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
...
find the mean and standard deviation for the diameters of the ball bearings from supplier a.
mean = 16.35
s = 0.07
(round to the nearest hundredth as needed.)
find the mean and standard deviation for the diameters of the ball bearings from supplier b.
mean =
s =
(round to the nearest hundredth as needed.)

Explanation:

Step1: Calculate the mean for Supplier B

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 B, the data points are 16.17, 16.22, 16.24, 16.33, 16.37, 16.42, 16.47. The sum of these values is $16.17 + 16.22 + 16.24 + 16.33 + 16.37 + 16.42 + 16.47 = 114.22$. There are $n = 7$ data points. So the mean is $\frac{114.22}{7} = 16.32$ (rounded to the nearest hundredth).

Step2: Calculate the standard deviation for Supplier B

The formula for the sample standard deviation $s$ is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. First, we calculate the squared differences from the mean:

• $(16.17 - 16.32)^{2}=(- 0.15)^{2}=0.0225$
• $(16.22 - 16.32)^{2}=(-0.1)^{2}=0.01$
• $(16.24 - 16.32)^{2}=(-0.08)^{2}=0.0064$
• $(16.33 - 16.32)^{2}=(0.01)^{2}=0.0001$
• $(16.37 - 16.32)^{2}=(0.05)^{2}=0.0025$
• $(16.42 - 16.32)^{2}=(0.1)^{2}=0.01$
• $(16.47 - 16.32)^{2}=(0.15)^{2}=0.0225$

The sum of these squared differences is $0.0225+0.01 + 0.0064+0.0001+0.0025+0.01+0.0225 = 0.074$. Then we divide by $n - 1=6$: $\frac{0.074}{6}\approx0.0123$. Taking the square root, we get $s=\sqrt{0.0123}\approx0.11$ (rounded to the nearest hundredth).

Answer:

mean = 16.32
s = 0.11