basic measures of center -2\nscore: 5.94/23 answered: 0/4\nprogress saved done √0\nquestion 1\n7 pts 1…

basic measures of center -2\nscore: 5.94/23 answered: 0/4\nprogress saved done √0\nquestion 1\n7 pts 1 details\nuse the appropriate formulas and methods to answer the following questions. do not use technology as an aid, unless otherwise stated.\npart 1 of 6\nhint 1 hint 2 hint 3 hint 4 hint 5\nbelow are iq scores from 30 randomly selected adults.\n{ 83, 85, 88, 89, 91, 92, 93, 95, 96, 96, 97, 100, 100, 101, 102, 103, 104, 106, 106, 109, 110, 110, 111, 111, 112, 120, 122, 122, 122, 130 }.\nfirst, give the mean of the data set.

basic measures of center -2\nscore: 5.94/23 answered: 0/4\nprogress saved done √0\nquestion 1\n7 pts 1 details\nuse the appropriate formulas and methods to answer the following questions. do not use technology as an aid, unless otherwise stated.\npart 1 of 6\nhint 1 hint 2 hint 3 hint 4 hint 5\nbelow are iq scores from 30 randomly selected adults.\n{ 83, 85, 88, 89, 91, 92, 93, 95, 96, 96, 97, 100, 100, 101, 102, 103, 104, 106, 106, 109, 110, 110, 111, 111, 112, 120, 122, 122, 122, 130 }.\nfirst, give the mean of the data set.

Answer

Explanation:

Step1: Recall the mean formula

The mean $\bar{x}$ of a data - set $x_1,x_2,\cdots,x_n$ is given by $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$, where $n$ is the number of data points and $\sum_{i=1}^{n}x_i$ is the sum of the data points. Here, $n = 30$.

Step2: Calculate the sum of the data points

$83+85+88+89+91+92+93+95+96+96+97+100+100+101+102+103+104+106+106+109+110+110+111+111+112+120+122+122+122+130$ $=(83 + 85+88+89)+(91+92+93+95)+(96+96+97)+(100+100+101)+(102+103+104)+(106+106+109)+(110+110+111)+(111+112+120)+(122+122+122)+130$ $=345+371+289+301+309+321+331+343+366+130$ $=(345+371)+(289+301)+(309+321)+(331+343)+(366+130)$ $=716+590+630+674+496$ $=(716+590)+(630+674)+496$ $=1306+1304+496$ $=2610+496$ $=3106$.

Step3: Calculate the mean

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{3106}{30}=\frac{1553}{15}\approx103.53$.

Answer:

$\frac{1553}{15}\approx103.53$