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Chapter 6: Problem 46

Assume that the test scores from a college admissions test are normally distributed, with a mean of 450 and a standard deviation of 100 a. What percentage of the people taking the test score between 400 and \(500 ?\) b. Suppose someone receives a score of \(630 .\) What percentage of the people taking the test score better? What percentage score worse? c. If a particular university will not admit anyone scoring below \(480,\) what percentage of the persons taking the test would be acceptable to the university?

Short Answer

Expert verified
a) 38.30%, b) 3.59% score better and 96.41% score worse, c) 61.79% are acceptable.

Step by step solution

01

Understand the Normal Distribution

The test scores are normally distributed with a mean (\(\mu\)) of 450 and a standard deviation (\(\sigma\)) of 100. We're dealing with problems that involve finding probabilities or percentages of scores within a certain range of this distribution.
02

Calculate the Z-Scores

The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \). For the range between 400 and 500: - Calculate the Z-score for 400: \( Z = \frac{400 - 450}{100} = -0.5 \) - Calculate the Z-score for 500: \( Z = \frac{500 - 450}{100} = 0.5 \)
03

Find the Percentage for Part (a)

Using a standard normal distribution table or calculator, find the probabilities for the calculated Z-scores: - Probability (Z < 0.5) is about 0.6915- Probability (Z < -0.5) is about 0.3085- The percentage between Z = -0.5 and Z = 0.5 is: \(0.6915 - 0.3085 = 0.3830\) or 38.30% of people score between 400 and 500.
04

Calculate Z-Score for Part (b)

Calculate the Z-score for 630: \( Z = \frac{630 - 450}{100} = 1.8 \)
05

Find the Percentage Above 630

Using the Z-table, find the probability (percentage) for Z=1.8:- Probability (Z < 1.8) is about 0.9641- Percentage scoring better (above 630) is: \(1 - 0.9641 = 0.0359\) or 3.59%.
06

Find the Percentage Scoring Worse Than 630

The percentage of people scoring worse than 630 is simply the cumulative probability for Z=1.8, which is 96.41%.
07

Calculate Z-Score for Part (c)

Calculate the Z-score for 480:\(Z = \frac{480 - 450}{100} = 0.3 \)
08

Find the Acceptance Percentage for the University

Using the Z-table, find the probability for Z=0.3: - Probability (Z < 0.3) is about 0.6179 - Therefore, 61.79% of the people meet the university's score requirement of 480.

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Key Concepts

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

Z-Scores and Their Significance
To understand how individual scores relate to the average, we use z-scores. A z-score tells us how many standard deviations a particular value is from the mean. This is crucial when comparing scores within a normal distribution.
The formula for calculating a z-score is:\[ Z = \frac{X - \mu}{\sigma} \]
  • \(X\) is the score you are examining.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
Z-scores help us standardize scores and compare them, even if they come from different scenarios or distributions. In the context of test scores, they help determine what percentage of scores fall above or below given values.
Understanding Probability in Normal Distribution
Probability in a normal distribution represents the likelihood that a given outcome will occur. It's one of the most vital aspects when analyzing normally distributed data.
When we calculate probabilities with a normal distribution, we are essentially looking for the area under the curve. This is often done using z-scores and a standard normal distribution table.
For example, in our exercise, when calculating the percentage of students scoring between two values, we determine the probability of scores falling within this range.
  • The probability for scores < 500 represented by a z-score encompasses a larger cumulative probability compared to scores < 400.
  • The range is found by subtracting these probabilities to find the exact probability of scoring between these values.
This approach is used to capably solve diverse real-world problems requiring probability estimates.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a tool used to describe the probability that a random variable will take on a value less than or equal to a particular value. It is crucial for interpreting data in the context of normal distribution.
The CDF helps us understand and visualize the area under the curve from the left up to a point on the normal distribution curve. This area corresponds to probability.
  • If you're given a z-score, the CDF will tell you the probability of a variable falling below or equal to a specific score.
  • Knowing the CDF allows you to find the likelihood of exceeding a certain threshold, or falling below it, by recognizing the complement of these probabilities.
The cumulative distribution is vital for determining percentiles and probabilities which help in making informed decisions.
Standard Deviation: A Measure of Variability
Standard deviation is a metric that indicates the spread or dispersion of a dataset. Specifically, it measures how far the scores deviate from the mean of the distribution.
In a normal distribution, it plays a critical role in determining the shape of the curve. A smaller standard deviation indicates that data points tend to be close to the mean, leading to a steeper curve. Conversely, a larger standard deviation indicates a flatter curve, with scores spread over a wider range.
  • For our exercise, knowing the standard deviation helps to calculate z-scores, which in turn aids in finding probabilities.
  • Understanding standard deviation is essential for accurately describing how scores are distributed around the mean.
In essence, standard deviation is a cornerstone concept for comprehending the spread within a dataset, providing insights into the variability of scores.

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Most popular questions from this chapter

The lifetime (hours) of an electronic device is a random variable with the following exponential probability density function. \\[f(x)=\frac{1}{50} e^{-x / 50} \quad \text { for } x \geq 0\\] a. What is the mean lifetime of the device? b. What is the probability that the device will fail in the first 25 hours of operation? c. What is the probability that the device will operate 100 or more hours before failure?

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