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Assume college women's heights are approximately Normally distributed with a mean of 65 inches and a standard deviation of \(2.5\) inches. Choose the StatCrunch output for finding the percentage of college women who are taller than 67 inches and report the correct percentage. Round to one decimal place.

Short Answer

Expert verified
Approximately 21.2% of college women are taller than 67 inches.

Step by step solution

01

Calculate Z-score

To calculate the z-score, subtract the mean from the given value and divide by the standard deviation. The formula for the z-score is \(z = (X - μ) / σ\). Here, \(X\) is 67 inches (the given height), \(μ\) is 65 inches (the mean height) and \(σ\) is 2.5 inches (the standard deviation). Thus, the z-score is \(z = (67 - 65) / 2.5 = 0.8.\)
02

Use a Z-table

Subsequently, one looks up this z-value in a standard normal distribution table (also known as a z-table). The value for 0.8 in the z-table is 0.7881. This means that 78.81% of women are shorter than 67 inches.
03

Calculate the percentage of women who are taller than 67 inches

Since we know that 78.81% of women are shorter than 67 inches (from above), we subtract this percentage from 100% to get the percentage of women who are taller than 67 inches. This is calculated as \(100% - 78.81% = 21.19%.\)

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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 crucial in statistics, as it provides a way to measure the distance between a particular value and the population mean in units of the standard deviation. In other words, it tells us how many standard deviations an element is from the mean.

Let's break down the steps of calculating a z-score using the example from the exercise: we have college women's heights, with a mean of 65 inches and a standard deviation of 2.5 inches. To find the z-score for a woman who is 67 inches tall, we subtract the mean from the observed value, 67 inches, and divide by the standard deviation. The formula for this calculation looks like this:
\[ z = \frac{(X - \mu)}{\sigma} \]
By plugging in the values, we get:
\[ z = \frac{(67 - 65)}{2.5} = 0.8 \]
This result tells us the woman's height is 0.8 standard deviations above the mean. This step is a fundamental part of statistical data analysis because it standardizes different data points, allowing for comparison across different distributions or datasets.
Standard Normal Distribution Table
Once we have the z-score, the next step is to interpret this score. This is where the standard normal distribution table, often referred to as the z-table, comes into play. This table provides the probability or area under the curve to the left of a specified z-score in a standard normal distribution. The standard normal distribution is a special case of the normal distribution that has a mean of 0 and a standard deviation of 1.

To consult the table, one would find the row for the first two digits of the z-score and the column corresponding to the second decimal place, if available. In the case of a z-score of 0.8, the table gives us a value of 0.7881. This number represents the probability that a randomly selected woman's height is less than 67 inches, telling us that roughly 78.81% of college women fall below that height.

Indeed, the z-table is an invaluable tool for statisticians and analysts alike, letting them ascertain probabilities and percentiles without complex calculations. If improvement is needed, practice using the z-table with varying scores to build familiarity and fluency.
Statistical Data Analysis
The final stage in our process is the application of our findings from the z-score and z-table to perform the data analysis. The goal is to make a real-world interpretation of the statistical results. In our exercise, we needed to find out what percentage of college women are taller than 67 inches.

We already discovered that about 78.81% are shorter, thanks to the z-table. To determine the remaining percentage, we simply subtract from 100%, as the total probability of any event (such as being taller or shorter than a given height) must equal 100%. This yields:
\[ 100\% - 78.81\% = 21.19\% \]
Thus, 21.19% of college women are taller than 67 inches. Through statistical data analysis, we transform abstract numbers into meaningful information, such as understanding height distributions, performance metrics, or financial risk assessments. Mastery of this practice is critical for students and professionals as it supports data-driven decision-making processes across various fields.

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Most popular questions from this chapter

According to dogtime .com, the mean weight of an adult St. Bernard dog is 150 pounds. Assume the distribution of weights is Normal with a standard deviation of 10 pounds. a. Find the standard score associated with a weight of 170 pounds. b. Using the Empirical Rule and your answer to part a, what is the probability that a randomly selected St. Bernard weighs more than 170 pounds? c. Use technology to confirm your answer to part \(\mathrm{b}\) is correct. d. Almost all adult St. Bernard's will have weights between what two values?

Rule with z-Scores The Empirical Rule applies rough approximations to probabilities for any unimodal, symmetric distribution. But for the Normal distribution we can be more precise. Use the figure and the fact that the Normal curve is symmetric to answer the questions. Do not use a Normal table or technology. According to the Empirical Rule, a. Roughly what percentage of \(z\) -scores are between \(-2\) and 2 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) b. Roughly what percentage of \(z\) -scores are between \(-3\) and 3 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) c. Roughly what percentage of \(z\) -scores are between \(-1\) and 1 . i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) d. Roughly what percentage of \(z\) -scores are greater than \(0 ?\) i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) e. Roughly what percentage of \(z\) -scores are between 1 and \(2 ?\) i. almost all iii. \(50 \%\) ii. \(13.5 \%\) iv. \(2 \%\)

According to a study by the Colorado Department of Transportation, \(25 \%\) of Colorado drivers admit to using their cell phones to send texts while driving. Suppose two Colorado drivers are randomly selected. a. If the driver texts while driving, record a \(\mathrm{T}\). If not, record an \(\mathrm{N}\). List all possible sequences of Ts and Ns for the two drivers. b. For each sequence, find the probability that it will occur by assuming independence. c. What is the probability that both drivers text while driving? d. What is the probability that neither driver texts while driving? e. What is the probability that exactly one of the drivers texts while driving?

Quantitative S \(\Lambda\) T scores are approximately Normally distributed with a mean of 500 and a standard deviation of 100 . Choose the correct StatCrunch output for finding the probability that a randomly selected person scores less than 450 on the quantitative SAT and report the probability as a percentage rounded to one decimal place.

For each situation, identify the sample size \(n\), the probability of a success \(p\), and the number of success \(x .\) When asked for the probability, state the answer in the form \(b(n, p, x)\). There is no need to give the numerical value of the probability. Assume the conditions for a binomial experiment are satisfied. A 2017 Gallup poll found that \(53 \%\) of college students were very confident that their major will lead to a good job. a. If 20 college students are chosen at random, what's the probability that 12 of them were very confident their major would lead to a good job? b. If 20 college students are chosen at random, what's the probability that 10 of them are not confident that their major would lead to a good job?

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