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(a) The area to the left of the endpoint on a \(N(5,2)\) curve is about 0.10 . (b) The area to the right of the endpoint on a \(N(500,25)\) curve is about 0.05

Short Answer

Expert verified
The score in the \(N(5,2)\) curve corresponding to an area of 0.10 to the left of the endpoint is approximately 2.44. The score in the \(N(500,25)\) curve corresponding to an area of 0.05 to the right of the endpoint is approximately 541.125.

Step by step solution

01

Determine the Z-scores

Using the standard normal distribution table or a Z-score calculator, determine the Z-scores corresponding to the given areas. For part a, the Z-score corresponding to an area of 0.10 to the left of the endpoint (which is equivalent to a cumulative probability of 0.1) is approximately -1.28. For part b, since the area is to the right of the endpoint, subtract the given area from 1 to get the cumulative probability (1-0.05 = 0.95). The corresponding Z-score is approximately 1.645.
02

Convert Z-scores back to scores in the normal distributions

To convert the Z-scores back to scores in the normal distributions, use the formula \[X = \mu + Z\sigma\] where \(\mu\) is the mean, \(\sigma\) is the standard deviation, and \(Z\) is the Z-score. For part a: \(X = 5 + (-1.28)(2) = 2.44\) For part b: \(X = 500 + (1.645)(25) = 541.125\) So the scores corresponding to the Z-scores in the \(N(5,2)\) and \(N(500,25)\) distributions are approximately 2.44 and 541.125, respectively.

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Key Concepts

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

Understanding Z-score
The Z-score is a crucial concept in statistics, particularly when dealing with normal distributions. It tells us how far away a particular data point is from the mean, in terms of standard deviations. If you imagine the normal distribution curve as a mountain, the Z-score helps you locate any specific point on this mountain.

Z-scores are calculated using the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
In this formula, \(X\) is our data point, \(\mu\) represents the mean of the distribution, and \(\sigma\) is the standard deviation.

These scores are valuable because they allow us to compare different data sets by standardizing scores, even if the datasets have different means and standard deviations. For example, a Z-score of 1 means the value is one standard deviation above the mean, whereas a Z-score of -1.28 like in our exercise indicates the value is just over one standard deviation below the mean.
Unveiling Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. It is often denoted by \(\sigma\). A low standard deviation indicates that the data points are closely clustered around the mean, whereas a high standard deviation indicates that the data points are more spread out.

Understanding this concept is essential because it helps us gauge the consistency of the data.

In a normal distribution, standard deviation determines the shape of the curve:
  • A smaller standard deviation results in a steeper, narrower curve.
  • A larger standard deviation leads to a flatter, wider curve.
For instance, in our exercise, we had two different normal distributions: \(N(5,2)\) and \(N(500,25)\). The standard deviations (2 and 25, respectively) tell us how "spread out" the data is from the means 5 and 500. Understanding standard deviation helps us interpret the distribution's spread and variability.
Exploring Cumulative Probability
Cumulative probability is the likelihood that a randomly selected value from a probability distribution will be less than or equal to a specified value. It is a crucial element when working with normal distribution as it helps in finding the probability of a data point being within a certain range.

For any given Z-score, the cumulative probability can be found using Z-tables or calculators which provide the area under the curve to the left of the Z-score.

An understanding of cumulative probability is essential for interpreting our original exercise. In part (a), a cumulative probability of 0.10 indicates that approximately 10% of values fall below the endpoint, corresponding to a Z-score of about -1.28. In part (b), since the area to the right was given, we subtracted the probability from 1 to find the left area or cumulative probability, resulting in 0.95 and a Z-score of approximately 1.645.

The concept allows us to assess the probability of variables falling within specific ranges on a normal distribution, which is fundamental for statistical analysis and decision-making.

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Most popular questions from this chapter

Exercises 5.21 to 5.28 ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves.] The lower \(10 \%\) for a standard normal distribution converted to a \(N(500,80)\) distribution.

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Writing on the SAT Exam In Table 5.1 with Exercise \(5.30,\) we see that scores on the Writing portion of the SAT (Scholastic Aptitude Test) exam are normally distributed with mean 492 and standard deviation 111. Use the normal distribution to answer the following questions: (a) What is the estimated percentile for a student who scores 450 on Writing? (b) What is the approximate score for a student who is at the 90 th percentile for Writing?

To Study Effectively, Test Yourself! Cognitive science consistently shows that one of the most effective studying tools is to self-test. A recent study 12 reinforced this finding. In the study, 118 college students studied 48 pairs of Swahili and English words. All students had an initial study time and then three blocks of practice time. During the practice time, half the students studied the words by reading them side by side, while the other half gave themselves quizzes in which they were shown one word and had to recall its partner. Students were randomly assigned to the two groups, and total practice time was the same for both groups. On the final test one week later, the proportion of items correctly recalled was \(15 \%\) for the reading-study group and \(42 \%\) for the self-quiz group. The standard error for the difference in proportions is about 0.07 . Test whether giving self-quizzes is more effective and show all details of the test. The sample size is large enough to use the normal distribution.

In Exercises 5.44 to 5.49 , find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(95 \%\) confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) if the samples have \(n_{1}=100\) with \(\bar{x}_{1}=256\) and \(s=51\) and \(n_{2}=120\) with \(\bar{x}=242\) and \(s=47\). and the standard error is \(S E=6.70\)

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