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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

Babies in the United States have a mean birth length of \(20.5\) inches with a standard deviation of \(0.90\) inch. The shape of the distribution of birth lengths is approximately Normal. a. Find the birth length at the \(2.5\) th percentile. b. Find the birth length at the \(97.5\) th percentile. c. Find the \(z\) -score for the length at the \(2.5\) th percentile. d. Find the \(z\) -score for the length at the \(97.5\) th percentile.

According to a survey conducted by OnePoll, a marketing research company, \(10 \%\) of Americans have never traveled outside their home state. Assume this percentage is accurate. Suppose a random sample of 80 Americans is taken. a. Find the probability that more than 12 have never travelled outside their home state. b. Find the probability that at least 12 have never travelled outside their home state. c. Find the probability that at most 12 have never travelled outside their home state.

Scores on the 2017 MCAT, an exam required for all medical school applicants, were approximately Normal with a mean score of 505 and a standard deviation of \(9.4\). a. Suppose an applicant had an MCAT score of 520 . What percentile corresponds with this score? b. Suppose to be considered at a highly selective medical school an applicant should score in the top \(10 \%\) of all test takers. What score would place an applicant in the top \(10 \%\) ?

Standard Normal Use technology or a Normal table to find each of the following. Include an appropriately labeled sketch of the Normal curve for each part with the appropriate area shaded. a. Find the probability that a \(z\) -score will be \(2.12\) or greater. b. Find the probability that a \(z\) -score will be less than \(-0.74\). c. Find the probability that a \(z\) -score will between \(1.25\) and \(2.37\).

The Normal model \(N(150,10)\) describes the distribution of scores on the LSAT, a standardized test required by most law schools. Which of the following questions asks for a probability, and which asks for a measurement? Identify the type of problem and then answer the given question. a. A law school applicant scored at the 60 th percentile on the LSAT. What was the applicant's LSAT score? b. A law school applicant scored 164 on the LSAT. This applicant scored higher than what percentage of LSAT test takers?

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