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Chapter 1: Problem 28

The discharge of suspended solids from a phosphate mine is normally distributed with mean daily discharge 27 milligrams per liter \((\mathrm{mg} / \mathrm{L})\) and standard deviation \(14 \mathrm{mg} / \mathrm{L}\). In what proportion of the days will the daily discharge be less than \(13 \mathrm{mg} / \mathrm{L} ?\)

Short Answer

Expert verified
About 15.87% of days have a discharge less than 13 mg/L.

Step by step solution

01

Understand the Problem

The problem involves finding the proportion of days that have a discharge less than a specific value. Here, we are looking for the probability that the discharge is less than 13 mg/L, given that it follows a normal distribution with a mean (\(\mu\)) of 27 mg/L and a standard deviation (\(\sigma\)) of 14 mg/L.
02

Standardize the Value

To find this probability, we first need to standardize the discharge level 13 mg/L using the Z-score formula:\[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the value we're interested in, \(\mu = 27\), and \(\sigma = 14\).Substitute the values:\[ Z = \frac{13 - 27}{14} = \frac{-14}{14} = -1 \]
03

Use the Standard Normal Distribution Table

With the Z-score calculated, \(Z = -1\), we can look this value up in the standard normal distribution table (or use a calculator with normal distribution functions) to find the proportion of the distribution that lies to the left of \(Z = -1\).
04

Find the Proportion

Checking the standard normal distribution table, we find that the probability corresponding to \(Z = -1\) is approximately 0.1587. This means that about 15.87% of the days will have a discharge less than 13 mg/L.

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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, also known as Gaussian distribution, is one of the most important probability distributions in statistics. It's the familiar bell-shaped curve that many real-world datasets follow closely. In a normal distribution:
  • The mean, median, and mode are all equal and located at the center of the distribution.
  • The curve is symmetric around the mean.
  • The total area under the curve is equal to 1, signifying total probability.
This distribution is important because it describes how data behaves in many situations. In our example, the discharge of suspended solids from a phosphate mine follows a normal distribution with a specified mean and standard deviation. This helps in predicting or understanding probabilities related to the discharge levels.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. The formula for calculating a Z-score is:\[Z = \frac{X - \mu}{\sigma}\]where:
  • \(X\) is the value being evaluated.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
In our exercise, we used the Z-score to standardize the discharge value of 13 mg/L. By converting this to a Z-score, we can easily use standard normal distribution tables to find probabilities. When calculating, if the Z-score is negative, it means the value is below the mean. A positive Z-score indicates a value above the mean.
Standard Normal Distribution Table
A standard normal distribution table, also known as the Z-table, is a tool that provides the probability that a statistic is below a Z-score in a standard normal distribution. The table makes it possible to understand the percentage of values that lie below a certain Z-score:
  • The table only contains positive Z-scores, so for negative scores, the properties of symmetry are used.
  • Each entry represents an area under the curve to the left of a specific Z-score.
In our example, by looking up a Z-score of -1 in the table, we found that approximately 15.87% of the days show a discharge less than 13 mg/L. This area corresponds to the probability needed.
Mean and Standard Deviation
The mean of a dataset is the average value and serves as the central point of a normal distribution. The standard deviation measures the amount of variation or dispersion in a set of values. Together, these two parameters shape the normal distribution:
  • The mean \(\mu\) shifts the center of the curve left or right on the graph.
  • The standard deviation \(\sigma\) determines the width of the curve; larger values indicate a wider spread.
For the phosphate mine's discharge problem, the mean was 27 mg/L, indicating most days are around this number. The standard deviation was 14 mg/L, showing how much the daily discharge typically differs from the mean. These values allowed us to calculate the Z-score for comparison purposes.

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