Answer / sattwik kumar

data b;
set a end=lastobs;
age_temp+age;
count+1;
if lastobs then
mean=age_temp/count;
run;

Answer / rg

data temp;
input age @@;
cards;
8 7 7 2 67 9 7 7 1 2 45 34 22 45 23 8 34 23
run;


proc sort data=temp; by age ;run;

data a(keep = median mode_age mean);
last_age=age;
retain max_cnt mode_age median;
set temp end=lastobs nobs=n;
by age;
age_temp + age;

/*Median */
if mod(n,2) = 1 and _n_ = (n+1)/2 then do ;median=age; end;
if mod(n,2) = 0 and _n_ = (n+2)/2 then do ;median=(age + last_age)/2; end;

/*Mean*/
if lastobs then do mean=age_temp/n ; end;

/*Mode */
if first.age then cnt =1; else cnt+1;
if last.age then lst_cnt=cnt;
if lst_cnt > max_cnt then do; max_cnt = lst_cnt; mode_age = age;end;


if lastobs then output;
run;

Answer / chandrakanth

options firstobs=1 obs=26;
data a;
input age @@;
datalines;
22 32 32 32 43 23 24 56 45 54 28 29 27 26
25 28 30 20 18 37 36 47 46
56 19 20
;
run;

proc sort data=a out=ab;
by age;
run;

options firstobs=13 obs=14;
data c;
set ab end=last;
sum_age+age;
if last then median=sum_age/2;
run;

we need observation 13 and 14 because total number of observations is n=26 when n is even we need to do mean of two observations n/2 and (n/2)+1 which is 13th and 14 th observations in this case.
let me know if you need more help... I'm still thinking about mode...will send once I get it

Answer / paul

data b;
set a end=lastobs;
age_temp+age;
count+1;
if lastobs then
mean=age_temp/count;
if _n_=13 then
Median=Age; /*median is the mid value */
run;
proc sort data=b out=c;
by age;
run;
data d;
set c;
if first.age=0 and last.age=0 then
Mode=age;/*mode is the most repeated value*/
by age;
run;

Answer / ravikumar

Answer:

data new;
set a end = last;
count+1;
age_n+age;
if last then do;
mean = round(((age_n)/(count)),.8);
median = ((count+1)/2);
end;
run;

By the above program we can calculate the mean and median.
lets us know me, if anybody find the logic for find mode

Answer / +59+

651

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