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    <p>Just some minor modification to your code will do (with some var renaming for clarity) :</p> <pre><code>double sum = 0; //average will have decimal point for(int i=0; i &lt; args.length; i++){ //parse string to double, note that this might fail if you encounter a non-numeric string //Note that we could also do Integer.valueOf( args[i] ) but this is more flexible sum += Double.valueOf( args[i] ); } double average = sum/args.length; System.out.println(average ); </code></pre> <p>Note that the loop can also be simplified:</p> <pre><code>for(String arg : args){ sum += Double.valueOf( arg ); } </code></pre> <p>Edit: the OP seems to want to use the <code>args</code> array. This seems to be a String array, thus updated the answer accordingly.</p> <p><strong>Update</strong>:</p> <p>As zoxqoj correctly pointed out, integer/double overflow is not taken care of in the code above. Although I assume the input values will be small enough to not have that problem, here's a snippet to use for really large input values:</p> <pre><code>BigDecimal sum = BigDecimal.ZERO; for(String arg : args){ sum = sum.add( new BigDecimal( arg ) ); } </code></pre> <p>This approach has several advantages (despite being somewhat slower, so don't use it for time critical operations):</p> <ul> <li>Precision is kept, with double you will gradually loose precision with the number of math operations (or not get exact precision at all, depending on the numbers)</li> <li>The probability of overflow is practically eliminated. Note however, that a <code>BigDecimal</code> might be bigger than what fits into a <code>double</code> or <code>long</code>.</li> </ul>
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    1. COIf i was to use args.lengths instead of nums.length what would i do to "sum += nums[i];" ?
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    2. CO@syncoroll Rereading your updated question I also updated my answer.
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    3. COinteger overflow is not taken care of
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