Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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.)


Sagot :

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.