Try this:
$(document).ready(function() {
$(".colorchg").each(function() {
$(this).css("background", $(this).val());
});
$(".colorchg").change(function() {
$(this).css("background", $(this).val());
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<label class="col-lg-6">37.sample 1 </label>
<select class="form-control colorchg">
<option></option>
<option value="green">YES</option>
<option value="red">NO</option>
<option value="gray">N/A</option>
</select>
<label class="col-lg-6">38. sample 2</label>
<select class="form-control colorchg">
<option></option>
<option value="green">YES</option>
<option value="red">NO</option>
<option value="gray">N/A</option>
</select>
<label class="col-lg-6">39. sample 3</label>
<select class="form-control colorchg">
<option></option>
<option value="green">YES</option>
<option value="red">NO</option>
<option value="gray">N/A</option>
</select>
By using classes, jQuery will find more than one element to use. ID's need to be unique, so it assumes it's only one element.
Instead of searching for the value again in the functions, you should use this
, otherwise the backgrounds will change to whatever the first option is set to.