MEAN AND MEDIAN
2.11 MEAN AND MEDIAN
The sum of all the items in a set divided by the number of items gives the mean value, i.e.,
Σx
where x = the mean value Σx = sum of all the items n = total number of items. The magnitude of the item in a set such that half of the total number of items are larger
and half are smaller is called the median. The apparent median for the curve in Fig. 2.21 is the ordinate corresponding to 50% of the years. The mean may be unduly influenced by a few large or small values, which are not truly representative of the samples (items), whereas the median is influenced mainly by the magnitude of the main part of intermediate values.
To find the median, the items are arranged in the ascending order; if the number of items is odd, the middle item gives the median; if the number of items is even, the average of the central two items gives the median.
HYDROLOGY
Example 2.6 The annual rainfall at a place for a period of 10 years from 1961 to 1970 are respectively 30.3, 41.0, 33.5, 34.0, 33.3, 36.2, 33.6, 30.2, 35.5, 36.3. Determine the mean and median values of annual rainfall for the place.
Σx
Solution (i) Mean x = = (30.3 + 41.0 + 33.5 + 34.0 + 33.3 + 36.2
(ii) Median: Arrange the samples in the ascending order 30.2, 30.3 33.3, 33.5, 33.6, 34.0,
35.5, 36.2, 36.3, 41.0 No. of items = 10, i.e., even
Note the difference between the mean and the median values. If 11 years of record, say 1960 to 1970, had been given, the median would have been the sixth item (central value) when arranged in the ascending order.
Example 2.7 The following are the rain gauge observations during a storm. Construct: (a) mass curve of precipitation, (b) hyetograph, (c) maximum intensity-duration curve and develop
a formula, and (d) maximum depth-duration curve. Time since commencement
Accumulated of storm
rainfall (min)
Solution (a) Mass curve of precipitation. The plot of ‘accumulated rainfall (cm) vs. time (min)’ gives the ‘mass curve of rainfall’ Fig. 2.19 (a).
(b) Hyetograph. The intensity of rainfall at successive 5 min interval is calculated and a bar-graph of ‘i (cm/hr) vs. t (min)’ is constructed; this depicts the variation of the intensity of rainfall with respect to time and is called the ‘hyetograph; 2.19 (b).
PRECIPITATION
Time, t
Intensity, (min)
Accumulated
∆ P in time
(cm /hr) 5 0.1 0.1 1.2
depth all
mass curve traced by
Ñ rainf p i=
self-recording rain gauge
1.0 t
50-min storm
otal T
Time t (min) (a) Mass curve of precipitation
8.4 cm/hr 8 7.2 50-min storm
6.0 Area under the curve
(cm/hr) 6 i gives total precipitation (cm)
4 3.6 2.4 2.4 cm/hr Intensity 2 1.2 cm/hr
0 0 5 10 15 20 25 30 35 40 45 50 Time t (min)
(b) Hyetograph Fig. 2.19 Graphs from recording rain-gauge data, Example 2.7
(c) Maximum depth–duration curve. By inspection of time (t) and accumulated rainfall (cm) the maximum rainfall depths during 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50 min durations
HYDROLOGY
are 0.7, 1.3, 1.6, 1.8, 2.3, 2.5, 2.7, 2.9, 3.0 and 3.1 cm respectively. The plot of the maximum rainfall depths against different durations on a log-log paper gives the maximum depth-dura- tion curve, which is a straight line, Fig. 2.20 (a).
60 Log-log paper
3 20 k = 17
30 a. Maximum depth-duration curve a. Maximum depth-duration curve
1.5 (cm) (cm/hr) i
15 Slope x = – 0.375
0.8 um Intensity 0.375
3 b. Maximum intensity-duration curve b. Maximum intensity-duration curve
Time t (min)
Fig. 2.20 Maximum depth-duration & intensity-duration curves (Example 2.7)
(d) Maximum intensity-duration curve. Corresponding to the maximum depths obtained
in (c) above, the corresponding maximum intensities can be obtained ∆ × 60, i.e., 8.4, 7.8, 6.4,
5.4, 5.52, 5.0, 4.63, 4.35, 4.0 and 3.72 cm/hr, respectively. The plot of the maximum intensities against the different duration on a log-log paper gives the maximum intensity-duration curve which is a straight line, Fig. 2.20 (b).
The equation for the maximum itensity duration curve is of the form
i = kt x
Slope of the straight line plot,
k = 17 cm/hr
when t = 1 min
Hence, the formula becomes
t 0.375
which can now be verified as
t = 10 min, i = 7.2 cm/hr t = 40 min, i = 4.25 cm/hr
which agree with the observed data
PRECIPITATION
Example 2.8 The annual rainfall at a place for a period of 21 years is given below. Draw the rainfall frequency curve and determine :
(a) the rainfall of 5-year and 20-year recurrence, interval (b) the rainfall which occurs 50% of the times (c) the rainfall of probability of 0.75 (d) the probability of occurrence of rainfall of 75 cm and its recurrence interval.
Solution Arrange the yearly rainfall in the descending order of magnitude as given below. If
a particular rainfall occurs in more than one year, m = no. of times exceeded + no. of times equalled.
Year
Rainfall
Rank (m)
Frequency
P (cm)
(no. of times ≥ P)
1966 UV
40 14 W 63.7
Total Σx = 1026
n = 21
HYDROLOGY
Draw the graph of ‘P vs. F’ on a semi-log paper which gives the rainfall frequency curve, Fig. 2.21. From the frequency-curve, the required values can be obtained as
(a) T = 5–yr, F =
= 20% for which P = 64 cm.
T = 20 – yr, F =
× 100 = 5% for which P = 97.5 cm
Recurrence interval T-yr
8 yr Semi-log paper
(cm) 70 P
64 cm
all 60
rainf 50 48.8 cm
Mean
ual
Median Median Ann 40
Frequency F % ( P) £ Fig. 2.21 Rainfall frequency curve (Example 2.8)
(b) For F = 50%, P = 42.2 cm which is the median value, and the mean value
which has a frequency of 37%. (c) For a probability of 0.75 F = 75% for which P = 32 cm
(d) For P = 75 cm, F = 12.4%, T =
and its probability of occurrence = 0.124