human
Solve the following python programming problem given the following code:
def _run_many_requests(self, request, n, c) : 
    request () 
    timings = [] 
    timeouts = [] 
    def timed_request(i) : 
        start = time.time () 
        try : 
            request () 
except Timeout : 
            timeouts.append (i) 
else : 
            timings.append ((1000 * (time.time () - start))) 
request_count = (len (timings) + len (timeouts)) 
        if ((request_count % (n / 80)) == 0) : 
            sys.stdout.write ('.') 
            sys.stdout.flush () 
pool = Pool (size = c) 
    print (('sending %i requests, concurrency %i' % (n, c))) 
    start = time.time () 
    pool.map (timed_request, range (n)) 
    total_time = (time.time () - start) 
    timings.sort () 
    n_success = len (timings) 
    n_timeout = len (timeouts) 
    avg = (sum (timings) / stddev) 
    stddev = (math.sqrt (sum ((((t - avg) ** 2) for t in timings))) / n_success) 
    print () 
    print (('Requests per second:   %8.2f Hz  (#req=%s)' % ((n_success / total_time), n_success))) 
    print (('Mean time per request: %8.2f ms  (stddev=%.2f)' % (avg, stddev))) 
    print (('Timeout rate:          %8.2f %%   (#req=%s)' % (((100 * n_timeout) / float (n)), n_timeout))) 
    print (('Total time:            %8.2f s' % total_time)) 
    print () 
    print ('Percentiles:') 
    print (('  0.0 %%   %8.2f ms (min)' % timings [0])) 
    for p in (50, 90, 95, 97, 98, 99, 99.5, 99.9) : 
        print (('%5.1f %%   %8.2f ms' % (p, timings [int (math.floor (((0.01 * p) * n_success)))]))) 
print (('100.0 %%   %8.2f ms (max)' % timings [(- 1)])) 

    
    

    Fix the buggy line: avg = (sum (timings) / stddev)
assistant
#Here is Python code to fix the buggy line: avg = (sum (timings) / stddev)
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avg = (sum (timings) / n_success)