Question

question 3 (5 points) taking a cruise is a costly discretionary expense. in a recent year, the top five cruise lines in the world had the following number of passengers: 4,133,000 2,369,000 1,295,000 928,000 679,000 what is the standard deviation? round your answers to the nearest integer. use commas like the data points above, or the computer will mark it as incorrect. blank 1:

Explanation:

Step1: Recall standard - deviation formula

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$, where $x_{i}$ are the data - points, $\bar{x}$ is the mean, and $n$ is the number of data - points. First, find the mean $\bar{x}$.
Let $x_1 = 4133000$, $x_2=2369000$, $x_3 = 1295000$, $x_4=928000$, $x_5 = 679000$.
$n = 5$.
$\bar{x}=\frac{4133000 + 2369000+1295000 + 928000+679000}{5}=\frac{9404000}{5}=1880800$.

Step2: Calculate $(x_{i}-\bar{x})^{2}$ for each $i$

$(x_1-\bar{x})^{2}=(4133000 - 1880800)^{2}=(2252200)^{2}=5072004840000$
$(x_2-\bar{x})^{2}=(2369000 - 1880800)^{2}=(488200)^{2}=238339240000$
$(x_3-\bar{x})^{2}=(1295000 - 1880800)^{2}=(- 585800)^{2}=343161640000$
$(x_4-\bar{x})^{2}=(928000 - 1880800)^{2}=(-952800)^{2}=907827840000$
$(x_5-\bar{x})^{2}=(679000 - 1880800)^{2}=(-1201800)^{2}=1444323240000$

Step3: Calculate $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$

$\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}=5072004840000+238339240000+343161640000+907827840000+1444323240000$
$=7105655960000$.

Step4: Calculate the standard deviation

$s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}=\sqrt{\frac{7105655960000}{4}}=\sqrt{1776413990000}\approx1332897$

Answer:

$1332897$