Answer:
(a) X(in '000s) = 4133, 2369, 1295, 928, 679.
(b) The mean number of passengers for these five cruise lines is 1880.8.
(c) The range of the data is 3454.
(d) The standard deviation of the data is 1414.75.
Step-by-step explanation:
We are given that in a recent year, the top five cruise lines in the world had this many passengers:
4,133,000, 2,369,000, 1,295,000, 928,000, 679,000
(a) Let X = Number of passengers in the top five cruise lines in the world
Representing each number in terms of thousands of passengers below;
X(in '000s) = 4133, 2369, 1295, 928, 679
(b) The mean of the data is given by the following formula;
Mean, [tex]\bar X[/tex] = [tex]\frac{\sum X}{n}[/tex]
Here, n = number of observations in data = 5
So, [tex]\bar X[/tex] = [tex]\frac{4133+2369+1295+928+679}{5}[/tex]
= [tex]\frac{9404}{5}[/tex] = 1880.8
Hence, the mean number of passengers for these five cruise lines is 1880.8.
(c) The range of the data is calculated by the following formula;
Range = Highest value - Lowest value
= 4133 - 679 = 3454
Hence, the range of the data is 3454.
(d) The standard deviation of the data is given by the following formula;
Standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1}[/tex]
= [tex]\frac{(4133-1880.8)^{2}+(2369-1880.8)^{2}+.......+(679-1880.8)^{2} }{5-1}[/tex]
= 1414.75
Hence, the standard deviation of the data is 1414.75.
