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A machine at Katz Steel Corporation makes 3 -inch-long nails. The probability distribution of the lengths of these nails is approximately normal with a mean of 3 inches and a standard deviation of \(.1\) inch. The quality control inspector takes a sample of 25 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than \(2.95\) inches or greater than \(3.05\) inches, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 25 nails, the inspector will conclude that the machine needs an adjustment?

Short Answer

Expert verified
The probability that based on a sample of 25 nails, the inspector will conclude that the machine needs an adjustment is approximately \(2.484\%.\)

Step by step solution

01

Identify the parameters of the normal distribution

In this case, the mean (\(\mu\)) is 3 inches, and the standard deviation (\(\sigma\)) is \(.1\) inches.
02

Establish the parameters of the desired range

The quality control inspector will only approve nails that are between \(2.95\) inches and \(3.05\) inches. We should convert these values to standard scores or Z scores. A Z score tells us how many standard deviations an element is from the mean. To find the Z score, subtract the mean from the element in question and divide by the standard deviation. Therefore, the Z scores for \(2.95\) and \(3.05\) inches would be \((2.95-3)/(.1/\sqrt{25})=-2.5\) and \((3.05-3)/(.1/\sqrt{25})=2.5\) respectively.
03

Find the area within the established range

The Z score tells us the area to the left of it. We need to find the area between -2.5 and 2.5. This area is \(.98758 - .01242 = .97516\) which corresponds to a probability of \(97.516\%\).
04

Find the probability that the machine needs an adjustment

The question asks us to find the probability that the machine needs an adjustment. The machine will need an adjustment if the nails are not within the 2.95 to 3.05 inches range. Therefore, we need to subtract our result from Step 3 from 1: \(1-.97516=.02484\). This corresponds to a probability of \(2.484\%\).

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Key Concepts

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

Probability
Probability helps us understand how likely an event is to occur, by assigning it a number between 0 and 1. In this nail-making context, probability gives us a measure of confidence in concluding whether the machine needs an adjustment.
The process involves calculating the likelihood of an undesirable event — in this case, the average nail length being outside the acceptable range. Let's break this down simply:
  • A probability of 0 means the event will not happen.
  • A probability of 1 means the event certainly happens.
In practical terms, if the probability of the machine needing an adjustment is 0.02484 (as calculated), there’s a 2.484% chance that the average nail length from the weekly sample falls outside the specified range of 2.95 to 3.05 inches.
This small probability suggests a low chance of faults but is still crucial for quality control in ensuring product standards are consistently met. In quality control, even small probabilities matter to maintain high standards.
Standard Deviation
Standard deviation is a measure of how spread out the values in a dataset are. It tells us how much variation or dispersion exists from the mean or average. In the problem, a standard deviation of 0.1 inch is given for the length of the nails.
Having a small standard deviation implies that the nail lengths are very close to the mean of 3 inches. This is important for quality because it indicates consistency in production.
Let's make this concept easier to grasp:
  • If the values (or nail lengths) are spread out, the standard deviation is larger.
  • If they are close to the mean, the standard deviation is smaller.
In this exercise, the small standard deviation highlights the precision of the machine in producing nails of the desired length. Quality control depends heavily on maintaining such precision to minimize defective products.
Z-score
The Z-score is a statistical measure that describes a value's relation to the mean of a group of values. It is expressed in terms of standard deviations from the mean. For the given exercise, understanding Z-scores allows us to convert nail lengths into standard units for comparison.
Here's the breakdown:
  • A Z-score tells us how many standard deviations away a particular data point is from the mean.
  • A Z-score of 0 indicates that the data point's score is identical to the mean.
  • Positive or negative Z-scores represent distances above or below the mean, respectively.
In our context, the Z-scores of -2.5 and 2.5 for 2.95 and 3.05 inches show how far these boundary lengths are from the mean of 3 inches. Calculating Z-scores helps in determining the probability that a set of sampled data falls within or outside these given points, thus aiding in quality decisions.

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Most popular questions from this chapter

A machine at Keats Corporation fills 64 -ounce detergent jugs. The probability distribution of the amount of detergent in these jugs is approximately normal with a mean of 64 ounces and a standard deviation of \(.4\) ounce. The quality control inspector takes a sample of 16 jugs once a week and measures the amount of detergent in these jugs. If the mean of this sample is either less than \(63.75\) ounces or greater than \(64.25\) ounces, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 16 jugs, the inspector will conclude that the machine needs an adjustment when actually it does not?

In a January 2014 survey conducted by the Associated PressWe TV, \(68 \%\) of American adults said that owning a home is the most important thing or \(a\) very important but not the most important thing (opportunityagenda.org). Assume that this percentage is true for the current population of American adults. Let \(\hat{p}\) be the proportion in a random sample of 1000 American adults who hold the above opinion. Find the mean and standard deviation of the sampling distribution of \(\hat{p}\) and describe its shape.

The living spaces of all homes in a city have a mean of 2300 square feet and a standard deviation of 500 square feet. Let \(\bar{x}\) be the mean living space for a random sample of 25 homes selected from this city. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\).

A certain elevator has a maximum legal carrying capacity of 6000 pounds. Suppose that the population of all people who ride this elevator have a mean weight of 160 pounds with a standard deviation of 25 pounds. If 35 of these people board the elevator, what is the probability that their combined weight will exceed 6000 pounds? Assume that the 35 people constitute a random sample from the population.

A population of \(N=1400\) has a population proportion equal to .47. In each of the following cases, which formula will you use to calculate \(\sigma_{\hat{p}}\) and why? Using the appropriate formula, calculate \(\sigma_{\hat{p}}\) for each of these cases. a. \(n=90\) b. \(n=50\)

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