Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 4

The diameters of ball bearings produced in a factory follow a normal distribution with mean \(6 \mathrm{~mm}\) and standard deviation \(0.04 \mathrm{~mm}\). Calculate the probability that a diameter is (a) more than \(6.05 \mathrm{~mm}\), (b) less than \(5.96 \mathrm{~mm}\), (c) between \(5.98\) and \(6.01 \mathrm{~mm}\).

Short Answer

Expert verified
Answer: The probability that a ball bearing has a diameter between 5.98 mm and 6.01 mm is 0.2902 or 29.02%.

Step by step solution

01

Identify the given values

The problem states that the ball bearing diameters follow a normal distribution with mean (µ) of \(6 \mathrm{~mm}\) and a standard deviation (σ) of \(0.04 \mathrm{~mm}\).
02

Calculate the z-score

For each part of the problem, we need to first calculate the z-score, which is given by the formula: \(z = \frac{x - \mu}{\sigma}\). We can plug in the given values for each of the parts: (a) For diameter more than \(6.05 \mathrm{~mm}\): \(z_a = \frac{6.05 - 6}{0.04} = 1.25\) (b) For diameter less than \(5.96 \mathrm{~mm}\): \(z_b = \frac{5.96 - 6}{0.04} = -1\) (c) For diameter between \(5.98\) and \(6.01 \mathrm{~mm}\): \(z_{c1} = \frac{5.98 - 6}{0.04} = -0.5\) and \(z_{c2} = \frac{6.01 - 6}{0.04} = 0.25\)
03

Calculate probabilities using the standard normal table

Now that we have the z-scores, we can use the standard normal table to look up the probabilities: (a) For diameter more than \(6.05 \mathrm{~mm}\): P(Z > 1.25) = 1 - P(Z ≤ 1.25) = 1 - 0.8944 = 0.1056 (b) For diameter less than \(5.96 \mathrm{~mm}\): P(Z < -1) = 0.1587 (c) For diameter between \(5.98\) and \(6.01 \mathrm{~mm}\): P(-0.5 < Z < 0.25) = P(Z ≤ 0.25) - P(Z ≤ -0.5) = 0.5987 - 0.3085 = 0.2902
04

Write the answers

Now we have the probabilities for the different diameter ranges: (a) The probability that a diameter is more than \(6.05 \mathrm{~mm}\) is 0.1056, or 10.56%. (b) The probability that a diameter is less than \(5.96 \mathrm{~mm}\) is 0.1587, or 15.87%. (c) The probability that a diameter is between \(5.98 \mathrm{~mm}\) and \(6.01 \mathrm{~mm}\) is 0.2902, or 29.02%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Components are manufactured by machines A and B. Machine A makes \(55 \%\) of the components. Of those components made by machine \(\mathrm{A}, 7 \%\) are unacceptable; of those made by machine \(\mathrm{B}, 5 \%\) are unacceptable. A component is picked at random. Calculate the probability that it is (a) made by machine \(\mathrm{B}\) (b) acceptable (c) acceptable and made by machine \(\mathrm{A}\) (d) unacceptable given it is made by machine \(\mathrm{B}\) (e) made by machine A given it is unacceptable.

Find the variance and standard deviation of the following frequency distribution: $$ \begin{array}{rr} \hline x & f \\ \hline 6 & 7 \\ 7 & 3 \\ 8 & 2 \\ 9 & 4 \\ 10 & 2 \\ \hline \end{array} $$

State two properties that a function must have in order to be a probability density function.

The diameters of bearings have a normal distribution with a mean of \(8 \mathrm{~mm}\) and a standard deviation of \(0.04 \mathrm{~mm}\). In a batch of 6000 bearings how many would you expect to have a diameter of (a) more than \(8.03 \mathrm{~mm}\) (b) less than \(7.95 \mathrm{~mm}\) (c) between \(8.01 \mathrm{~mm}\) and \(8.06 \mathrm{~mm}\) (d) more than \(2.5\) standard deviations from the mean?

Find the variance and standard deviation of the following sets of data: (a) 611109789 (b) \(\begin{array}{lllll}5.3 & 7.2 & 9.1 & 8.6 & 5.9 & 7.3\end{array}\) (c) \(-6-6-5-10 &{2} &{1} &{0} &{-2} \end{array}\) Which set has the greatest variation?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Force

Read Explanation

Energy Physics

Read Explanation

Thermodynamics

Read Explanation

Famous Physicists

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free