91Ó°ÊÓ

Ball bearings are manufactured with a mean diameter of 5 millimeter (mm). Because of variability in the manufacturing process, the diameters of the ball bearings are approximately normally distributed with a standard deviation of \(0.02 \mathrm{mm}\) (a) What proportion of ball bearings has a diameter more than \(5.03 \mathrm{mm} ?\) (b) Any ball bearings that have a diameter less than 4.95 mm or greater than \(5.05 \mathrm{mm}\) are discarded. What proportion of ball bearings will be discarded? (c) Using the results of part (b), if 30,000 ball bearings are manufactured in a day, how many should the plant manager expect to discard? (d) If an order comes in for 50,000 ball bearings, how many bearings should the plant manager manufacture if the order states that all ball bearings must be between \(4.97 \mathrm{mm}\) and \(5.03 \mathrm{mm} ?\)

Step by step solution

01

- Understanding the normal distribution

The diameters of the ball bearings are normally distributed with a mean \(\bar{x} = 5 \, \text{mm}\) and standard deviation \(\sigma = 0.02 \, \text{mm}\). We'll use this to calculate the required proportions using Z-scores.

02

- Calculating Z-score for part (a)

To find the proportion of ball bearings with a diameter more than \(5.03 \, \text{mm}\), calculate the Z-score: \[ Z = \frac{X - \mu}{\sigma} = \frac{5.03 - 5}{0.02} = 1.5 \]

03

- Finding the proportion for part (a)

Using Z-tables or normal distribution calculators, find the area to the right of Z=1.5. This gives the proportion of ball bearings with diameter more than \(5.03 \, \text{mm} (P(Z > 1.5) = 0.0668) \).

04

- Calculating Z-scores for part (b)

For ball bearings less than \(4.95 \, \text{mm}\) or greater than \(5.05 \, \text{mm}\), calculate: \[ Z_{4.95} = \frac{4.95 - 5}{0.02} = -2.5 \] and \[ Z_{5.05} = \frac{5.05 - 5}{0.02} = 2.5 \]

05

- Finding proportion for part (b)

Find the areas to the left of \(Z = -2.5\) and to the right of \(Z = 2.5\). From Z-tables, \(P(Z < -2.5) = 0.0062\, \) and \(P(Z > 2.5) = 0.0062\). Adding these gives \(0.0124 \).

06

- Calculating number of discarded bearings for part (c)

With 30,000 bearings manufactured daily, the expected number discarded: \[(0.0124) \times 30,000 = 372 \]

07

- Calculating required Z-scores for part (d)

For bearings between \(4.97 \, \text{mm}\) and \(5.03 \, \text{mm}\): \[ Z_{4.97} = \frac{4.97 - 5}{0.02} = -1.5 \] and \[ Z_{5.03} = \frac{5.03 - 5}{0.02} = 1.5 \]

08

- Finding proportion for bearing range in part (d)

Find the area between \(Z = -1.5\) and \(Z = 1.5\): \[ P(-1.5 < Z < 1.5) = 0.9332 \]

09

- Calculating required manufacture number for part (d)

To fill an order of 50,000 bearings within diameters \(4.97 \, \text{mm}\) to \(5.03 \, \text{mm}\), considering \(0.9332\) pass rate: \[(\frac{50,000}{0.9332}) \approx 53,575\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score

The Z-score is a crucial concept in statistics to understand how data points relate to the mean of a dataset. It measures how many standard deviations a point is from the mean. The formula for calculating the Z-score is:
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Proportion calculation

Proportion calculation is used to determine the fraction of occurrences within a specified range or interval for a dataset. This involves determining how many data points fall within a particular segment. For example, in part (b) of our exercise, we calculated the proportions of ball bearings falling below 4.95 mm and above 5.05 mm.
Firstly, we calculated the Z-scores: * For 4.95 mm: 'A-barcode.jpg' {Download Image}' => '/refs/.arsc_type_b1/batch9002.jpg'}.