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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

Determine the following standard normal (z) curve areas: a. The area under the \(z\) curve to the left of 1.75 b. The area under the \(z\) curve to the left of -0.68 c. The area under the \(z\) curve to the right of 1.20 d. The area under the \(z\) curve to the right of -2.82 e. The area under the \(z\) curve between -2.22 and 0.53 f. The area under the \(z\) curve between -1 and 1 g. The area under the \(z\) curve between -4 and 4

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the random variable \(x\) as \(x=\) the number of people who actually show up for a sold-out flight on this plane From past experience, the probability distribution of \(x\) is given in the following table: $$ \begin{array}{|cc|} \hline \boldsymbol{x} & \boldsymbol{p}(\boldsymbol{x}) \\ \hline 95 & 0.05 \\ 96 & 0.10 \\ 97 & 0.12 \\ 98 & 0.14 \\ 99 & 0.24 \\ 100 & 0.17 \\ 101 & 0.06 \\ 102 & 0.04 \\ 103 & 0.03 \\ 104 & 0.02 \\ 105 & 0.01 \\ 106 & 0.005 \\ 107 & 0.005 \\ 108 & 0.005 \\ 109 & 0.0037 \\ 110 & 0.0013 \\ \hline \end{array} $$ a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

State whether each of the following random variables is discrete or continuous: a. The number of defective tires on a car b. The body temperature of a hospital patient c. The number of pages in a book d. The number of draws (with replacement) from a deck of cards until a heart is selected e. The lifetime of a light bulb

6.81 FlightView surveyed 2600 North American airline passengers and reported that approximately \(80 \%\) said that they carry a smartphone when they travel. Suppose that the actual percentage is \(80 \% .\) Consider randomly selecting six passengers and define the random variable \(x\) to be the number of the six selected passengers who travel with a smartphone. The probability distribution of \(x\) is the binomial distribution with \(n=6\) and \(p=0.8\). a. Calculate \(p(4),\) and interpret this probability. b. Calculate \(p(6),\) the probability that all six selected passengers travel with a smartphone. c. Calculate \(P(x \geq 4)\).

6.99 A symptom validity test (SVT) is sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for Posttraumatic Stress Disorder: Application of the Binomial Distribution" (Journal of Anxiety Disorders [2008]: 1297-1302) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is 0.7 (compared to 0.96 for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=0.7\) to calculate various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: $$ \text { i. } \quad P(x=42) $$ ii. \(P(x<42)\) $$ \text { iii. } P(x \leq 42) $$ c. Explain why the probabilities calculated in Part (b) are not all equal. d. The authors calculated the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=0.96,\) they found $$ P(x \leq 42)=.000000000013 $$ Explain why the authors calculated this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.
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