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Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 78 and a standard deviation of 7 . The top \(15 \%\) of all scores have been designated A's. Your score is \(89 .\) Did you earn an \(\mathrm{A}\) ? Explain.

Short Answer

Expert verified
The student's score of 89 is greater than the cutoff score of approximately 85.28, which corresponds to the top 15% of scores on this normally distributed midterm exam. Therefore, the student did earn an A.

Step by step solution

01

Calculate the z-score corresponding to the top 15% of scores

To find the cutoff score for the top 15% of scores, we will first find the z-score that corresponds to this percentage. We can use a standard normal table, a calculator with percentile-to-z-score functionality, or online tools for this purpose. Look for the z-score corresponding to the area to the left (or percentile) of 85% (100% - 15%), which is approximately equal to 1.04.
02

Convert the z-score to the corresponding exam score

Now that we have the z-score, we can convert it to the corresponding exam score using the following formula for a normal distribution: \(Score = \mu + (z \cdot \sigma)\) Where, \(\mu\) is the mean of the distribution, \(z\) is the z-score, and \(\sigma\) is the standard deviation of the distribution. Plugging in the values, we get: \(Score = 78 + (1.04 \cdot 7)\)
03

Calculate the exam score corresponding to the z-score

Using the formula from Step 2, we can calculate the cutoff score for the top 15% of exam scores: \(Score = 78 + (1.04 \cdot 7) \approx 85.28\)
04

Compare the student's score to the cutoff score

We are given that the student's score is 89. To determine if the student earned an A, compare their score to the calculated cutoff score: \(Student\_score = 89 > 85.28 = Cutoff\_score\) Since the student's score (89) is greater than the cutoff score (85.28), they earned an A on the midterm exam.

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Key Concepts

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

Understanding Z-Score
The concept of a 'z-score' is fundamental in statistics, particularly when working with normal distribution. Simply put, a z-score is a measurement that describes a value's relationship to the mean of a group of values. It is expressed in terms of standard deviations from the mean. When you calculate the z-score of a data point, you are finding out how many standard deviations away from the mean it is.

Mathematically, it is represented by the equation: \( z = \frac{x - \mu}{\sigma} \), where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. A z-score can be positive or negative, indicating whether the score is above or below the mean, respectively. In the example of the midterm exam, by calculating the z-score, we can determine whether a given exam score falls within the top 15% of all scores, a necessary step to find out if a student earned an A grade.
Insights into Standard Deviation
Standard deviation, denoted as \( \sigma \), measures the amount of variation or dispersion of a set of values. A low standard deviation means that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

Understanding standard deviation is critical when interpreting z-scores since it forms the denominator of the z-score formula. It gives context to the z-score value, helping to understand how extreme a data point is in relation to the overall distribution. For instance, in the context of the midterm exam, a standard deviation of 7 points tells us that most students scored within 7 points above or below the average score of 78.
From Percentile to Z-Score
The percentile-to-z-score conversion is a powerful tool when working with the standard normal distribution. A percentile rank is a value below which a given percentage of observations fall. To find the z-score associated with a certain percentile, one often uses a z-score table or statistical software.

In our exercise, finding the z-score for the top 15% of scores (or the 85th percentile) allows us to set a cutoff point for grading. Typically, the 85th percentile z-score is around 1.04, meaning that a score 1.04 standard deviations above the mean marks the lower bound of the top 15%. Transforming a percentile to a z-score is an essential step when setting benchmarks such as passing grades or qualification limits.
Properties of Normal Distribution
Normal distribution is a key concept in statistics with remarkable properties, making it invaluable in many fields. It has a bell-shaped curve which is symmetrical about the mean. Here are a few crucial properties:
  • The mean, median, and mode of a normal distribution are all equal.
  • Approximately 68% of the data within a normal distribution falls within one standard deviation of the mean. Similarly, about 95% falls within two standard deviations, and about 99.7% falls within three standard deviations.
  • The tails of the distribution curve extend infinitely, theoretically covering every possible value of x.

For example, when we apply this understanding to the midterm exam scores, we anticipate that most students' scores will cluster around the mean, with few extreme high or low scores.

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Most popular questions from this chapter

A box contains five slips of paper, marked \(\$ 1, \$ 1, \$ 1, \$ 10,\) and \(\$ 25 .\) The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable \(w\) by \(w=\) amount awarded. Determine the probability distribution of \(w\). (Hint: Think of the slips as numbered \(1,2,3,4,\) and \(5 .\) An outcome of the experiment will consist of two of these numbers.)

State whether each of the following random variables is discrete or continuous: a. The number of defective tires on a car b. The body temperature of a hospital patient c. The number of pages in a book d. The number of draws (with replacement) from a deck of cards until a heart is selected e. The lifetime of a light bulb

Determine each of the following areas under the standard normal (z) curve: a. To the left of -1.28 b. To the right of 1.28 c. Between -1 and 2 d. To the right of 0 e. To the right of -5 f. Between -1.6 and 2.5 g. To the left of 0.23

The probability distribution of \(x,\) the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: \(\begin{array}{lccccc}x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20\end{array}\) a. Calculate the mean value of \(x\). b. Interpret the mean value of \(x\) in the context of a long sequence of observations of number of defective tires. c. What is the probability that \(x\) exceeds its mean value? d. Calculate the standard deviation of \(x\).

Purchases made at small "corner stores" were studied by the authors of the paper "Changes in Quantity, Spending, and Nutritional Characteristics of Adult, Adolescent and Child Urban Corner Store Purchases After an Environmental Intervention" (Preventive Medicine [2015]: 81-85). Corner stores were defined as stores that are less than 200 square feet in size, have only one cash register, and primarily sell food. After observing a large number of corner store purchases in Philadelphia, the authors reported that the average number of grams of fat in a corner store purchase was \(21.1 .\) Suppose that the variable \(x=\) number of grams of fat in a corner store purchase has a distribution that is approximately normal with a mean of 21.1 grams and a standard deviation of 7 grams. a. What is the probability that a randomly selected corner store purchase has more than 30 grams of fat? b. What is the probability that a randomly selected corner store purchase has between 15 and 25 grams of fat? c. If two corner store purchases are randomly selected, what it the probability that both of these purchases will have more than 25 grams of fat?

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