Question

1. the table on the right, from the statistical abstract of the united states, shows amusement - park attendance at the top 15 amusement parks for 5 recent years.

a. find the mean attendance, in thousands, for the 5 years shown, to the nearest integer.

b. find the standard deviation, in thousands, for the years shown. round to the nearest integer.

Explanation:

Step1: Calculate the mean

The mean $\bar{x}$ of a set of data $x_1,x_2,\cdots,x_n$ is given by $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here, $n = 5$, $x_1=107345$, $x_2 = 108321$, $x_3=112509$, $x_4=110420$, $x_5=119951$.
$\sum_{i=1}^{5}x_i=107345 + 108321+112509+110420+119951=558546$.
$\bar{x}=\frac{558546}{5}=111709.2\approx111709$.

Step2: Calculate the standard - deviation

The formula for the sample standard deviation $s$ is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$.
First, calculate $(x_1-\bar{x})^2=(107345 - 111709)^2=(-4364)^2 = 19044496$.
$(x_2-\bar{x})^2=(108321-111709)^2=(-3388)^2 = 11478544$.
$(x_3-\bar{x})^2=(112509-111709)^2=(800)^2 = 640000$.
$(x_4-\bar{x})^2=(110420-111709)^2=(-1289)^2 = 1661521$.
$(x_5-\bar{x})^2=(119951-111709)^2=(8242)^2 = 67930564$.
$\sum_{i = 1}^{5}(x_i-\bar{x})^2=19044496+11478544+640000+1661521+67930564=100755125$.
$s=\sqrt{\frac{100755125}{4}}=\sqrt{25188781.25}\approx5018.84\approx5019$.

Answer:

A. Mean: 111709
B. Standard - deviation: 5019