Question

1. taking a cruise is a costly discretionary expense. 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 1. what is the mean number of passengers for these five cruise lines? (give the full number.) 2. what is the range? (give the full number.) 3. what is the standard deviation? (give the full number.)

Explanation:

Step1: Calculate the sum of passengers

$4133000 + 2369000+1295000 + 928000+679000=9404000$

Step2: Calculate the mean

The mean $\bar{x}=\frac{9404000}{5}=1880800$

Step3: Find the maximum and minimum values

The maximum value $x_{max}=4133000$, the minimum value $x_{min}=679000$

Step4: Calculate the range

The range $R=x_{max}-x_{min}=4133000 - 679000=3454000$

Step5: Calculate the variance

First, calculate the deviations from the mean:
$(4133000 - 1880800)^2+(2369000 - 1880800)^2+(1295000 - 1880800)^2+(928000 - 1880800)^2+(679000 - 1880800)^2$
$=(2252200)^2+(488200)^2+(- 585800)^2+(-952800)^2+(-1201800)^2$
$=5072004840000 + 238339240000+343161640000+907827840000+1444323240000$
$=7910856960000$
The variance $s^{2}=\frac{7910856960000}{5 - 1}=\frac{7910856960000}{4}=1977714240000$

Step6: Calculate the standard deviation

The standard deviation $s=\sqrt{1977714240000}\approx1406309$

Answer:

1. $1880800$
2. $3454000$
3. $1406309$