Find the Frequency, Relative frequency, Mean, median, Modal Class, variance, and
ID: 3171881 • Letter: F
Question
Find the Frequency, Relative frequency, Mean, median, Modal Class, variance, and standard deviation.
Hawaii Ocean Temperatures
Find the Frequency, Relative frequency, Mean, median, Modal Class, variance, and standard deviation.
Hawaii Ocean Temperatures
76.6 74.8 74.9 73.1 75.7 75.0 71.0 73.3 77.7 77.2 77.0 77.0 75.8 75.9 76.9 75.7 78.2 76.0 75.4 76.2 73.7 73.3 74.1 74.7 73.8 74.5 75.9 77.0 74.2 72.1 77.5 75.3 74.4 75.6 77.7 73.0 76.1 73.8 73.9 76.0 79.3 77.4 77.2 76.5 72.4 72.8 77.8 75.8 76.6 76.4 75.3 75.6 76.7 75.5 78.6 78.8 78.3 73.1 73.2 75.1 77.7 73.0 80.7 75.2 74.6 74.0 75.9 74.7 73.8 78.4 75.1 75.4 77.4 80.5 73.9 76.8 74.8 75.6 78.1 78.9 75.5 74.5 78.2 76.4 74.6 74.8 73.1 76.7 77.7 75.3 77.5 77.4 78.2 78.7 76.0 77.4 77.6 74.6 76.6 76.0 75.5 75.0 73.5 74.7 74.6 75.5 73.6 77.7 75.0 80.5 72.8 83.2 78.8 80.1 76.2 73.4 75.9 74.4 81.1 77.1 74.7 75.4 73.7 79.7 77.3Explanation / Answer
Solution
Back-up Theory
Let X = temperature.
The range of the given observations: 71.0 to 83.2………………………………….(1)
So, we take the class limits as
(70.0 – 71.9), (72.0 – 73.9), (74.0 – 75.9), …., (82.0 – 83.9) ……………………….(2)
Since temperature is a continuous variable, we convert the class limits into class boundaries as:
(69.95 – 71.95], (71.95 – 73.95], (73.95 – 75.95], …., (81.95 – 83.95], where ( excludes and ] includes…………………………………………………………………………(3)
The mid-point or what is also known as class mark
= (lower boundary + upper boundary)/2. …………………………………………….(4)
Relative frequency = class frequency/total frequency. ……………………………….(5)
Mean = X bar = {sum(xi.fi)}/{sum(fi)}, where xi and fi are respectively mid-point and class frequency of the ith class and sum is over I = 1 to k, k being the number of classes…..(6)
Variance = V(X) = [sum{(xi – X bar)2.fi)}]/{sum(fi)}………………………………..(7)
Standard Deviation = SD(X) = sq.rt of V(X)………………………………………….(8)
Median = L + [{(n/2) – C}(W/f)] ……………………………………………………..(9)
where L, W and f are respectively the Lower boundary, Class width and frequency of the median class, C is the cumulative frequency upto the median class and n = total frequency;
Median class is that class up to which the cumulative frequency is less than n/2, but cumulative frequency including the class frequency is greater than n/2.
Modal class is that class which has the maximum frequency ………………………… (10).
Now, to work out the solution,
All calculations and workings are presented in the following table:
Class
Boundary
Frequency
Relative
Frequency
Cumulative
Frquency
Class
Mark (x)
(69.95-71.95]
1
0.008
1
70.95
(71.95-73.95]
21
0.168
22
72.95
(73.95-75.95]
46
0.368
68
74.95
(75.95-77.95]
38
0.304
106
76.95
(77.95-79.95]
13
0.104
119
78.95
(79.95-81.95]
5
0.040
124
80.95
(81.95-83.95]
1
0.008
125
82.95
Total
125
1.000
Mean = 75.91
Variance = 4.7104
SD = 2.17
[all the above calculations are done using Excel Functions]
Median = 73.95 + {(62.5 - 22)/(2/68)} = 75.14
Modal class = (73.95 – 75.95)
DONE
Class
Boundary
Frequency
Relative
Frequency
Cumulative
Frquency
Class
Mark (x)
(69.95-71.95]
1
0.008
1
70.95
(71.95-73.95]
21
0.168
22
72.95
(73.95-75.95]
46
0.368
68
74.95
(75.95-77.95]
38
0.304
106
76.95
(77.95-79.95]
13
0.104
119
78.95
(79.95-81.95]
5
0.040
124
80.95
(81.95-83.95]
1
0.008
125
82.95
Total
125
1.000
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