/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 GRE scores, Part III. \(151, \si... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

GRE scores, Part III. \(151, \sigma=7\) ) for the verbal part of the exam and \(N(\mu=153, \sigma=7.67)\) for the quantitative part. Suppose performance on these two sections is independent. Use this information to compute each of the following: (a) The probability of a combined (verbal + quantitative) score above 320 . (b) The score of a student who scored better than \(90 \%\) of the test takers overall.

Short Answer

Expert verified
(a) 0.0606; (b) 317 (approximately).

Step by step solution

01

Define the Scores

The verbal part of the exam is normally distributed with a mean of 151 and a standard deviation of 7. That is, we have a distribution \( N(\mu = 151, \sigma = 7) \). The quantitative part is normally distributed with a mean of 153 and a standard deviation of 7.67, which can be described by \( N(\mu = 153, \sigma = 7.67) \).
02

Find Combined Statistics

Since the scores on the two sections are independent, the combined score \( X = V + Q \) will also be normally distributed with mean \( \mu_{X} = \mu_{V} + \mu_{Q} = 151 + 153 = 304 \) and variance \( \sigma_{X}^2 = \sigma_{V}^2 + \sigma_{Q}^2 = 7^2 + 7.67^2 \). Convert the variance to standard deviation \( \sigma_{X} = \sqrt{49 + 58.8289} \approx 10.34 \).
03

Calculate Probability for Part (a)

We are looking for \( P(X > 320) \). First, convert 320 to a z-score using \( z = \frac{X - \mu_{X}}{\sigma_{X}} = \frac{320 - 304}{10.34} \approx 1.55 \). Use the standard normal distribution table or calculator to find \( P(Z > 1.55) \), which is about 0.0606.
04

Determine 90th Percentile for Part (b)

The score that is higher than 90% of test-takers corresponds to the 90th percentile of the distribution. Find the z-score for the 90th percentile, which is approximately 1.28 in the standard normal distribution. Convert using \( X = \mu_{X} + z \cdot \sigma_{X} = 304 + 1.28 \times 10.34 \approx 317.2 \). A score of about 317 is needed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
The normal distribution is a continuous probability distribution that is symmetrical, meaning most observations cluster around the central peak, and the probabilities for values taper off symmetrically on both sides. It's also known as a bell curve due to its shape. Normal distributions are defined by two parameters:
  • Mean (\( \mu \)): This is the average or expected value of the data points.
  • Standard Deviation (\( \sigma \)): This measures the spread of the data points around the mean.
In the context of the GRE scores mentioned in the exercise, the verbal and quantitative scores are modeled as normal distributions with given means and standard deviations. This allows us to make predictions about where a student's scores are likely to fall within the distribution of all test-takers.
Independent Events
Independent events are those whose outcomes do not affect one another. In probability, two events are independent if the occurrence of one does not influence the probability of the other occurring.

For example, when rolling two dice, the result of one die does not affect the result of the other die. In our exercise, the verbal and quantitative sections of the GRE are independent. This means a student's performance on one section does not impact their performance on the other.

This independence allows for simplified calculations when combining the scores from these sections. For the combined score's variance, we add the variances of each section (since they are independent), thereby simplifying the calculations.
Z-score
A z-score is a statistical measurement that describes a data point's relation to the mean of a group of values. It shows how many standard deviations a particular value is from the mean, capturing whether it is above or below the average.

The formula for calculating a z-score is:\[ z = \frac{X - \mu}{\sigma} \] where \( X \) is the value being assessed, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • A positive z-score indicates the value is above the mean.
  • A negative z-score indicates it is below the mean.
Using z-scores, you can determine probabilities and percentiles within the normal distribution. In the original problem, the z-score helps us find the probability of a combined GRE score exceeding a certain value and identifying percentile ranks.
Percentile
A percentile is a measure used in statistics to indicate the value below which a given percentage of observations fall. For example, the 90th percentile is the value below which 90% of the observations can be found.

Percentiles are useful in comparing scores and are often used to understand where an individual's score stands relative to others.

To find the percentile of a particular score in a normal distribution, you first find the z-score and then use the standard normal distribution to determine the corresponding percentile.
  • For example, a z-score of approximately 1.28 corresponds to the 90th percentile, meaning that if a student scores this high or higher, they performed better than 90% of test-takers.
Percentiles are hugely valuable in educational testing and other fields to evaluate comparative performance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

CFLs. A manufacturer of compact fluorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1.000 hours. (a) What is the probability that a randomly chosen light bulb lasts more than 10.500 hours? (b) Describe the distribution of the mean lifespan of 15 light bulbs. (c) What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10.500 hours? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

Social network use. The Pew Research Center estimates that as of January \(2014,89 \%\) of \(18-29\) year olds in the United States use social networking sites. \(^{\text {. }}\) Calculate the probability that at least \(95 \%\) of 500 randomly sampled \(18-29\) year olds use social networking sites.

Is it Bernoulit? Determine if each trial can be considered an independent Bernoulli trial for the following sitmations. (a) Cards dealt in a hand of poker. (b) Outcome of each roll of a die.

Triathlon times, Part L. In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages \(30-34\) group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 1: 22: 28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is same information on the performance of their groups: \- The finishing times of the Men, Ages \(30-34\) group has a mean of 4313 seconds with a standard deviation of 583 seconds. \- The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. \- The distributions of finishing times for both groups are approximately Normal. Remember: a better performance corresponds to a faster finish. (a) What are the Z-ncores for Leo's and Mary's finishing times? What do these Z-scores tell you? (b) Did Leo or Mary rank better in their respective groups? Explain your reasoning. (c) What percent of the triathletes did Leo finish faster than in his group? (d) What percent of the triathletes did Mary finish faster than in her group? (e) If the distributions of finishing times are not nearly normal, would your answers to parts (a)change? Explain your reasoning.

Eye color, Part I. A husband and wife both have brown eyes but carry genes that make it possible for their children to have brown eyes (probability 0.75\()\), blue eyes \((0.125),\) or green eyes (0.125) . (a) What is the probability the first blue-eyed child they have is their third child? Assume that the eye colors of the children are independent of each other. (b) On average, how many children would such a pair of parents have before having a blue-eyed child? What is the standard deviation of the number of children they would expect to have until the first blue-eyed child?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.