An auto transmission manufacturer receives ball bearings from two different suppliers. The ball bearings must have a specified diameter of 16.30 mm with a tolerance of [tex]$\pm 0.1$[/tex] mm. Recent shipments from the two suppliers had ball bearings with the following diameters:

\begin{tabular}{llllllll}
Supplier A: & 16.23 & 16.27 & 16.32 & 16.33 & 16.36 & 16.42 & 16.45 \\
Supplier B: & 16.18 & 16.21 & 16.25 & 16.32 & 16.39 & 16.43 & 16.45
\end{tabular}

a. Find the mean and standard deviation for each of the two data sets.

Find the mean and standard deviation for the diameters of the ball bearings from Supplier A.
[tex]$
\begin{array}{l}
\text { mean }=16.34 \\
s=0.08
\end{array}
$[/tex]
(Round to the nearest hundredth as needed.)

Find the mean and standard deviation for the diameters of the ball bearings from Supplier B.
[tex]$
\begin{array}{l}
\text { mean }=\square \\
s =\square
\end{array}
$[/tex]
(Round to the nearest hundredth as needed.)



Answer :

Let's analyze the ball bearings' diameters from each supplier.

Supplier A:
Diameter measurements: 16.23, 16.27, 16.32, 16.33, 16.36, 16.42, 16.45

For Supplier A, the mean diameter is:
[tex]\[ \text{mean}_A = 16.34 \, \text{mm} \][/tex]

The standard deviation for Supplier A is:
[tex]\[ s_A = 0.08 \, \text{mm} \][/tex]

Supplier B:
Diameter measurements: 16.18, 16.21, 16.25, 16.32, 16.39, 16.43, 16.45

For Supplier B, the mean diameter is:
[tex]\[ \text{mean}_B = 16.32 \, \text{mm} \][/tex]

The standard deviation for Supplier B is:
[tex]\[ s_B = 0.11 \, \text{mm} \][/tex]

Summarizing both findings, we have:

Supplier A:
[tex]\[ \begin{array}{l} \text{mean} = 16.34 \, \text{mm} \\ s = 0.08 \, \text{mm} \end{array} \][/tex]

Supplier B:
[tex]\[ \begin{array}{l} \text{mean} = 16.32 \, \text{mm} \\ s = 0.11 \, \text{mm} \end{array} \][/tex]