Summary stats:
i) Mean µ = 1nx
= 113 (6 + 26 + 28 + 30 + 32 +34 + 39 + 41 + 46 + 46 + 68 + 100 + 124 + 132)
= 113 (706)
= 54.3077
ii) Median = Middle (x)
= 39
iii) Standard deviation σ = 1n i=1n(x₁–µ)²
Standard deviation σ = 113 (18976.7695)
= 38.2067
iv) Inter quartile Range = Q₃ - Q₁
6, 26, 28, 30, 32, 34, 39, 41, 46, 68, 100, 124, 132
Q₃ = 34n = 34 13 = 9.75th term = 12 (68 + 100)
= 84
Q₁ = 14n = 14 13 =3.25th term = 12 (28 + 30)
= 29
IQR= 84 – 29 = 55
v) Range = Max x – Min x
= 132 - 6 = 126
Distribution of the data
The mean is more than the median indicating that the data is not normally distributed and is positively skewed. This implies that there are more locations with few breeding pairs and few locations with many breeding pairs thus pushing the mean above the median.
20th percentile= ( 20%of nth term)
= 20100 (13) = 2.6th term
= 12 (26 + 28)
= 27
Z score of 26 breeding pairs
Z score= 1σ (x–µ)
= 138.2067 (26 – 54.3077)
= -0.7409
The p-value for the above Z-score is 0.22965 implying that the probability that a location would have 26 or less breeding pairs is 0.22965.
Z score= 1σ (x–µ)
1.18 = 138.2067 (x – 54.3077)
x = 99.3916
Outliers
The data appears to have outliers. Outliers are extremely low or high values relative to other values in the data. 6 appears to be an extremely low value in the dataset since the next value is 26. Besides, 100, 124 and 132 appear to be outliers since they are extremely far from other values in the data.
Summary stats when the lowest and highest values are removed
i) Mean µ = 1nx
= 111 (26 + 28 + 30 + 32 +34 + 39 + 41 + 46 + 46 + 68 + 100 + 124)
= 111 (568)
= 51.6364
ii) Median = 39
iii) Standard deviation σ = 1n i=1n(x₁–µ)²
Standard deviation σ = 111 (10528.5454)
= 30.9377
iv) Inter quartile Range = Q₃ - Q₁
26, 28, 30, 32, 34, 39, 41, 46, 68, 100, 124
Q₃ = 34n = 34 11 = 8.25th term = 12 (46 + 68)
= 57
Q₁= 14n = 14 11 =2.75th term = 12 (28 + 30)
= 29
IQR= 57 – 29 = 28
v) Range = Max x – Min x
= 124 - 26 = 98
Why the median has not changed?
The median has not changed since the two values have been removed from either end of the data. The median can only change if the numbers removed from the lower end is not equal to that removed from the upper end.
Measures of variability
Measures of variability decrease since the removal leaves values that are close to the mean. The lowest and highest values are outliers hence their removal reduces the variability of values from the mean.
Decrease in mean
The mean decreases due to the effect of the two outliers. The highest value pushed the mean up while the lowest value had a less than proportionate effect on the mean. This explains why the mean decreases slightly when the outliers are removed.
