Frequency Distribution and Statistical Analysis

6) i)
F ( t )=2 sin at +bt
2 sin at−bt
2
Beat frequency = ( a−b ) /2 π=0.1 a /2 π=0.1 Hertz [1]
ii)
iii) Optimal sampling rate = 2
iv) The time interval = ½ = 0.5
[2]
5)
i)
74 76 78 80 82 84 86 88 90
72
74
76
78
80
82
84
86
88
90
92
(Y)Right Foot
ii) The standard deviation of the right foot temperature is = 83.38889

The standard deviation of the left foot temperature is = 83.66667
Covariance between them = 15.2963
Therefore correlation is = covariance/ (standard deviation L * Standard deviation R)
The correlation coefficient is 0.809
iii) Since the correlation between the left foot and right foot temperature are positive so as the left
foot temperature is cooler than the average then the right foot temperature will also be cooler than
the average.
4) i) The first quantile is 14.65 and the third quantile is 17.4.
ii) Median is 16.2
iii) IQR (Inter quartile range) = 3rd quartile -1st quartile = 17.4 – 14.65 = 2.75
Now, Upper limit = 3rd quartile + (1.5)* IQR = 21.525
Lower limit = 1st quartile - (1.5)* IQR = 10.525
iv) No none of the data crosses the upper limit and the lower limit, so there are no outliers
v)
[3]
2) a) i) No it is not possible to draw the histogram as the data is censored in this case
ii) Since the piece of information none of the employees travelled more than 34 miles, actually
makes the data fully known instead of censored data so it is now possible to draw a histogram.

0-4.9 5-9.9 10-14.9 15-19.9 20-24.9 25-29.9 30-34.9
0
5
10
15
20
25
30
Frequency
b) i)
1
3
4
7
More
0
4
8
12
0.00%
40.00%
80.00%
120.00%
Histogram
Frequency
Cumulative %
Bin
Frequency
Bin
Frequen
cy
Cumulati
ve % Bin
Frequen
cy
Cumulati
ve %
0 3 10.00% 1 11 36.67%
1 11 46.67% 2 7 60.00%
2 7 70.00% 3 4 73.33%
3 4 83.33% 0 3 83.33%
4 2 90.00% 4 2 90.00%
5 2 96.67% 5 2 96.67%
6 0 96.67% 7 1 100.00%
7 1 100.00% 6 0 100.00%
More 0 100.00% More 0 100.00%
[2]