/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 If bolt thread length is normall... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is a. Within \(1.5 \mathrm{SDs}\) of its mean value? b. Farther than \(2.5 \mathrm{SD}\) from its mean value? c. Between 1 and \(2 \mathrm{SDs}\) from its mean value?

Short Answer

Expert verified
a. 86.64% b. 0.62% c. 13.59%

Step by step solution

01

Understand the Empirical Rule

For a normally distributed variable, the Empirical Rule (68-95-99.7 Rule) states that approximately 68% of the data falls within one standard deviation (SD) of the mean, 95% within two SDs, and 99.7% within three SDs. This will help in estimating probabilities for the questions.
02

Calculate Probability Within 1.5 SDs

Since 68% of the data lies within 1 SD and 95% within 2 SDs, we recognize that within 1.5 SDs (less than 2 SDs), we need to find the exact value using the standard normal distribution table. Approximately, this is in line with using 86.64%. This implies a probability of 0.8664 (86.64%) for being within 1.5 SDs of the mean.
03

Calculate Probability Farther Than 2.5 SDs

From the normal distribution table, find the cumulative probability for 2.5 SDs, which is about 0.9938. To find the probability of being farther than 2.5 SDs, subtract this from 1 (1 - 0.9938 = 0.0062). Thus, the probability is 0.0062 (0.62%).
04

Calculate Probability Between 1 and 2 SDs

Using the normal distribution table, find the cumulative probabilities at 1 SD (0.8413) and 2 SDs (0.9772). Subtract the cumulative probability for 1 SD from that for 2 SDs to find the probability of being between these two limits (0.9772 - 0.8413 = 0.1359). Thus, the probability is 0.1359 (13.59%).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Empirical Rule
The Empirical Rule, also known as the 68-95-99.7 Rule, is a handy tool when working with normally distributed data. It provides a way to understand how data is spread in relation to the mean. Essentially, this rule tells us:
  • About 68% of the data will fall within one standard deviation (SD) from the mean.
  • Roughly 95% will be within two SDs.
  • An overwhelming 99.7% will lie within three SDs.
This rule helps simplify probability calculations by focusing on these critical thresholds. Imagine a symmetrical bell-shaped curve (that’s the normal distribution). The Empirical Rule tells us how much of the dataset lies close to the center. Therefore, when interpreting probabilities, this rule gives an immediate glimpse into the spread and concentration of the data within these SD ranges.
In our exercise, understanding this rule allows us to quickly gauge how probability calculations shift as we move away from the mean. It becomes a foundational step in solving problems related to normally distributed datasets.
Standard Deviation
The standard deviation is a measure of how spread out numbers in a dataset are. When working with normally distributed datasets, standard deviation (SD) becomes crucial to understanding the variability of data points around the mean. Smaller standard deviations indicate that data points are close to the mean, while larger ones suggest they're spread out more widely.
Imagine a dataset of bolt thread lengths that follows a normal distribution. Here's what happens:
  • If the standard deviation is small, most thread lengths are quite similar, clustering around the average length.
  • If it's large, thread lengths vary significantly, with some much longer or shorter than the average.
In our exercise, the standard deviation helps us define the zones discussed in the Empirical Rule (like within 1.5 SDs or between 1 and 2 SDs). It gives scale to the distances from the mean and is indispensable for precise probability calculations. By knowing the SD, we can better predict the range and typicality of how values are distributed relative to the mean of the dataset. Understanding SD is key to applications like quality control and risk assessment in various fields.
Probability Calculation
Probability can be thought of as the likelihood or chance of an event occurring. In the context of a normal distribution, probability calculations help us determine how likely it is for a data point to fall within a specific range of values. To solve the problems in our exercise, we use the cumulative probabilities from a standard normal distribution table. This table tells us the probability that a data point will fall below a certain number of standard deviations from the mean. Here's how we use it:
  • Within 1.5 SDs: We know from calculations that approximately 86.64% of data is within 1.5 SDs of the mean. This uses the table for precise percentile reads.
  • Farther Than 2.5 SDs: For this, we subtract the cumulative probability beyond 2.5 SDs from 1 (total probability). This tells us only about 0.62% of data falls here.
  • Between 1 and 2 SDs: By finding the cumulative probabilities for both 1 SD (84.13%) and 2 SDs (97.72%), we calculate that 13.59% of data lies between these two points.
These calculations guide us in predicting quantities such as how often a bolt's length might fall outside the norm. Mastering such probability calculations is vital for planning, quality control, and statistical analysis.

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

The paper "Microwave Observations of Daily Antarctic Sea-Ice Edge Expansion and Contribution Rates" (IEEE Geosci. and Remote Sensing Letlers, 2006: 54-58) states that "The distribution of the daily sea-ice advance/retreat from each sensor is similar and is approximately double exponential." The proposed double exponential distribution has density function \(f(x)=.5 \lambda e^{-\lambda|x|}\) for \(-\infty

There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean \(3 \mathrm{~cm}\) and standard deviation \(.1 \mathrm{~cm}\). The second machine produces corks with diameters that have a normal distribution with mean \(3.04 \mathrm{~cm}\) and standard deviation \(.02 \mathrm{~cm}\). Acceptable corks have diameters between \(2.9 \mathrm{~cm}\) and \(3.1 \mathrm{~cm}\). Which machine is more likely to produce an acceptable cork?

Stress is applied to a 20 -in. steel bar that is clamped in a fixed position at each end. Let \(Y=\) the distance from the left end at which the bar snaps. Suppose \(Y / 20\) has a standard beta distribution with \(E(Y)=10\) and \(V(Y)=\frac{100}{7}\). a. What are the parameters of the relevant standard beta distribution? b. Compute \(P(8 \leq Y \leq 12)\). c. Compute the probability that the bar snaps more than 2 in. from where you expect it to.

The special case of the gamma distribution in which \(\alpha\) is a positive integer \(n\) is called an Erlang distribution. If we replace \(\beta\) by \(1 / \lambda\) in Expression (4.8), the Erlang pdf is $$ f(x, \lambda, n)=\left\\{\begin{array}{cl} \frac{\lambda(\lambda x)^{x-1} e^{-\lambda x}}{(n-1) !} & x \geq 0 \\ 0 & x<0 \end{array}\right. $$ It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter \(\lambda\), then the total time \(X\) that elapses before all of the next \(n\) events occur has pdf \(f(x ; \lambda, n)\). a. What is the expected value of \(X\) ? If the time (in minutes) between arrivals of successive customers is exponentially distributed with \(\lambda=.5\), how much time can be expected to elapse before the tenth customer arrives? b. If customer interarrival time is exponentially distributed with \(\lambda=.5\), what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next 30 min? c. The event \(\\{X \leq t\\}\) occurs iff at least \(n\) events occur in the next \(t\) units of time. Use the fact that the number of events occurring in an interval of length \(t\) has a Poisson distribution with parameter \(\lambda t\) to write an expression (involving Poisson probabilities) for the Erlang cdf \(F(t, \lambda, n)=P(X \leq t)\).

The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean value \(40 \mathrm{~V}\) and standard deviation \(1.5 \mathrm{~V}\). a. What is the probability that the voltage of a single diode is between 39 and 42 ? b. What value is such that only \(15 \%\) of all diodes have voltages exceeding that value? c. If four diodes are independently selected, what is the probability that at least one has a voltage exceeding 42 ?

See all solutions

Recommended explanations on Math Textbooks

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.