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Suppose your statistics professor reports test grades as \(z\) -scores, and you got a score of 2.20 on an exam. a. Write a sentence explaining what that means. b. Your friend got a z-score of -1 . If the grades satisfy the Nearly Normal Condition, about what percent of the class scored lower than your friend?

Short Answer

Expert verified
A Z-score of 2.20 suggests your grade was 2.20 standard deviations above the class average. Assuming a nearly normal distribution of scores, about 34.13% of the class scored lower than your friend who got a Z-score of -1.

Step by step solution

01

Interpret the Z-score of 2.20

A Z-score measures how many standard deviations an element is from the mean. Here, a score of 2.20 indicates that the grade is 2.20 standard deviations above the mean. That is, the score is greater than the average score by 2.20 times the standard deviation.
02

Understand the Nearly Normal Condition

The Nearly Normal Condition states that if a graph of the data appears unimodal and symmetric, the data may be treated as if it came from a nearly normal distribution. In this context, it is inferred that the class scores are distributed nearly normally.
03

Estimate the proportion of class with lower grades using the Z-score of -1

A Z-score of -1 means the grade is 1 standard deviation below the mean. According to the standard normal distribution, about 15.87% of data falls more than 1 standard deviation below the mean - However, since the distribution is symmetric around the mean, we calculate by using either side of the mean. Hence, we get 50% - 15.87% = 34.13% of the data falls within -1 standard deviation on the left side from the mean. This means that approximately 34.13% of the class scored lower than the friend.

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Key Concepts

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

Standard Deviations
Understanding standard deviations is essential in the field of statistics as it gives insights into the variability or spread of a data set. The standard deviation is a measure that quantifies the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.

A Z-score, sometimes known as a standard score, provides a way to quantify how many standard deviations an observation is from the mean. For instance, a Z-score of 2.20, as in the exercise, means the test score was 2.20 standard deviations above the class mean. This represents a score that is substantially higher than what is considered average, assuming the test scores follow a normal distribution pattern.
Nearly Normal Condition
The 'Nearly Normal Condition' refers to a situation where a set of data is not perfectly normal but is close enough to assume it follows a normal distribution for purposes of analysis. When working with real-world data, perfect normality is rare, so this assumption allows us to apply various statistical techniques that assume normality. To claim that a data set satisfies the Nearly Normal Condition, the distribution should be unimodal (one peak) and symmetric.

In the context of this exercise, we are inferring that class grades approximate a normal distribution. They are spread out in a way where most of the scores cluster around the average, and there are progressively fewer grades the further you go from the mean. Under this condition, we can use the properties of the standard normal distribution to draw conclusions about student performance, like estimating the percentage of students who scored below a certain Z-score.
Normal Distribution
The normal distribution, often symbolized by a bell-curve, is a foundational concept in statistics representing a continuous probability distribution that is symmetric around the mean. Its shape denotes that data near the mean are more frequent in occurrence than data far from the mean.

A fundamental feature of the normal distribution is that it's determined by two parameters: the mean and the standard deviation. The mean indicates where the center of the distribution lies, while the standard deviation shows how spread out the data is. The area under the curve corresponds to probability, and for any given range of values, we can calculate the likelihood of a random observation falling within that range. The total area under the curve is equal to 1 (or 100%), representing the entire range of possible outcomes.
Standard Normal Distribution
When we talk about the standard normal distribution, we're referring to a special case of the normal distribution where the mean is 0 and the standard deviation is 1. This standardized version is used as a reference to understand other normal distributions, making it easier to compare scores from different sets of data.

In terms of Z-scores, any normal distribution can be transformed into a standard normal distribution using Z-score calculations. This involves subtracting the mean of the original distribution from a value and then dividing by the standard deviation of the original distribution. The resulting Z-scores tell us how many standard deviations away from the mean an observation lies, and it can be positioned on the standard normal distribution accordingly. This process enables statisticians to use standard normal distribution tables (or software) to find probabilities associated with specific Z-scores, as demonstrated in the given exercise.

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Most popular questions from this chapter

Based on the model \(N(1152,84)\) describing Angus steer weights from Exercise 29 ?, what are the cutoff values for a. the highest \(10 \%\) of the weights? b. the lowest \(20 \%\) of the weights? c. the middle \(40 \%\) of the weights?

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A tire manufacturer believes that the treadlife of its snow tires can be described by a Normal model with a mean of 32,000 miles and standard deviation of 2500 miles. a. If you buy one of these tires, would it be reasonable for you to hope it will last 40,000 miles? Explain. b. Approximately what fraction of these tires can be expected to last less than 30,000 miles? c. Approximately what fraction of these tires can be expected to last between 30,000 and 35,000 miles? d. Estimate the IQR of the treadlives. e. In planning a marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer does not want to take too big a risk. If the dealer is willing to give refunds to no more than 1 of every 25 customers, for what mileage can he guarantee these tires to last?

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