diff -r 5a9f2de4dc16 Lib/statistics.py
--- a/Lib/statistics.py	Sun Jan 03 19:37:07 2016 +0530
+++ b/Lib/statistics.py	Sun Jan 03 19:38:53 2016 +0530
@@ -105,6 +105,7 @@
 from fractions import Fraction
 from decimal import Decimal
 from itertools import groupby
+from bisect import bisect_left, bisect_right
 
 
 
@@ -305,6 +306,21 @@
     return table
 
 
+def _find_lteq(a, x):
+    'Locate the leftmost value exactly equal to x'
+    i = bisect_left(a, x)
+    if i != len(a) and a[i] == x:
+        return i
+    raise ValueError
+
+
+def _find_rteq(a, l, x):
+    'Locate the rightmost value exactly equal to x'
+    i = bisect_right(a, x, lo=l)
+    if i != (len(a)+1) and a[i-1] == x:
+        return i-1
+    raise ValueError
+
 # === Measures of central tendency (averages) ===
 
 def mean(data):
@@ -442,9 +458,15 @@
     except TypeError:
         # Mixed type. For now we just coerce to float.
         L = float(x) - float(interval)/2
-    cf = data.index(x)  # Number of values below the median interval.
-    # FIXME The following line could be more efficient for big lists.
-    f = data.count(x)  # Number of data points in the median interval.
+
+    # Uses bisection search to search for x in data with log(n) time complexity
+    # Find the position of leftmost occurence of x in data
+    l1 = _find_lteq(data, x)
+    # Find the position of rightmost occurence of x in data[l1...len(data)]
+    # Assuming always l1 <= l2
+    l2 = _find_rteq(data, l1, x)
+    cf = l1
+    f = l2 - l1 + 1
     return L + interval*(n/2 - cf)/f