For higher precision with (GNU) awk (with bignum compiled in) use:

$ echo '0.4970436865354813' | awk -M -v PREC=100 '{printf("%.18f\n", $1)}'
0.497043686535481300

The PREC=100 means 100 bits instead of the default 53 bits.
If that awk is not available, use bc

$ echo '0.4970436865354813*1.1' | bc -l
.54674805518902943

Or you will need to learn to live with the inherent imprecision of floats.

In your original lines there are several issues:

• A factor of 1.1 is 10% increase, not 1% (should be a 1.01 multiplier). I'll use 10%.

• The conversion format from a string to a (floating) number is given by CONVFMT. Its default value is %.6g. That limits the values to 6 decimal digits (after the dot). That is applied to the result of the gsub change of $1.

  $ a='0.4970436865354813'
  $ echo "$a" | awk '{printf("%.16f\n", $1*1.1)}'
  0.5467480551890295
  
  $ echo "$a" | awk '{gsub(1*1.1)}; {printf("%.16f\n", $1)}'
  0.5467480000000000
  

• The printf format g removes trailing zeros:

  $ echo "$a" | awk '{gsub(1*1.1)}; {printf("%.16g\n", $1)}'
  0.546748
  
  $ echo "$a" | awk '{gsub(1*1.1)}; {printf("%.17g\n", $1)}'
  0.54674800000000001
  

  Both issues could be solved with:

  $ echo "$a" | awk '{printf("%.17g\n", $1*1.1)}'
  0.54674805518902947
  

  Or

  $ echo "$a" | awk -v CONVFMT=%.30g '{gsub(1*1.1)}; {printf("%.17f\n", $1)}'
  0.54674805518902947 
  

But don't get the idea that this means higher precision. The internal number representation is still a float in double size. That means 53 bits of precision and with that you could only be sure of 15 correct decimal digits, even if many times up to 17 digits look correct. That's a mirage.

$ echo "$a" | awk -v CONVFMT=%.30g '{gsub(1*1.1}; {printf("%.30f\n", $1)}'
0.546748055189029469325134868996

The correct value is:

$ echo "scale=18; 0.4970436865354813 * 1.1" | bc
.54674805518902943

Which could be also calculated with (GNU) awk if the bignum library has been compiled in:

$ echo "$a" | awk -M -v PREC=100 -v CONVFMT=%.30g '{printf("%.30f\n", $1)}'
0.497043686535481300000000000000

To complement karakfa's helpful answer:

In addition to explicit number formatting with printf and sprintf, you can also change awk's default floating-point number formatting, via built-in variable OFMT.
Note that it does not apply to integers.

OFMT defaults to %.6g, which means that any floating-point number is rounded to 6 significant digits and for exponents starting with 7 is represented in scientific notation.

Calculation result 1887436.8 - which has 8 significant digits - is therefore represented as 1.88744e+06, i.e., in scientific notation with 6 significant digits.

The following example sets OFMT to %1.f in order to output all floating-point numbers with 1 decimal place by default:

$ awk -v OFMT='%.1f' 'BEGIN {ram=(1.8 * 1024) * 1024; print ram}'
1887436.8

Note, however, that OFMT does not apply in the following scenarios:

• If the floating-point number is used in a string concatenation:

  $ awk -v OFMT='%.1f' 'BEGIN { print "result: " 1 / 3 }'
  result: 0.333333
  
  # Workaround: Use `sprintf()` with OFMT
  awk -v OFMT='%.1f' 'BEGIN { print "result: " sprintf(OFMT, 1 / 3) }'
  result: 0.3
  

• If a literal can be parsed as an integer - even if it looks like a floating-point number:

  $ awk -v OFMT='%.1f' 'BEGIN { print 1.000 }'
  1
  

Caveat: There are many subtleties around number conversion and formatting in awk, not least because of the limited precision of floating-point numbers (which in awk are always of the ISO C double type).

I think you are closer than you think, but your problems stem from not handling putting the pieces together in the right order. By default awk handles floating-point math. Your summation of sum[$1] += $2 will be done with floating point numbers if your input numbers contains decimal places.

Why you are using a Command Substitution, e.g. $(..) is unclear. That isn't needed to fill temp.txt with the sorted sums from your input. Let's take for example your input file in a file named input. Then to sum and output the values you need nothing more than:

awk '{sum[2} END {for (i in sum) print i, sum[i]}' input

Which results in the output of:

Bruce 83.1
Dave 42.7
Adrian 51
Janick 65.4

To sort the results, simply pipe the awk output to sort, e.g.

awk '{sum[2} END {for (i in sum) print i, sum[i]}' input | sort

Result:

Adrian 51
Bruce 83.1
Dave 42.7
Janick 65.4

Now, to fill temp.txt, you simply need to redirect the output to temp.txt, e.g.

awk '{sum[2} END {for (i in sum) print i, sum[i]}' input | sort > temp.txt

temp.txt now contains the sorted output. To nicely tabularize the output, you can format the output by saving the max length for $1 and setting the name field-width to max and then outputting the floating-point number in a fixed format (total floating-point field-width 10 with 1 digit following the decimal point), e.g.

{sum[$1] += $2; len=length($1); if(len>max)max=len} ...  printf "%-*s%10.1f\n", max, i, sum[i]

Or in full-form:

awk '{sum[2; len=length($1); if(len>max)max=len} END {for (i in sum) printf "%-*s%10.1f\n", max, i, sum[i]}' input | sort > temp.txt

With the results in temp.txt now being:

$ cat temp.txt
Adrian      51.0
Bruce       83.1
Dave        42.7
Janick      65.4

I suggest:

result="123456"
var=2            # decimal places

awk -v res="$result" -v dp="$var" 'BEGIN {printf "%." dp "f", res / 1000}'

or

awk -v res="$result" -v CONVFMT="%.${var}f" 'BEGIN {printf res / 1000}'

Output:

123.46

From man awk:

CONVFMT: The conversion format for numbers, "%.6g", by default.