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

plus or minus±0.1

mm. Recent shipments from the two suppliers had ball bearings with the following diameters. Complete parts​ (a) through​ (c).

Supplier​ A:

16.2316.23

16.2616.26

16.3116.31

16.3516.35

16.3816.38

16.4116.41

16.4616.46

Supplier​ B:

16.1816.18

16.2116.21

16.2616.26

16.3416.34

16.3816.38

16.4216.42

16.4616.46

Question content area bottom

Part 1

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.

meanequals=16.3416.34

sequals=0.080.08

​(Round to the nearest hundredth as​ needed.)

Part 2

Find the mean and standard deviation for the diameters of the ball bearings from Supplier B.

meanequals=enter your response here

sequals=enter your response here

​(Round to the nearest hundredth as​ needed.)

To calculate the mean and standard deviation of the ball bearing diameters for Supplier B, we'll follow these steps:

1. **Calculate the Mean (Average)**: The mean is calculated by summing all the values and dividing by the number of values.

2. **Calculate the Standard Deviation (s)**: Standard deviation is calculated by taking the square root of the average of the squared differences from the Mean.

### Supplier B's Data
The diameters from Supplier B are:
\[
16.18, 16.21, 16.26, 16.34, 16.38, 16.42, 16.46
\]

#### Step 1: Calculate the Mean
\[
\text{Mean} = \frac{16.18 + 16.21 + 16.26 + 16.34 + 16.38 + 16.42 + 16.46}{7}
\]

Let's perform this calculation:

1. Sum of diameters for Supplier B:
   \[
   16.18 + 16.21 + 16.26 + 16.34 + 16.38 + 16.42 + 16.46 = 113.25
   \]

2. Divide by the number of items (7):
   \[
   \text{Mean} = \frac{113.25}{7} \approx 16.32
   \]

So, the mean for Supplier B is approximately **16.32 mm**.

#### Step 2: Calculate the Standard Deviation
The formula for the standard deviation (for a sample) is:
\[
s = \sqrt{\frac{\sum (x_i - \text{Mean})^2}{n-1}}
\]
where \( x_i \) are the individual values, and \( n \) is the number of values.

1. Calculate each \( (x_i - \text{Mean})^2 \):
   - \( (16.18 - 16.32)^2 = 0.0196 \)
   - \( (16.21 - 16.32)^2 = 0.0121 \)
   - \( (16.26 - 16.32)^2 = 0.0036 \)
   - \( (16.34 - 16.32)^2 = 0.0004 \)
   - \( (16.38 - 16.32)^2 = 0.0036 \)
   - \( (16.42 - 16.32)^2 = 0.0100 \)
   - \( (16.46 - 16.32)^2 = 0.0196 \)

2. Sum these squared deviations:
   \[
   0.0196 + 0.0121 + 0.0036 + 0.0004 + 0.0036 + 0.0100 + 0.0196 = 0.0689
   \]

3. Divide by \( n-1 = 6 \):
   \[
   \frac{0.0689}{6} \approx 0.0115
   \]

4. Take the square root to find \( s \):
   \[
   s \approx \sqrt{0.0115} \approx 0.11
   \]

So, the standard deviation for Supplier B is approximately **0.11 mm**.

### Summary of Results for Supplier B:
- **Mean** = 16.32 mm
- **Standard Deviation** = 0.11 mm

Would you like to see these calculations in further detail, or do you have other questions? 

Here are some related questions to consider:
1. How do the variances of Supplier A and B compare?
2. What is the probability of a ball bearing falling within the acceptable range?
3. How would an increased standard deviation affect the tolerance?
4. Could these means and standard deviations suggest a preference for one supplier over the other?
5. How would you test if these means are significantly different?

**Tip:** Consistency in manufacturing (lower standard deviation) often implies higher precision, which is crucial for parts with tight tolerances.

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 ±0.1 mm. Recent shipments from the two suppliers had ball bearings with the following diameters. Complete parts (a) through (c).

Supplier A:
16.23, 16.26, 16.31, 16.35, 16.38, 16.41, 16.46

Supplier B:
16.18, 16.21, 16.26, 16.34, 16.38, 16.42, 16.46

Part 1
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.
meanequals=16.34
sequals=0.08
(Round to the nearest hundredth as needed.)

Part 2
Find the mean and standard deviation for the diameters of the ball bearings from Supplier B.
meanequals=enter your response here
sequals=enter your response here
(Round to the nearest hundredth as needed.)

Learn how to find the mean and standard deviation for ball bearings from two suppliers with specified diameters. This problem involves descriptive statistics and provides a step-by-step guide to calculating mean and standard deviation, useful for students in Grades 10-12.

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