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Suppose that the \(\mathrm{pH}\) of soil samples taken from a certain geographic region is normally distributed with a mean \(\mathrm{pH}\) of \(6.00\) and a standard deviation of \(0.10 .\) If the \(\mathrm{pH}\) of a randomly selected soil sample from this region is determined, answer the following questions about it: 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. What value will be exceeded by only \(5 \%\) of all such pH values?

Short Answer

Expert verified
a. The probability that the pH is between 5.90 and 6.15 is approximately 0.5328. b. The probability that the pH exceeds 6.10 is about 0.1587. c. The probability that the pH is at most 5.95 is approximately 0.0668. d. The pH value that will be exceeded by only 5% of all such pH values is approximately 6.13.

Step by step solution

01

Calculate Z-Scores

Convert the given pH values into z-scores. The formula to calculate z-scores is \(Z = \frac{X - \mu}{\sigma}\), where X is the given pH value, \(\mu = 6.00\) is the mean and \(\sigma = 0.10\) is the standard deviation.
02

Find Probability for part (a)

Substitute X=5.90 and X=6.15 into the Z-score formula to calculate z-scores. Look up these z-scores in a standard normal distribution table or use a calculator to find the respective probabilities. Then subtract the smaller probability from the larger one to get the probability that the pH value lies between these two values.
03

Find Probability for part (b)

Substitute X=6.10 into the Z-score formula. Use the normal distribution table or calculator to find the probability for this z-score. As the question is about exceeding this value, subtract the obtained probability from 1.
04

Find Probability for part (c)

Substitute X=5.95 into the z-score formula. Use the normal distribution table or calculator to find the probability for this z-score. Here we are interested in the pH value being at most 5.95, so the obtained probability is the solution.
05

Find pH value for part (d)

Here we are asked for the pH value that will be exceeded by only 5% of all such pH values. This translates to finding a z-score such that the area to its right under the standard normal curve is 0.05. Use the standard normal distribution table or calculator to find this z-score. Once you have the z-score, use the formula \(X = \mu + Z \cdot \sigma\) to find the pH value.

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Key Concepts

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

Z-score calculation
Understanding the Z-score is essential in statistics, particularly when dealing with normal distributions. A Z-score represents how many standard deviations away a particular data point is from the mean. To calculate it, use the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value you're considering, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In the case of pH level measurement, if the soil sample's pH level is 6.10 and the average pH is 6.00 with a standard deviation of 0.10, applying the formula \( Z = \frac{6.10 - 6.00}{0.10} \) gives us a Z-score of 1. This means the soil sample's pH is one standard deviation above the average pH level for the region. Calculating the Z-score is the first step in quantifying how unusual or typical a specific value is within the normal distribution.
Probability of a range in normal distribution
To find the probability of a variable falling within a certain range in a normal distribution, we use the Z-scores of the range's boundaries. For instance, with the soil sample pH levels, we can calculate the probability that a sample will have a pH between 5.90 and 6.15.

We first convert these pH values to Z-scores using the formula mentioned in the previous section. Then we consult a standard normal distribution table or a calculator to find the probabilities associated with each Z-score. The final probability is the difference between these two probabilities, which effectively gives us the likelihood that the pH level will lie within that specified range.
Standard deviation in statistics
Standard deviation is a fundamental concept in statistics, measuring the amount of variance or dispersion present in a set of data values. It is the square root of the variance and provides insight into the typical distance between the data points and the mean.

In practical terms, a low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation suggests that the data points are spread out over a larger range of values. In the context of the soil pH levels, a standard deviation of 0.10 means that the vast majority of pH readings are expected to be within a narrow band around the average value, reflecting minor variability in acidity or alkalinity across the samples.
pH level statistical analysis
Statistical analysis of pH levels involves interpreting the acidity or alkalinity measurements within a context, such as a geographic region. Typically, the pH scale ranges from 0 to 14, with values below 7 being acidic and above 7 being alkaline. The mean pH and standard deviation help us understand how concentrated the pH values are around a central value.

For example, if only 5% of the values are expected to exceed a certain pH level, this can have significant implications for environmental studies, agriculture, and land management. Understanding the dispersion of pH levels using statistical analysis tools like the Z-score and normal distribution allows scientists and researchers to make informed decisions and predictions based on the acidity or alkalinity of the soil in a given area.

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Most popular questions from this chapter

A new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(80 \%\) of all batteries have acceptable voltages, and let \(y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(y=2)\) ? b. What is \(p(3) ?\) (Hint: There are two different outeomes that result in \(y=3 .\) ) c. In order to have \(y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(y=5\), and then determine \(p(5)\). d. Use the pattern in your answers for Parts (a)-(c) to obtain a general formula for \(p(y)\).

The Los Angeles Times (December 13,1992 ) reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost \(80 \%\) did so. Suppose that for a particular route the actual percentage is exactly \(80 \%\), and consider randomly selecting six passengers. Then \(x\), the number among the selected six who rested or slept, is a binomial random variable with \(n=6\) and \(\pi=.8\). a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6)\), the probability that all six selected passengers rested or slept. c. Determine \(P(x \geq 4)\).

Let \(x\) denote the amount of gravel sold (in tons) during a randomly selected week at a particular sales facility. Suppose that the density curve has height \(f(x)\) above the value \(x\), where $$ f(x)=\left\\{\begin{array}{ll} 2(1-x) & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ The density curve (the graph of \(f(x)\) ) is shown in the following figure: Use the fact that the area of a triangle \(=\frac{1}{2}\) (base)(height) to calculate each of the following probabilities: a. \(P\left(x<\frac{1}{2}\right)\) b. \(P\left(x \leq \frac{1}{2}\right)\) c. \(P\left(x<\frac{1}{4}\right)\) d. \(P\left(\frac{1}{4}

Starting at a particular time, each car entering an intersection is observed to see whether it turns left (L) or right (R) or goes straight ahead (S). The experiment terminates as soon as a car is observed to go straight. Let \(y\) denote the number of cars observed. What are possible \(y\) values? List five different outcomes and their associated \(y\) values.

Suppose that fuel efficiency for a particular model car under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this type is between 29 and \(31 \mathrm{mpg}\) ? b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg}\) ? c. If three cars of this model are randomly selected, what is the probability that all three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(c\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(c\) (i.e., \(P(x>c)=.95\) ).

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