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Chapter 7: Problem 8

Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. Customer Service A random sample of weekly work logs at an automobile repair station was obtained, and the average number of customers per day was recorded. $$\begin{array}{lllll} \hline 26 & 24 & 22 & 25 & 23 \\ \hline 24 & 25 & 23 & 25 & 22 \\ \hline 21 & 26 & 24 & 23 & 24 \\ \hline 25 & 24 & 25 & 24 & 25 \\ \hline 26 & 21 & 22 & 24 & 24 \\ \hline \end{array}$$

Short Answer

Expert verified
The data is likely normally distributed if the points in the normal probability plot roughly form a straight line.

Step by step solution

01

- Organize the Data

List all the given values in a single sorted list: 21, 21, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 26, 26.
02

- Calculate Percentiles

Assign a rank to each value. For each rank (i), compute its percentile using the formula: Percentile Rank = \[ \frac{i-0.5}{n} \], where n is the total number of data points (25 in this case).
03

- Determine Z-scores

Convert each percentile rank to a corresponding z-score using a standard normal distribution table.
04

- Construct the Normal Probability Plot

Plot the sorted data values on the y-axis and the corresponding z-scores on the x-axis.
05

- Assess Linearity

Evaluate the plot for linearity. If the points closely follow a straight line, it suggests that the data could have come from a normally distributed population.

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Key Concepts

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

normal distribution
A normal distribution, also known as the Gaussian distribution, is a symmetric, bell-shaped curve where most of the data points cluster around the mean (average). This type of distribution is important in statistics because it naturally describes a variety of random variables.
Key characteristics:
  • The mean, median, and mode of a normal distribution are all equal and located at the center.
  • It has a single peak (unimodal).
  • The tails of the distribution extend infinitely but never touch the x-axis.
  • It is fully defined by its mean and standard deviation.
In the context of the exercise, understanding whether the sample data follows a normal distribution helps in making reliable predictions and decisions based on that data. Observing the weekly work logs through this lens allows us to determine patterns and irregularities in customer service metrics.
z-scores
A z-score tells us how many standard deviations a data point is from the mean. Z-scores are used to understand the position of an individual data point within the distribution.
Formula: The z-score of a value, x, in a dataset is calculated using the formula:
\[ z = \frac{x - \text{mean}}{\text{standard deviation}} \]
Importance of Z-scores:
  • They allow for standardization of data, making it possible to compare values from different distributions.
  • Z-scores help in identifying outliers (data points that are significantly higher or lower than the rest).
  • In the normal probability plot, z-scores help us evaluate how closely our data aligns with the normal distribution.
In the provided solution, calculating z-scores for the data points at each percentile aids in constructing the normal probability plot.
percentiles
Percentiles measure the relative standing of a value within a dataset 鈥� specifically, the value below which a given percentage of data falls.
Use in Statistics:
  • If a score is in the 90th percentile, it means that it is higher than 90% of the other scores.
  • They are useful in comparing datasets and understanding the distribution of data.
Formula: To find the percentile rank of a value, we use:
\[ \text{Percentile Rank} = \frac{i-0.5}{n} \] where 'i' is the rank order of a data point and 'n' is the total number of data points. In our exercise, calculating the percentile ranks for the sorted data points was an essential step preceding the conversion to z-scores. This helps maintain the sequence and establish a foundation for the normal probability plot.
linearity assessment
Linearity assessment involves evaluating the relationship between two variables to see if they form a straight line when plotted. In a normal probability plot, linearity assessment helps determine if the data follows a normal distribution.
Steps:
  • Plot the sorted data values on the y-axis.
  • Plot the corresponding z-scores on the x-axis.
  • Evaluate the alignment of the points.
Interpretation:
  • If the points form a straight line, the data likely follows a normal distribution.
  • If the points deviate significantly from a straight line, the data may not be normally distributed.
Understanding this helps in verifying assumptions about the distribution of your data, which is crucial for many statistical methods and analyses.

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Most popular questions from this chapter

Assume that the random variable \(X\) is normally distributed, with mean \(\mu=50\) and standard deviation \(\sigma=7 .\) Compute the following probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded. \(P(38

The reading speed of sixth-grade students is approximately normal, with a mean speed of 125 words per minute and a standard deviation of 24 words per minute. (a) Draw a normal model that describes the reading speed of sixth-grade students. (b) Find and interpret the probability that a randomly selected sixth-grade student reads less than 100 words per minute. (c) Find and interpret the probability that a randomly selected sixth-grade student reads more than 140 words per minute. (d) Find and interpret the probability that a randomly selected sixth-grade student reads between 110 and 130 words per minute. 0.3189 (e) Would it be unusual for a sixth grader to read more than 200 words per minute? Why?

Ball bearings are manufactured with a mean diameter of 5 millimeters \((\mathrm{mm})\). Because of variability in the manufacturing process, the diameters of the ball bearings are approximately normally distributed, with a standard deviation of \(0.02 \mathrm{~mm}\) (a) What proportion of ball bearings has a diameter more than \(5.03 \mathrm{~mm} ?\) (b) Any ball bearings that have a diameter less than \(4.95 \mathrm{~mm}\) or greater than \(5.05 \mathrm{~mm}\) are discarded. What proportion of ball bearings will be discarded? (c) Using the results of part (b), if 30,000 ball bearings are manufactured in a day, how many should the plant manager expect to discard? (d) If an order comes in for 50,000 ball bearings, how many bearings should the plant manager manufacture if the order states that all ball bearings must be between \(4.97 \mathrm{~mm}\) and \(5.03 \mathrm{~mm} ?\)

Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. O-Ring Thickness A random sample of O-rings was obtained, and the wall thickness (in inches) of each was recorded. $$\begin{array}{lllll} \hline 0.276 & 0.274 & 0.275 & 0.274 & 0.277 \\ \hline 0.273 & 0.276 & 0.276 & 0.279 & 0.274 \\ \hline 0.273 & 0.277 & 0.275 & 0.277 & 0.277 \\ \hline 0.276 & 0.277 & 0.278 & 0.275 & 0.276 \\ \hline \end{array}$$

According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe 300 randomly selected individuals' habits as they sneeze. Use the normal approximation to the binomial to approximate the probability that of the 300 randomly observed individuals: (a) exactly 100 do not cover the mouth when sneezing. (b) fewer than 75 do not cover the mouth. (c) Would you be surprised if, after observing 300 individuals, more than 100 did not cover the mouth when sneezing? Why?
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