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 \text{ mm}[/tex]. Recent shipments from the two suppliers had ball bearings with the following diameters. Complete parts (a) through (c).

\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]\[
\begin{array}{l}
\text{mean} = 16.34 \\
s = 0.07
\end{array}
\][/tex]
(Round to the nearest hundredth as needed.)



Answer :

To find the mean and standard deviation for the diameter of ball bearings from Supplier A, let's follow a step-by-step approach:

### Step-by-Step Solution:

#### Step 1: List the Diameters
The diameters of ball bearings from Supplier A are:
- 16.23
- 16.27
- 16.32
- 16.33
- 16.36
- 16.42
- 16.45

#### Step 2: Calculate the Mean
The mean (average) is calculated by summing all the diameters and then dividing by the total number of measurements.

Mathematically, it’s given by the formula:
[tex]\[ \text{Mean} = \frac{1}{n}\sum_{i=1}^{n} x_i \][/tex]

For Supplier A:
[tex]\[ \text{Mean} = \frac{16.23 + 16.27 + 16.32 + 16.33 + 16.36 + 16.42 + 16.45}{7} \][/tex]
[tex]\[ \text{Mean} \approx 16.34 \][/tex]

So, the mean diameter for Supplier A is [tex]\( 16.34 \, \text{mm} \)[/tex].

#### Step 3: Calculate the Standard Deviation
The sample standard deviation measures the amount of variation or dispersion in a set of values. It is calculated using the formula:
[tex]\[ s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2} \][/tex]
where [tex]\( \bar{x} \)[/tex] is the mean, and [tex]\( x_i \)[/tex] are the individual measurements.

For Supplier A:
[tex]\[ s \approx 0.08 \][/tex]

So, the standard deviation of the diameters for Supplier A is [tex]\( 0.08 \, \text{mm} \)[/tex].

### Summary of Results for Supplier A:
- Mean Diameter: [tex]\( 16.34 \, \text{mm} \)[/tex]
- Standard Deviation: [tex]\( 0.08 \, \text{mm} \)[/tex]

Therefore, the mean diameter is [tex]\( 16.34 \)[/tex] mm and the standard deviation is [tex]\( 0.08 \)[/tex] mm when rounded to the nearest hundredth.