Question

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: 16.24 16.29 16.32 16.34 16.37 16.42 16.44
supplier b: 16.17 16.22 16.24 16.33 16.37 16.42 16.47

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.
mean =
s =
(round to the nearest hundredth as needed.)

Explanation:

Step1: Calculate the mean for Supplier A

The formula for the mean $\bar{x}$ is $\bar{x} = \frac{\sum_{i = 1}^{n} x_{i}}{n}$, where $x_{i}$ are the data points and $n$ is the number of data points. For Supplier A, the data points are 16.24, 16.29, 16.32, 16.34, 16.37, 16.42, 16.44. $n = 7$.
First, sum the data points: $16.24 + 16.29 + 16.32 + 16.34 + 16.37 + 16.42 + 16.44 = 114.42$.
Then, the mean is $\frac{114.42}{7} = 16.3457\approx16.35$ (rounded to the nearest hundredth).

Step2: Calculate the standard deviation for Supplier A

The formula for the sample standard deviation $s$ is $s = \sqrt{\frac{\sum_{i = 1}^{n} (x_{i}-\bar{x})^{2}}{n - 1}}$.
First, find the deviations from the mean for each data point:

• $16.24 - 16.35 = -0.11$, squared: $(-0.11)^{2}=0.0121$
• $16.29 - 16.35 = -0.06$, squared: $(-0.06)^{2}=0.0036$
• $16.32 - 16.35 = -0.03$, squared: $(-0.03)^{2}=0.0009$
• $16.34 - 16.35 = -0.01$, squared: $(-0.01)^{2}=0.0001$
• $16.37 - 16.35 = 0.02$, squared: $(0.02)^{2}=0.0004$
• $16.42 - 16.35 = 0.07$, squared: $(0.07)^{2}=0.0049$
• $16.44 - 16.35 = 0.09$, squared: $(0.09)^{2}=0.0081$

Sum these squared deviations: $0.0121 + 0.0036 + 0.0009 + 0.0001 + 0.0004 + 0.0049 + 0.0081 = 0.0301$.
Then, divide by $n - 1 = 6$: $\frac{0.0301}{6}\approx0.005017$.
Take the square root: $s = \sqrt{0.005017}\approx0.07$ (rounded to the nearest hundredth).

Answer:

mean = $\boxed{16.35}$
s = $\boxed{0.07}$