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Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets the following results: 24.8, 23.9, 26.1, 25.1, 24.5 seconds. Joe uses the standard deviation of these numbers as the "margin of error" on his measurement.

Does Joe's average time agree with the accepted value of 25.9 seconds, within his margin of error? Can anyone help me with this?

User Siasmj
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1 Answer

1 vote

Step by step solution :

standard deviation is given by :


\sigma = \sqrt\frac{{\sum (x-\bar{x})^2}}{n}

where,
\sigma is standard deviation


\bar{x} is mean of given data

n is number of observations

From the above data,
\bar{x}=24.88

Now, if
x=24.8, then
(x-\bar{x})^2=0.0064

If
x=23.9, then
(x-\bar{x})^2=0.9604

if
x=26.1, then
(x-\bar{x})^2=1.4884

If
x=25.1, then
(x-\bar{x})^2=0.0484

If
x=24.5, then
(x-\bar{x})^2=0.1444

so,
\sum (x-\bar{x})^2 =(0.0064+0.9604+1.4884+0.0484+0.1444)/(5)


\sum (x-\bar{x})^2 =2.648


\sqrt{\sum \frac{(x-\bar{x})^2}{n}}


\sigma =0.7277

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

User PrasathBabu
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5.7k points