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Chapter 4: Problem 86

Measurement error that is normally distributed with a mean of 0 and a standard deviation of 0.5 gram is added to the true weight of a sample. Then the measurement is rounded to the nearest gram. Suppose that the true weight of a sample is 165.5 grams. (a) What is the probability that the rounded result is 167 grams? (b) What is the probability that the rounded result is 167 grams or more?

Short Answer

Expert verified
(a) 0.0227; (b) 0.0228.

Step by step solution

01

Understand the Measurement Error

The measurement error is a random variable following a normal distribution with mean \( \mu = 0 \) and a standard deviation \( \sigma = 0.5 \) grams. This error is added to the true weight of 165.5 grams.
02

Define the Measured Weight

The measured weight \( X \) is given by \( X = 165.5 + E \), where \( E \) is the measurement error. Thus, \( X \) is distributed normally with a mean of 165.5 grams and a standard deviation of 0.5 grams.
03

Calculate Rounding Range for Part (a)

The rounded result is 167 grams if the actual measured weight is between 166.5 and 167.5 grams. So, we need to calculate \( P(166.5 \leq X < 167.5) \).
04

Standardize the Normal Variable for Part (a)

Convert the measured weights to a standard normal distribution: \( Z = \frac{X - 165.5}{0.5} \). This gives us \( P\left( \frac{166.5 - 165.5}{0.5} \leq Z < \frac{167.5 - 165.5}{0.5} \right) = P(2 \leq Z < 4) \).
05

Calculate Probability for Part (a) Using Z-table

Using the Z-table, find the probabilities: \( P(Z < 4) \approx 0.9999 \) and \( P(Z < 2) \approx 0.9772 \). Thus, \( P(2 \leq Z < 4) = 0.9999 - 0.9772 = 0.0227 \).
06

Calculate Rounding Threshold for Part (b)

For the result to be 167 grams or more, \( X \) must be at least 166.5 grams. We calculate \( P(X \geq 166.5) \).
07

Standardize and Calculate for Part (b)

Convert to standard normal: \( Z = \frac{166.5 - 165.5}{0.5} = 2 \). So, we need \( P(Z \geq 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 \).
08

Express Final Probabilities

The probability that the rounded result is 167 grams exactly is 0.0227, and the probability that the result is 167 grams or more is 0.0228.

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Key Concepts

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

Normal Distribution
The normal distribution is a continuous probability distribution that is often represented as a bell curve. It's crucial in statistics because many variables naturally follow this type of distribution.

Key characteristics of a normal distribution include:
  • Its shape: Symmetrical around the mean, making both sides equally likely.
  • Two parameters: Mean (\( \mu \)) and standard deviation (\( \sigma \)).
  • Most values are clustered around the mean, with probabilities decreasing as you move away from it.
In our example, the measurement error follows a normal distribution with a mean (\( \mu = 0 \)) and standard deviation (\( \sigma = 0.5 \)). This means errors are equally likely to be above or below zero and are generally small. Understanding this concept helps predict measurement probabilities accurately.
Standard Deviation
Standard deviation (\( \sigma \)) measures the amount of variation or spread in a set of data. In a normal distribution, it helps determine how much the values deviate from the mean (average).

Here's what you need to know:
  • A small standard deviation indicates the data values are concentrated close to the mean.
  • A larger standard deviation shows that values are more spread out over a range.
  • 68% of data in a normal distribution fall within one standard deviation of the mean, 95% within two, and about 99.7% within three.
In our exercise, the standard deviation of 0.5 grams tells us that most of the measurement errors lie within 0.5 grams of the true weight. This precision allows us to calculate the probability of rounding outcomes accurately. When we standardized the measurement (\( Z \)), we could easily use the Z-table to find probabilities.
Rounding Numbers
Rounding is a mathematical technique where numbers are simplified, usually to make them easier to work with or interpret. When dealing with measurements, rounding can impact the final results significantly.

Let's break it down:
  • Rounding to the nearest integer involves adjusting numbers so that they are closer to one of the two nearest whole numbers.
  • For example, if a measured weight is 166.7 grams, it's rounded to 167 grams.
  • Deciding on the rounding cut-off involves understanding probability ranges, such as determining between which values a measurement will round to 167 grams.
In our exercise context, rounding applies to measurement outcomes. By identifying the probability of a measured value falling within these precise ranges, like from 166.5 to 167.5 grams, we determine what the final rounded result will likely be. This allows us to accurately calculate the chances of specific rounding outcomes, adding statistical depth to what might seem like a simple process.

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Most popular questions from this chapter

The sick-leave time of employees in a firm in a month is normally distributed with a mean of 100 hours and a standard deviation of 20 hours. (a) What is the probability that the sick-leave time for next month will be between 50 and 80 hours? (b) How much time should be budgeted for sick leave if the budgeted amount should be exceeded with a probability of only \(10 \% ?\)

The distance between major cracks in a highway follows an exponential distribution with a mean of five miles. (a) What is the probability that there are no major cracks in a 10 -mile stretch of the highway? (b) What is the probability that there are two major cracks in a 10 -mile stretch of the highway? (c) What is the standard deviation of the distance between major cracks? (d) What is the probability that the first major crack occurs between 12 and 15 miles of the start of inspection? (e) What is the probability that there are no major cracks in two separate five-mile stretches of the highway? (f) Given that there are no cracks in the first five miles inspected, what is the probability that there are no major cracks in the next 10 miles inspected?

The length of an injection-molded plastic case that holds magnetic tape is normally distributed with a length of 90.2 millimeters and a standard deviation of 0.1 millimeter. (a) What is the probability that a part is longer than 90.3 millimeters or shorter than 89.7 millimeters? (b) What should the process mean be set at to obtain the highest number of parts between 89.7 and 90.3 millimeters? (c) If parts that are not between 89.7 and 90.3 millimeters are scrapped, what is the yield for the process mean that you selected in part (b)? Assume that the process is centered so that the mean is 90 millimeters and the standard deviation is 0.1 millimeter. Suppose that 10 cases are measured, and they are assumed to be independent. (d) What is the probability that all 10 cases are between 89.7 and 90.3 millimeters? (e) What is the expected number of the 10 cases that are between 89.7 and 90.3 millimeters?

Errors caused by contamination on optical disks occur at the rate of one error every \(10^{5}\) bits. Assume that the errors follow a Poisson distribution. (a) What is the mean number of bits until five errors occur? (b) What is the standard deviation of the number of bits until five errors occur? (c) The error-correcting code might be ineffective if there are three or more errors within \(10^{5}\) bits. What is the probability of this event?

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function \(f(x)=2 / x^{3}\) for \(x>1 .\) Determine the following: (a) \(P(X<2)\) (b) \(P(X>5)\) (c) \(P(48)\) (e) \(x\) such that \(P(X
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