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The Normal model \(N(65,2.5)\) describes the distribution of heights of college women (inches). Which of the following questions asks for a probability and which asks for a measurement (and is thus an inverse Normal question)? a. What is the probability that a random college woman has a height of 68 inches or more? b. To be in the Tall Club, a woman must have a height such that only \(2 \%\) of women are taller. What is this height?

Short Answer

Expert verified
For problem (a), the probability of a random college woman having a height of 68 inches or more needs to be calculated by finding the Z-score and referring to the Z-table. For problem (b), the inverse Normal distribution method is used to calculate the height above which only 2% of the women fall, by referring the Z-score of 0.98 in the Z-table and using the Z-score formula.

Step by step solution

01

Understand and Apply Probability of Normal Distribution

For the first part, we know that we are to find the probability that a random college woman has a height of 68 inches or more. Given a normal distribution with mean 65 and standard deviation 2.5, we can calculate the Z-score using the formula: \( Z = (X - \mu) / \sigma \), where X is 68, \( \mu \) is 65 and \( \sigma \) is 2.5. Then, we can find the Z-score in the Z-table to find the probability that a woman's height is less than 68 inches and subtract it from 1 to get the probability that a woman's height is 68 inches or more.
02

Apply Inverse Normal Calculation

For the second part of the exercise, which is an inverse normal question, we have to calculate the height (a measurement) for which only 2% of women are taller. We know that if only 2% are taller, then 98% are shorter. We can lookup the Z-score for 0.98 in the Z-table, which is around 2.05.Then we can use the formula for Z-score: \( Z = (X - \mu) / \sigma \) to find X (which represents the height). So, \( X = Z* \sigma + \mu \), where Z is 2.05, \( \mu \) is 65 and \( \sigma \) is 2.5.

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Key Concepts

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

Understanding Probability in Normal Distribution
Probability is a measure of how likely an event is to occur, and in the realm of statistics, it's crucial for interpreting the normal distribution. In the given exercise, we're looking at the probability that a college woman's height will be 68 inches or more. This involves looking at the entire distribution of heights, which follows a bell curve, meaning most heights cluster around the mean with fewer individuals being either much shorter or taller.

For the Normal model N(65, 2.5), which represents the mean height (65 inches) and standard deviation (2.5 inches), to find this probability, we compute the area under the curve to the right of 68 inches. We do this by first calculating the Z-score, which represents the number of standard deviations a particular value is from the mean. From the Z-score, we commonly use statistical tables (or software) to find the corresponding probability. Thus, the area to the right of the Z-score gives us the probability of a woman being 68 inches or taller, helping us answer part (a) of the exercise.
Inverse Normal Calculation for finding Specific Measurements
Inverse normal calculation is used when we need to find a value on the original scale of a distribution given a certain probability. It's essentially working backwards from the known probability to find an unknown measurement. In our textbook exercise, to be in the Tall Club, a height must be above that at which only 2 percent of women are taller (part b). The inverse normal calculation helps us find that specific height.

Firstly, we look up the Z-score that corresponds to the cumulative probability of 98% (since 2% are taller, 98% must be shorter). This Z-score tells us how many standard deviations away from the mean our measurement will be. Utilizing the Z-score formula in reverse, we can solve for the value that fits within this percentile, thus finding the height that is taller than 98% of the sample distribution.
Z-score and Its Role in Normal Distributions
The Z-score is a standard measure in statistics that indicates how many standard deviations an element is from the mean. In a normal distribution, Z-scores can be used to calculate the probability of a score occurring within or outside of a particular area of the distribution.

For instance, in the exercise provided, we use the Z-score formula Z = (X - µ) / σ to find how far in standard deviation terms the height of 68 inches is from the mean height of 65 inches. This Z-score can then be used to look up the probability associated with that Z-value on the standard normal distribution table. It's a powerful tool for finding probabilities and also for comparing scores from different normal distributions.
Importance of Standard Deviation in Normal Distributions
Standard deviation is a critical statistic that measures the amount of variation or dispersion in a set of values. In the context of normal distributions, it tells us how spread out the values are from the mean. A smaller standard deviation means the values are more tightly clustered around the mean, whereas a larger one signifies wider dispersion.

In both parts of our exercise, the standard deviation (2.5 inches) is key. It not only helps calculate the Z-score but also assists in understanding the variability of college women's heights. When using the inverse normal calculation, the standard deviation is part of the formula to back-calculate from our Z-score to the actual height measurement we are seeking. Standard deviation, therefore, is vital for interpreting and understanding the shape and spread of the normal distribution in relation to our data.

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Use a table or technology to answer each question. Include an appropriately labeled sketch of the Normal curve for each part. Shade the appropriate region. a. Find the area to the left of a \(z\) -score of \(-0.50\). h. Find the area to the right of a \(z\) -score of \(-0.50\).

Suppose there is a club for tall people that requires that women be at or above the 98 th percentile in height. Assume that women's heights are distributed as \(N(64,2.5)\). Find what women's height is the minimum required for joining the club, rounding to the nearest inch. Draw a well-labeled sketch to support your answer.

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