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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

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1\). Determine each of the following probabilities: a. \(P(z<0.10)\) b. \(P(z<-0.10)\) c. \(P(0.40-1.25)\) g. \(P(z<-1.50\) or \(z>2.50)\)

Determine each of the following areas under the standard normal (z) curve: a. To the left of \(-1.28\) b. To the right of \(1.28\) c. Between \(-1\) and 2 d. To the right of 0 e. To the right of \(-5\) f. Between \(-1.6\) and \(2.5\) g. To the left of \(0.23\)

Exercise \(7.9\) introduced the following probability distribution for \(y=\) the number of broken eggs in a carton: $$ \begin{array}{lrrrrr} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & .01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y}\) ? Does this surprise you? c. Why doesn't \(\mu_{y}=(0+1+2+3+4) / 5=2.0\). Explain.

You are to take a multiple-choice exam consisting of 100 questions with 5 possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Compute the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

The article "The Distribution of Buying Frequency Rates" (Journal of Marketing Research \([1980]: 210-216)\) reported the results of a \(3 \frac{1}{2}\) -year study of dentifrice purchases. The investigators conducted their research using a national sample of 2071 households and recorded the number of toothpaste purchases for each household participating in the study. The results are given in the following frequency distribution: $$ \begin{array}{cc} \begin{array}{l} \text { Number of } \\ \text { Purchases } \end{array} & \begin{array}{l} \text { Number of House- } \\ \text { holds (Frequency) } \end{array} \\ \hline 10 \text { to }<20 & 904 \\ 20 \text { to }<30 & 500 \\ 30 \text { to }<40 & 258 \\ 40 \text { to }<50 & 167 \\ 50 \text { to }<60 & 94 \\ 60 \text { to }<70 & 56 \\ 70 \text { to }<80 & 26 \\ 80 \text { to }<90 & 20 \\ 90 \text { to }<100 & 13 \\ 100 \text { to }<110 & 9 \\ 110 \text { to }<120 & 7 \\ 120 \text { to }<130 & 6 \\ 130 \text { to }<140 & 6 \\ 140 \text { to }<150 & 3 \\ 150 \text { to }<160 & 0 \\ 160 \text { to }<170 & 2 \\ \hline \end{array} $$ a. Draw a histogram for this frequency distribution. Would you describe the histogram as positively or negatively skewed? b. Does the square-root transformation result in a histogram that is more symmetric than that of the original data? (Be careful! This one is a bit tricky, because you don't have the raw data; transforming the endpoints of the class intervals will result in class intervals that are not necessarily of equal widths, so the histogram of the transformed values will have to be drawn with this in mind.)

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