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Chapter 4: Problem 30

Find the following percentiles for the standard normal distribution. Interpolate where appropriate. a. 91 st b. 9 th c. 75 th d. 25 th e. 6 th

Short Answer

Expert verified
Use the Z-table to find z-scores: 91st ~ 1.34, 9th ~ -1.34, 75th ~ 0.67, 25th ~ -0.67, 6th ~ -1.55.

Step by step solution

01

Understanding Percentiles and the Z-Table

Percentiles indicate the value below which a given percentage of observations in a dataset fall. For a standard normal distribution, we use the Z-table to determine the z-score that corresponds to a specific percentile.
02

Finding the Z-Score for the 91st Percentile

Locate 0.9100 in the cumulative probability section of the Z-table. The closest value is typically found, and its corresponding z-score is recorded. If 0.9100 falls between two values, interpolate between the two closest Z-values.
03

Finding the Z-Score for the 9th Percentile

Locate 0.0900 in the cumulative probability section of the Z-table. Record the corresponding z-score, interpolating if 0.0900 falls between two table values.
04

Finding the Z-Score for the 75th Percentile

Find 0.7500 in the cumulative probability section of the Z-table. The corresponding z-score is noted, interpolating if necessary.
05

Finding the Z-Score for the 25th Percentile

Identify 0.2500 in the cumulative probability section of the Z-table and note the z-score, using interpolation if 0.2500 falls between two values.
06

Finding the Z-Score for the 6th Percentile

Locate 0.0600 in the Z-table and note the corresponding z-score, interpolating if 0.0600 falls between two table entries.
07

Final Review and Conclusion

Verify all calculated z-scores with the Z-table to ensure accuracy, ensuring interpolation methods were applied correctly if needed.

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Key Concepts

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

Percentiles
Percentiles are a way to understand the distribution of a data set by dividing up the values into 100 equal parts. If you think of a percentile as a score on a test, a score like the 91st percentile indicates that 91% of scores are below it. Here's how percentiles work within any distribution:
  • The 50th percentile is the median. Half of the values are above it, and half are below.
  • The 25th and 75th percentiles are known as the first and third quartiles, framing the middle 50% of your data.
  • Low percentiles (e.g., 9th) are closer to the minimum, while high percentiles (e.g., 91st) approach the maximum.
In modeling data that follows a standard normal distribution (a bell curve), percentiles are translated into z-scores to pinpoint precise values on this curve.
Z-table
The Z-table is a crucial tool for statistics, detailing the cumulative probabilities of the standard normal distribution. It's a matrix that helps you find the area under the normal curve to the left of a given z-score. The Z-table makes it easy to convert percentiles into z-scores, which represent the number of standard deviations a given value is from the mean. To use it effectively:
  • Find your desired percentile in the form of a probability (e.g., 0.9100 for 91%).
  • Locate this probability in the Z-table, which is arranged in rows and columns representing z-scores.
  • Each cell in the table reveals the probability from the left-most end of the curve up to a specific z-score.
Remember, the Z-table only gives probabilities for the positive side of the curve, so understanding both sides is vital for complete data interpretation.
Z-score
The z-score is a standard score indicating how many standard deviations an element is from the mean of the distribution. In a standard normal distribution, the mean is zero, and the z-score provides a precise location on the bell curve:
  • A z-score of 0 is the mean value.
  • Positive z-scores fall above the mean, moving right on the bell curve.
  • Negative z-scores are below the mean, moving left on the curve.
Z-scores are crucial for converting raw values from any normal distribution into comparable scores across different datasets. They standardize the values, facilitating easier cross-comparison between different data sets or different attributes within the same set.
Interpolation
Interpolation is a mathematical method used to estimate unknown values that lie between known data points. In the context of a Z-table, interpolation comes in when the exact percentile you need does not directly match an entry in the table. Here's how to interpolate between two values:
  • Identify the two closest cumulative probabilities around your target percentile.
  • Determine the z-scores corresponding to these probabilities.
  • Calculate the interpolated z-score by averaging the differences proportionally based on how close the target percentile is to these values.
This tool enhances precision when dealing with tables that cannot list every possible probability or when working with very fine percentile points, ensuring your calculations remain accurate and meaningful.

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Most popular questions from this chapter

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research, 1984: 1169-1174) suggests the uniform distribution on the interval \((7.5,20)\) as a model for depth \((\mathrm{cm})\) of the bioturbation layer in sediment in a certain region. a. What are the mean and variance of depth? b. What is the cdf of depth? c. What is the probability that observed depth is at most 10 ? Between 10 and 15 ? d. What is the probability that the observed depth is within 1 standard deviation of the mean value? Within 2 standard deviations?

A theoretical justification based on a certain material failure mechanism underlies the assumption that ductile strength \(X\) of a material has a lognormal distribution. Suppose the parameters are \(\mu=5\) and \(\sigma=.1\). a. Compute \(E(X)\) and \(V(X)\). b. Compute \(P(X>125)\). c. Compute \(P(110 \leq X \leq 125)\). d. What is the value of median ductile strength? e. If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength of at least 125 ? f. If the smallest \(5 \%\) of strength values were unacceptable, what would the minimum acceptable strength be?

Suppose the reaction temperature \(X\) (in \({ }^{\circ} \mathrm{C}\) ) in a certain chemical process has a uniform distribution with \(A=-5\) and \(B=5 .\) a. Compute \(P(X<0)\). b. Compute \(P(-2.5

The automatic opening device of a military cargo parachute has been designed to open when the parachute is \(200 \mathrm{~m}\) above the ground. Suppose opening altitude actually has a normal distribution with mean value \(200 \mathrm{~m}\) and standard deviation \(30 \mathrm{~m}\). Equipment damage will occur if the parachute opens at an altitude of less than \(100 \mathrm{~m}\). What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

If \(X\) has an exponential distribution with parameter \(\lambda\), derive a general expression for the \((100 p)\) th percentile of the distribution. Then specialize to obtain the median.
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