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Suppose that the distribution of \(\mathrm{pH}\) readings for soil samples taken in a certain geographic region can be approximated by a normal distribution with mean \(6.00\) and standard deviation \(0.10 .\) The \(\mathrm{pH}\) of a randomly selected soil sample from this region is to be determined. a. What is the probability that the resulting \(\mathrm{pH}\) is between \(5.90\) and \(6.15 ?\) b. What is the probability that the resulting \(\mathrm{pH}\) exceeds \(6.10 ?\) c. What is the probability that the resulting \(\mathrm{pH}\) is at most \(5.95 ?\) d. Describe the largest \(5 \%\) of the \(\mathrm{pH}\) distribution.

Short Answer

Expert verified
a. The probability that the resulting pH is between 5.90 and 6.15 is 0.7745. b. The probability that the resulting pH exceeds 6.10 is 0.1587. c. The probability that the resulting pH is at most 5.95 is 0.3085. d. The largest 5% of the pH distribution corresponds to pH values greater than 6.1645.

Step by step solution

01

Find the probability that the resulting pH is between 5.90 and 6.15

First, convert the pH values to Z-scores using the formula Z = (X - μ) / σ where X is the value, μ is the mean, and σ is the standard deviation. For 5.90, Z = (5.90 - 6.00) / 0.10 = -1.00. For 6.15, Z = (6.15 - 6.00) / 0.10 = 1.50. Refer to a Z-table to find the probability associated with these Z-scores. The probability that Z ≤ 1.50 is 0.9332. The probability that Z ≤ -1.00 is 0.1587. The probability that pH is between 5.90 and 6.15 is therefore 0.9332 - 0.1587 = 0.7745.
02

Find the probability that the resulting pH exceeds 6.10

Again, convert the pH value to a Z-score. For 6.10, Z = (6.10 - 6.00) / 0.10 = 1.00. The probability that Z ≤ 1.00 is 0.8413. Since we want the probability that pH exceeds 6.10, we subtract this value from 1: 1 - 0.8413 = 0.1587.
03

Find the probability that the resulting pH is at most 5.95

Convert the pH value 5.95 to a Z-score: Z = (5.95 - 6.00) / 0.10 = -0.50. The probability that Z ≤ -0.50 is 0.3085. Therefore, the probability that the pH is at most 5.95 is 0.3085.
04

Describe the largest 5% of the pH distribution

The largest 5% of the pH distribution can be found by finding the Z-score associated with a probability of 0.95 (since the largest 5% would leave 95% below that). From the Z-table, this Z-score is approximately 1.645. Now convert this back to a pH value using the formula X = μ + Z * σ = 6.00 + 1.645 * 0.10 = 6.1645. Therefore, the largest 5% of the pH distribution corresponds to pH values greater than 6.1645.

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Key Concepts

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

Z-score Calculation
Understanding Z-score calculation is essential in statistics when dealing with the normal distribution. The Z-score, or standard score, indicates how many standard deviations an element is from the mean. The calculation is straightforward:
For any given value X, the Z-score is found using the formula
\[ Z = \frac{(X - \mu)}{\sigma} \]
where \( \mu \) is the mean and \( \sigma \) is the standard deviation of the distribution. In the context of our soil sample problem, if we want to calculate the Z-score for a \( \text{pH} \) reading of 5.90, and we know that the mean \( \text{pH} \) is 6.00 with a standard deviation of 0.10, the Z-score would be
\[ Z = \frac{(5.90 - 6.00)}{0.10} = -1.00 \]
This negative Z-score indicates that the value is below the mean. Similarly, a positive Z-score means the value is above the mean.
Normal Distribution Properties
The normal distribution is a continuous probability distribution that is symmetrical around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. One of the most remarkable properties of a normal distribution is its characteristic bell curve shape.
Some key properties include:
  • The mean, median, and mode of a normal distribution are equal.
  • The curve is symmetric at the center (around the mean, \( \mu \)).
  • Approximately 68% of the data falls within one standard deviation of the mean. Likewise, 95% falls within two standard deviations, and 99.7% falls within three standard deviations, known as the Empirical Rule.
  • The total area under the curve is 1, representing a probability of 100%.
Using these properties, we can determine various probabilities associated with the distribution, as was done in solving the pH example from our text.
Statistical Probability
The field of statistical probability allows us to quantify the likelihood of events occurring, under a given set of circumstances, using various probability distributions, of which the normal distribution is one. In a practical sense, statistical probabilities are frequently represented by areas under a probability curve.
For our soil sample problem, we can use the calculated Z-scores to determine the likelihood of different \( \text{pH} \) levels. For instance, the probability of obtaining a \( \text{pH} \) greater than 6.10 is found by subtracting the area under the curve to the left of the Z-score from one. Through the concept of statistical probability, we can translate a given Z-score into a percentile rank, which in turn corresponds to an actual probability.
To sum up, statistical probabilities help us to make predictions and decisions based on data. Understanding how to use and interpret Z-scores and the properties of the normal distribution, allows us to assess risks and probabilities in everyday decisions, scientific studies, and business endeavors.

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Most popular questions from this chapter

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean \(45 \mathrm{~min}\) and standard deviation \(5 \mathrm{~min}\). a. If \(50 \mathrm{~min}\) is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if we wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

Because \(P(z<0.44)=.67,67 \%\) of all \(z\) values are less than \(0.44\), and \(0.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the \(30 \mathrm{th} z\) percentile?

Let \(x\) denote the duration of a randomly selected pregnancy (the time elapsed between conception and birth). Accepted values for the mean value and standard deviation of \(x\) are 266 days and 16 days, respectively. Suppose that a normal distribution is an appropriate model for the probability distribution of \(x\). a. What is the probability that the duration of pregnancy is between 250 and 300 days? b. What is the probability that the duration of pregnancy is at most 240 days? c. What is the probability that the duration of pregnancy is within 16 days of the mean duration? d. A Dear Abby column dated January 20,1973 , contained a letter from a woman who stated that the duration of her pregnancy was exactly 310 days. (She wrote that the last visit with her husband, who was in the navy, occurred 310 days before the birth.) What is the probability that the duration of a pregnancy is at least 310 days? Does this probability make you a bit skeptical of the claim? e. Some insurance companies will pay the medical expenses associated with childbirth only if the insurance has been in effect for more than 9 months ( 275 days). This restriction is designed to ensure that the insurance company has to pay benefits only for those pregnancies for which conception occurred during coverage. Suppose that conception occurred 2 weeks after coverage began. What is the probability that the insurance company will refuse to pay benefits because of the 275 -day insurance requirement?

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of \(120 \mathrm{sec}\) and a standard deviation of \(20 \mathrm{sec}\). The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Accurate labeling of packaged meat is difficult because of weight decrease resulting from moisture loss (defined as a percentage of the package's original net weight). Suppose that the normal distribution with mean value \(4.0 \%\) and standard deviation \(1.0 \%\) is a reasonable model for the variable \(x=\) moisture loss for a package of chicken breasts. (This model is suggested in the paper "Drained Weight Labeling for Meat and Poultry: An Economic Analysis of a Regulatory Proposal," Journal of Consumer Affairs \([1980]: 307-325 .)\) a. What is the probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) ? b. What is the probability that \(x\) is at most \(4.0 \%\) ? c. What is the probability that \(x\) is at least \(7 \%\) ? d. Describe the largest \(10 \%\) of the moisture loss distribution.

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