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Chapter 6: Problem 138

Assume that the distribution of data in Exercise 6.137 was exactly normally distributed, with a mean of 0.00 and a standard deviation of 0.020. a. Find the bounds of the middle \(95 \%\) of the distribution. b. What percent of the data is actually within the interval found in part a? c. Using \(z\) -scores, determine the percentage of estimated conformance to specification. That is, what percentage of the measurements would be expected to fall within the specification range of \(0.000 \pm 0.030\) units?

Short Answer

Expert verified
a. The bounds for the middle 95% of the distribution are approximately -0.0392 and 0.0392 units. b. 95% of the data is within these bounds. c. Approximately 99.87% of the measurements are within the specification range of 卤0.030 units.

Step by step solution

01

Calculation of Bounds of the Middle 95% of the Distribution

To find the bounds of the middle 95% of a normally distributed data set, use the z-scores for 2.5% and 97.5% (since 5% is divided on either end of the distribution). For a normal distribution, these z-scores are -1.96 and 1.96. Multiply each z-score by the standard deviation (0.020) and add the mean (0) to calculate the bounds. Consequently, lower_bound = mean + (-1.96 * standard_deviation) and upper_bound = mean + (1.96 * standard_deviation)
02

Verification of the Percentage Data Within the Interval

Since the z-scores of -1.96 and 1.96 represent the bounds of the middle 95% of a normal distribution, it is confirmed that 95% of the data lies within these bounds.
03

Calculation of z-scores for Specification Range

To find the percentage of measurements expected to fall within the range of 卤0.030 units, first calculate the z-score of the limits of this range: minus_limit_z = (minus_limit - mean) / standard_deviation and plus_limit_z = (plus_limit - mean) / standard_deviation, where minus_limit = 0 - 0.030, plus_limit = 0 + 0.030. The z-scores correspond to areas under the normal distribution curve. By looking these areas up in a Standard Normal Table or using a calculator, find the proportion of the data within these bounds.

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Key Concepts

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

Understanding Z-scores
Z-scores are a handy way to see where a specific value stands in a normal distribution. They help us determine how far away a measurement is from the mean. When you have a value, you can find its corresponding z-score using the formula: \( z = \frac{(X - \text{mean})}{\text{standard deviation}} \).
  • Here, \(X\) is your data point, \(\text{mean}\) is the average value of your dataset, and \(\text{standard deviation}\) measures the spread of data.
  • A positive z-score means the data point is above the mean, while a negative score indicates it's below the mean.
  • Z-scores help us understand what percentage of the data falls within a specific range around the mean by allowing us to use standard normal distribution tables.

For example, to find the middle \(95\%\) of data in a normal distribution like the one in the exercise, we look up z-scores of \(-1.96\) and \(1.96\). This tells us the data between these z-scores accounts for \(95\%\) of the sample.
Exploring Standard Deviation
Standard deviation is a key concept in statistics that measures how spread out numbers are in a dataset. When data points are close to the mean, the standard deviation is low; when they are spread out, it is higher.
  • Mathematically, it is found using the square root of the variance.
  • It allows us to understand the variability or diversity of the data.
  • A low standard deviation in a normal distribution suggests less variability, while a high value indicates more spread.

In the exercise's context, a standard deviation of \(0.020\) tells us the typical distance each data point is from the mean. By multiplying z-scores by the standard deviation, we can calculate how extreme or typical a data point is.
Calculating the Mean
The mean, often referred to as the average, represents the central point of a given dataset. It is calculated by summing up all data points and then dividing by the total number of points.
  • The mean gives us a single value that summarizes the dataset.
  • It is sensitive to very high or very low values, which can skew the dataset.
  • Understanding the mean is essential when working with normal distributions as it acts as a balancing point, where data above and below are evenly distributed.

In the exercise provided, the mean is \(0\), which means our normal distribution is centered right at this point. Every calculation concerning z-scores and bounds into the distribution stems from this average value.

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Most popular questions from this chapter

Given that \(x\) is a normally distributed random variable with a mean of 60 and a standard deviation of 10 find the following probabilities. a. \(\quad P(x>60)\) b. \(\quad P(60< x<72)\) c. \(\quad P(57< x<83)\) d. \(\quad P(65< x<82)\) e. \(\quad P(38< x<78)\) f. \(\quad P(x<38)\)

Given that \(z\) is the standard normal variable, find the value of \(k\) such that a. \( P(|z|>1.68)=k\) b. \(P(|z|<2.15)=k\)

In order to see what happens when the normal approximation is improperly used, consider the binomial distribution with \(n=15\) and \(p=0.05 .\) since \(n p=0.75\) the rule of thumb \((n p>5 \text { and } n q>5)\) is not satisfied. Using the binomial tables, find the probability of one or fewer successes and compare this with the normal approximation.

In a study of the length of time it takes to play Major League Baseball games during the early 2008 season, the variable "time of game" appeared to be normally distributed, with a mean of 2 hours 49 minutes and a standard deviation of 21 minutes. a. Some fans describe a game as "unmanageably long" if it takes more than 3 hours. What is the probability that a randomly identified game was unmanageably long? b. Many fans describe a game lasting less than 2 hours, 30 minutes as "quick." What is the probability that a randomly selected game was quick? c. What are the bounds of the interquartile range for the variable time of game? d. What are the bounds for the middle \(90 \%\) of the variable time of game?

Find the \(z\)-score that forms the upper boundary for the lower \(20 \%\) of a normal distribution.
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