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In Exercises 5.15 to 5.17 , find the specified areas for a normal distribution. For a \(N(15,5)\) distribution (a) The area to the right of 28 (b) The area to the left of 12

Short Answer

Expert verified
(a) The area to the right of 28 under the normal distribution is approximately 0.0047. (b) The area to the left of 12 under the normal distribution is approximately 0.2743.

Step by step solution

01

Convert to Z-scores

To convert a score x to a Z-score, use the following formula: \( z = (x - \mu)/\sigma \). Here, \(\mu = 15\) and \(\sigma = 5\). So for (a) \(z = (28 - 15) / 5 = 2.6\), while for (b) \(z = (12 - 15) / 5 = -0.6\).
02

Finding the Probability using the Z-table

Refer to the standard normal distribution table, correlating to the calculated z-scores. One can use software or calculators typically available online, if not, use any Z-table. The values obtained from the table for \(z=2.6\) and \(z= -0.6\) are 0.9953 and 0.2743 respectively.
03

Determine the Area for Given Conditions

(a) The question asks for the area to the right of 28. For a Z-table, probabilities are given to the left of a Z value, so the area to the 'right' will be \(1 - P\). So, \(1 - 0.9953 = 0.0047\). (b) For the area to the left of 12, it's just the probability from the Z-table which is 0.2743.

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Key Concepts

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

Z-scores
A Z-score is a statistical measure that gives you an idea of how far a particular data point is from the mean, in terms of standard deviations. It helps in standardizing different data sets and comparing them using a common measure. By using the formula \( z = \frac{x - \mu}{\sigma} \), where
  • \( x \) is the value of interest.
  • \( \mu \) is the mean of the data set.
  • \( \sigma \) is the standard deviation of the data set.
You convert any normal distribution into a standard normal distribution, which has a mean of 0 and a standard deviation of 1. This transformation simplifies the process of finding probabilities, as it allows us to use the standard normal (Z) table. For example, in the given problem, calculating Z-scores for 28 and 12, with mean 15 and standard deviation 5, yields scores of 2.6 and -0.6, respectively.

The utility of Z-scores becomes evident when determining probabilities for the tails of the distribution curve. A higher absolute Z-score indicates a value further from the mean, making it more unusual if the data are normally distributed.
Probability
Probability in a normal distribution relates to the area under the curve, representing how likely different outcomes are. Once you calculate a Z-score for a specific value, you use a Z-table (also known as a standard normal distribution table) to find the probability of observing less than or more than your specified value.

In the context of the exercise:
  • For point (a) of the exercise, with a Z-score of 2.6, looking up this value in a Z-table provides a cumulative probability of 0.9953.
  • For point (b), a Z-score of -0.6 correlates to a cumulative probability of 0.2743.
These values help determine the proportion of the dataset falling to the left or right of a given point. Importantly, probabilities allow conclusions based upon real-world expectations, and predictively model likely future outcomes.

When seeking the area to the right of a particular Z-score, you subtract the table probability from 1. Conversely, the table provides the probability directly if you're looking to the left.
Standard Normal Distribution
A standard normal distribution is a special type of normal distribution that has been standardized: its mean is 0 and its standard deviation is 1. This is achieved through the transformation of raw scores (data points) to Z-scores.

This transformation is crucial because it allows for the use of the standard normal (Z) table, simplifying the process of determining probabilities. Each value in the table corresponds to a cumulative probability that a random variable with a standard normal distribution is less than or equal to a given Z-score.
  • The standard normal distribution is symmetric, and its total area is equal to 1.
  • The central point represents the mean, median, and mode, making it a powerful tool in statistics for evaluating data relative to a central point.
Understanding and applying the standard normal distribution allows statisticians and researchers to compare different data sets with ease. By converting any normal curve into the standard form, the complexity of calculations is reduced, and data analysis becomes more streamlined.

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Most popular questions from this chapter

Where Is the Best Seat on the Plane? A survey of 1000 air travelers \(^{21}\) found that \(60 \%\) prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is \(S E=0.015 .\) Use a normal distribution to find and interpret a \(99 \%\) confidence interval for the proportion of air travelers who prefer a window seat.

What Is Your Sleep Position? In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position when they fall asleep at night. The most common answer was the fetal position (on the side, with legs pulled up), with \(41 \%\) of the participants saying they start in this position. Use a normal distribution to find and interpret a \(95 \%\) confidence interval for the proportion of all UK adults who start sleep in this position. Use the fact that the standard error of the estimate is 0.016 .

How Do You Get Your News? A study by the Pew Research Center \(^{9}\) reports that in \(2010,\) for the first time, more adults aged 18 to 29 got their news from the Internet than from television. In a random sample of 1500 adults of all ages in the US, \(66 \%\) said television was one of their main sources of news. Does this provide evidence that more than \(65 \%\) of all adults in the US used television as one of their main sources for news in \(2010 ?\) A randomization distribution for this test shows \(S E=0.013\). Find a standardized test statistic and compare to the standard normal to find the p-value. Show all details of the test.

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu=10\) vs \(H_{a}: \mu \neq 10\) when the sample has \(n=75, \bar{x}=11.3,\) and \(s=0.85,\) with \(S E=0.10\).

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu=80\) vs \(H_{a}: \mu>80\) when the sample has \(n=20, \bar{x}=82.4,\) and \(s=3.5,\) with \(S E=0.8\).

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