number of data items. If they don’t, an error occurs.• The result of this operation is stored in ListAns Memory.Example To find whether List 1 (75, 16, 98, 46, 56) or List 2 (35, 59, 58, 72, 67)contains the greatest valueK1(LIST)6( g) 2(Max)6( g) 6( g) 1(List) b,1(List)c)wu To calculate the mean of data items [OPTN]-[LIST]-[Mean]K1(LIST) 6( g) 3(Mean) 6( g) 6( g) 1(List)
)wExample To calculate the mean of data items in List 1 (36, 16, 58, 46, 56)AK1(LIST)6( g) 3(Mean)6( g) 6( g) 1(List) b)wu To calculate the median of data items of specified frequency[OPTN]-[LIST]-[Med]This procedure uses two lists: one that contains values and one that indicates the frequency(number of occurrences) of each value. The frequency of the data in Cell 1 of the first list isindicated by the value in Cell 1 of the second list, etc.• The two lists must contain the same number of data items. If they don’t, an error occurs.K1(LIST) 6( g) 4(Med) 6( g) 6( g) 1(List)
,1(List)
)w
