Step by step solution :
standard deviation is given by :
[tex]\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}[/tex]
where, [tex]\sigma[/tex] is standard deviation
[tex]\bar{x}[/tex] is mean of given data
n is number of observations
From the above data, [tex]\bar{x}=24.88[/tex]
Now, if [tex]x=24.8[/tex], then [tex](x-\bar{x})^2=0.0064[/tex]
If [tex]x=23.9[/tex], then [tex](x-\bar{x})^2=0.9604[/tex]
if [tex]x=26.1[/tex], then [tex](x-\bar{x})^2=1.4884[/tex]
If [tex]x=25.1[/tex], then [tex](x-\bar{x})^2=0.0484[/tex]
If [tex]x=24.5[/tex], then [tex](x-\bar{x})^2=0.1444[/tex]
so, [tex]\sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}[/tex]
[tex]\sum (x-\bar{x})^2 =2.648[/tex]
[tex]\sqrt{\sum \frac{(x-\bar{x})^2}{n}}[/tex]
[tex]\sigma =0.7277[/tex]
No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.
