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

c. what percentage of ball bearings from each supplier meet specifications?
find the percentage of ball bearings from supplier a that meet specifications.
□% (round to the nearest integer as needed.)

Explanation:

Step1: Determine the acceptable range

The specified diameter is \( 16.30 \) mm with a tolerance of \( \pm 0.1 \) mm. So the lower limit is \( 16.30 - 0.1 = 16.20 \) mm and the upper limit is \( 16.30 + 0.1 = 16.40 \) mm. We need to find the number of ball bearings from Supplier A that have diameters between \( 16.20 \) mm and \( 16.40 \) mm (inclusive).

Step2: Count the valid diameters for Supplier A

Supplier A's diameters: \( 16.24, 16.29, 16.32, 16.34, 16.37, 16.42, 16.44 \)
Check which are between \( 16.20 \) and \( 16.40 \):

• \( 16.24 \): valid
• \( 16.29 \): valid
• \( 16.32 \): valid
• \( 16.34 \): valid
• \( 16.37 \): valid
• \( 16.42 \): \( 16.42 > 16.40 \), invalid
• \( 16.44 \): \( 16.44 > 16.40 \), invalid

So the number of valid diameters for Supplier A is \( 5 \).

Step3: Calculate the percentage

The total number of ball bearings from Supplier A is \( 7 \). The percentage is calculated as \( \frac{\text{Number of valid}}{\text{Total number}} \times 100 \). So \( \frac{5}{7} \times 100 \approx 71.43 \), which rounds to \( 71 \) (or \( 71\% \)). Wait, wait, let's recalculate: \( 5\div7\approx0.7143 \), times 100 is \( 71.43\% \), which rounds to \( 71\% \) or maybe \( 71 \) (but let's check again the diameters:

Wait, Supplier A's diameters: 16.24 (valid, 16.24 ≥16.20 and ≤16.40), 16.29 (valid), 16.32 (valid), 16.34 (valid), 16.37 (valid), 16.42 (16.42 >16.40, invalid), 16.44 (16.44>16.40, invalid). So that's 5 valid out of 7. So \( 5/7 \times 100 \approx 71.43 \), which rounds to 71% (or 71 when rounded to the nearest integer). Wait, but maybe I made a mistake. Wait, 16.40 is the upper limit. So 16.40 is included. So 16.40 is the max. So 16.42 is 16.42, which is more than 16.40, so invalid. 16.44 is also invalid. So 5 valid. So 5/7 ≈71.43, so 71% when rounded to the nearest integer.

Answer:

\( 71 \) (or \( 71\% \))