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Chapter 6: Problem 96

The distribution of the number of items produced by an assembly line during an 8 -hour shift can be approximated by a normal distribution with mean value 150 and standard deviation 10 . a. What is the approximate probability that the number of items produced is at most \(120 ?\) b. What is the approximate probability that at least 125 items are produced? c. What is the approximate probability that between 135 and 160 (inclusive) items are produced?

Short Answer

Expert verified
The approximate probabilities for the given scenarios are as follows: a. The probability that the number of items produced is at most 120 is approximately 0.0013. b. The probability that at least 125 items are produced is approximately 0.9938. c. The probability that between 135 and 160 (inclusive) items are produced is approximately 0.7745.

Step by step solution

01

(Part a: Probability of at most 120 items produced)

To find the probability that the production is at most 120 items, we first calculate the z-score: \(z=\frac{X-碌}{蟽}\) Plugging in the values: \(z=\frac{120-150}{10}\) \(z=-3\) Now, using the standard normal distribution table or calculator, we find the probability P(Z鈮�-3) which is approximately 0.0013. So, the approximate probability that the number of items produced is at most 120 is 0.0013.
02

(Part b: Probability of at least 125 items produced)

To find the probability that the production is at least 125 items, we calculate the z-score: \(z=\frac{X-碌}{蟽}\) Plugging in the values: \(z=\frac{125-150}{10}\) \(z=-2.5\) Now, using the standard normal distribution table or calculator, we find the probability P(Z鈮�-2.5) which is approximately 0.0062. Since we need the probability of at least 125 items produced, we calculate the complementary probability: P(Z>=-2.5) = 1- P(Z鈮�-2.5)= 1-0.0062= 0.9938 So, the approximate probability that at least 125 items are produced is 0.9938.
03

(Part c: Probability of between 135 and 160 items produced)

To find the probability that between 135 and 160 (inclusive) items are produced, we calculate the z-scores for 135 and 160: For X=135: \(z=\frac{135-150}{10}\) \(z=-1.5\) For X=160: \(z=\frac{160-150}{10}\) \(z=1\) Now, using the standard normal distribution table or calculator, we find the probabilities P(Z鈮�1) and P(Z鈮�-1.5) which are approximately 0.8413 and 0.0668, respectively. To find the probability between these z-values, we subtract the smaller probability from the larger one: P(-1.5鈮鈮�1) = P(Z鈮�1) - P(Z鈮�-1.5) = 0.8413 - 0.0668 = 0.7745 So, the approximate probability that between 135 and 160 (inclusive) items are produced is 0.7745.

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Key Concepts

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

Probability Calculations
Understanding probability calculations in a normal distribution is key to solving many statistical problems. Imagine the normal distribution as a bell curve where most occurrences take place around the mean, with fewer instances appearing as you move further from the mean on either side. This principle helps us predict the likelihood of various outcomes within a dataset.
  • To calculate probabilities, we first transform the data into a standard normal distribution format using z-scores.
  • By using z-scores and tables or statistical calculators, we can determine the probability of any given value occurring.
Given a normal distribution with a specific mean and standard deviation, calculate the z-score to convert a raw score into a standard score. Then, use the standard normal distribution table to determine the probability of the event.
Z-Score
A z-score provides a way to understand how far away a particular value is from the mean in terms of standard deviations. This standardization allows for easy comparison across different datasets. The formula to calculate a z-score is simple: \[ z = \frac{X - \mu}{\sigma} \]Where:
  • \( X \) is the value of interest.
  • \( \mu \) is the mean of the distribution.
  • \( \sigma \) is the standard deviation.
For instance, when calculating the z-score for 120 items produced, given that the mean is 150 and the standard deviation is 10, you substitute these values into the formula to get:\[ z = \frac{120 - 150}{10} = -3 \]This z-score tells us that 120 is three standard deviations below the mean.
Standard Deviation
Standard deviation is a crucial concept that reflects the amount of variability or spread in a set of data. A smaller standard deviation indicates that the data points are close to the mean, while a larger one suggests they are spread out over a wider range. In the context of a normal distribution:
  • The mean represents the center of the data or the average.
  • Standard deviations help determine the likelihood of data points appearing at various positions relative to the mean.
A standard deviation of 10, in our exercise, means that the numbers of items produced typically cluster within 10 items either above or below the mean of 150. This variability plays a vital role when determining probabilities for events occurring within the normal distribution.
Complementary Probability
Complementary probability is the concept of finding the probability of the opposite of a wanted event happening. To calculate this, you can subtract the probability of the event occurring from 1. This is helpful when determining probabilities such as "at least". When asked to find the probability of producing at least 125 items, calculate the probability of producing fewer than 125 items and subtract that from 1.For example, if the probability \( P(Z \leq -2.5) \) is 0.0062, then:\[ P(Z \geq -2.5) = 1 - P(Z \leq -2.5) = 1 - 0.0062 = 0.9938 \]This shows the effectiveness of using complementary probability to simplify complex probability calculations.

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Most popular questions from this chapter

A point is randomly selected on the surface of a lake that has a maximum depth of 100 feet. Let \(x\) be the depth of the lake at the randomly chosen point. What are possible values of \(x\) ? Is \(x\) discrete or continuous?

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