Question

find the mean standard deviation for the diameters of the ball bearings from 16.19, 16.21, 16.26, 16.36, 16.38, 16.42, 16.45 if the ball bearing must have a specific diameter of 16.30 mm with a tolerance of +-0.1mm

Answer 1

Given the measures for the diameters of ball beraings

16.19, 16.21, 16.26, 16.36, 16.38, 16.42, 16.45

You have to calculate the mean and standard deviation.

1) To calculate the mean you have to use the following formula:

`X\lbrack bar\rbrack=(\Sigma x_i)/(n)`

2) To calculate the standard deviation you have to calculate the variance first and the the square root of the variance using the formulas:

`S^2=(1)/(n-1)\lbrack\Sigma x^2_i-((\Sigma x_i)^2)/(n)\rbrack`
`S=\sqrt[]{S^2}`

As you see both formulas involve the summation of the observations and the summation of the squared observations, so those are the first calculations you have to do:

`\Sigma x_i=16.19+16.21+16.26+16.36+16.38+16.42+16.45=114.27`
`\Sigma x^2_i=16.19^2+16.21^2+16.26^2+16.36^2+16.38^2+16.42^2+16.45^2=1865.44`

There are 7 oobservations in the data set so n=7

1) The mean is

`X\lbrack bar\rbrack=(\Sigma x_i)/(n)=(114.27)/(7)=16.32`

2) The variance is

`\begin{gathered} S^2=(1)/(n-1)\lbrack\Sigma x^2_i-((\Sigma x_i)^2)/(n)\rbrack \\ S^2=(1)/(7-1)\lbrack1865.44-(((114.27)^2)/(7))\rbrack \\ S^2=(1)/(6)\lbrack1865.44-(13057.6329)/(7)\rbrack \\ S^2=(1)/(6)\lbrack1865.44-1865.376129\rbrack \\ S^2=(1)/(6)\cdot0.06387 \\ S^2=0.0106 \end{gathered}`

Now to determine the standard deviation you have to calculate the square root of the variance:

`\begin{gathered} S=\sqrt[]{S^2} \\ S=\sqrt[]{0.0106} \\ S=0.103\cong0.1 \end{gathered}`

The meand is X[bar]=16.32mm and the standard deviation is S=0.1 mm