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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

Classify each of the following random variables as either discrete or continuous: a. The fuel efficiency (miles per gallon) of an automobile b. The amount of rainfall at a particular location during the next year c. The distance that a person throws a baseball d. The number of questions asked during a 1-hour lecture e. The tension (in pounds per square inch) at which a tennis racket is strung f. The amount of water used by a household during a given month g. The number of traffic citations issued by the highway patrol in a particular county on a given day

6.89 Suppose that \(20 \%\) of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of size 2000 have (approximately) a binomial distribution? Explain.

Consider the random variable \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose the probability distribution of \(y\) is as follows: \(\begin{array}{cccccc}y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ?\end{array}\) a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than \(0 .\) What is \(p(4) ?\) b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

The paper referenced in Example 6.21 suggested that a normal distribution with mean 3500 grams and standard deviation 550 grams is a reasonable model for birth weights of babies born in Canada. a. One common medical definition of a large baby is any baby that weighs more than 4000 grams at birth. What is the probability that a randomly selected Canadian baby is a large baby? b. What is the probability that a randomly selected Canadian baby weighs either less than 2000 grams or more than 4000 grams at birth? c. What birth weights describe the \(10 \%\) of Canadian babies with the greatest birth weights? 6.52 The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

A company makes hardwood flooring, which it sells in boxes that will cover 500 square feet of floor. Let \(x=\) the number of boxes ordered by a randomly chosen customer. Suppose the probability distribution of \(x\) is as follows: \(x \quad 1\) \(2 \quad 3\) 4 \(p(x)\) 0.2 0.4 0.3 12 a. Calculate and interpret the mean value of \(x\). b. Calculate and interpret the variance and standard deviation of
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