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.