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The proportion of observations from a standard Normal distribution that take values greater than \(1.78\) is about (a) \(0.9554\). (b) \(0.0446 .\) (c) \(0.0375\).

Short Answer

Expert verified
The correct answer is (c) 0.0375.

Step by step solution

01

Understanding the Standard Normal Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. Values from this distribution are typically represented by the variable \(Z\), and a \(Z\)-score represents the number of standard deviations a point is from the mean.
02

Finding the Z-Score

We're asked to determine the proportion of values greater than 1.78 in a standard normal distribution. The Z-score here is simply 1.78, since we're dealing with a standard normal distribution where the mean \(\mu = 0\) and \(\sigma = 1\).
03

Using the Z-Table to Find Probabilities

We need to refer to a standard normal (Z) table or use a statistical tool to find the probability that \(Z < 1.78\). The Z-table provides the probability that a standard normal variable is less than a particular Z-score.
04

Calculating \(P(Z > 1.78)\)

From the Z-table or a statistical calculator, find \(P(Z < 1.78)\). This value is approximately 0.9625. To find the probability that \(Z > 1.78\), subtract this from 1, i.e., \(1 - 0.9625 = 0.0375\).
05

Verifying Your Answer

Compare the calculated probability \(P(Z > 1.78) \approx 0.0375\) with the options provided. The correct answer must match the closest option, which here is 0.0375.

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Key Concepts

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

Z-score
The Z-score is an essential concept when dealing with standard normal distributions. It helps us understand how a single data point relates to the rest of the dataset. The Z-score measures the number of standard deviations an observation is from the mean. In the context of a standard normal distribution, where the mean is zero (\(\mu = 0\)) and the standard deviation is one (\(\sigma = 1\)), the Z-score directly represents our observation's position along this distribution.

When you have a Z-score of 1.78, like in our exercise, it indicates that the value is 1.78 standard deviations away from the mean. This score helps in determining the likelihood of occurrence of this or more extreme values, which is a core aspect of probability calculations in statistics. Understanding Z-scores paves the way for systematic approaches to various statistical analyses, such as hypothesis testing and confidence intervals.
Probability calculation
In statistics, one of the key applications of the Z-score is in probability calculation. We calculate the probability to understand the likelihood of a certain event or observation occurring. For our current problem, the task was to calculate the probability of observing a value greater than 1.78 in a standard normal distribution.

Here's how this works:
  • First, you find the probability of a Z-score less than 1.78. This gives us the cumulative probability for all values up to 1.78 in the distribution.
  • Once we have this cumulative probability, calculating the probability for the complementary event (values greater than 1.78) is simple. Since total probability equals 1, you subtract the cumulative probability from 1.
Therefore, the calculation goes like this: if \(P(Z < 1.78) = 0.9625\), then \(P(Z > 1.78) = 1 - 0.9625 = 0.0375\). This step-by-step process ensures accuracy and helps compare the result to predefined answers, confirming the correct probability estimation.
Z-table usage
The Z-table is a fundamental tool in statistics for looking up probabilities associated with the standard normal distribution. Think of it as a reference guide existing to simplify our calculations when dealing with Z-scores.

To use a Z-table effectively, follow these steps:
  • Identify the Z-score you need to evaluate. In our problem, this was 1.78.
  • Locate this Z-score in the Z-table. The table is structured with Z-scores and their corresponding cumulative probabilities.
  • Read off the cumulative probability. In our example, for a Z-score of 1.78, the table shows approximately 0.9625, which is the probability that a value is less than 1.78.
Once you have this cumulative probability, you can use it to find probabilities for values greater than your Z-score by subtracting from 1. This makes the Z-table not just a helpful tool, but an indispensable resource for simplifying probability calculations in statistics.

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Most popular questions from this chapter

The distribution of hours of sleep per week night, among college students, is found to be Normally distributed, with a mean of \(6.5\) hours and a standard deviation of 1 hour. The percentage of college students that sleep at least eight hours per night is about (a) \(95 \%\). (b) \(6.7 \%\). (c) \(2.5 \%\)

The scores of adults on an IQ test are approximately Normal with mean 100 and standard deviation 15. Alysha scores 135 on such a test. Her z-score is about (a) \(1.33\). (b) \(2.33\) (c) 6.33.

Body mass index. Your body mass index (BMI) is your weight in kilograms divided by the square of your height in meters. Many online BMI calculators allow you to enter weight in pounds and height in inches. High BMI is a common but controversial indicator of overweight or obesity. A study by the National Center for Health Statistics found that the BMI of American young men (ages 20-29) is approximately Normal with mean \(26.8\) and standard deviation 5.2. 12 (a) People with BMI less than \(18.5\) are often classified as "underweight." What percent of men aged 20-29 are underweight by this criterion? (b) People with BMI over 30 are often classified as "obese." What percent of men aged \(20-29\) are obese by this criterion?

Osteoporosis. Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD \(2.5\) standard deviations below the mean for healthy young adults. BMD measurements in a population of people similar in age and sex roughly follow a Normal distribution. (a) What percent of healthy young adults have osteoporosis by the WHO criterion? (b) Women aged \(70-79\) are, of course, not young adults. The mean BMD in this age is about \(-2\) on the standard scale for young adults. Suppose the standard deviation is the same as for young adults. What percent of this older population have osteoporosis?

What's your percentile? Reports on a student's test score such as the SAT or a child's height or weight usually give the percentile as well as the actual value of the variable. The percentile is just the cumulative proportion stated as a percent: the percent of all values of the variable that were lower than this one. The upper arm lengths of females in the United States are approximately Normal with mean \(35.8 \mathrm{~cm}\) and standard deviation \(2.1 \mathrm{~cm}\), and those for males are approximately Normal with mean \(39.1 \mathrm{~cm}\) and standard deviation \(2.3 \mathrm{~cm}\). (a) Cecile, a 73-year-old female in the United States, has an upper arm length of \(33.9 \mathrm{~cm}\). What is her percentile? (b) Measure your upper arm length to the nearest tenth of a centimeter, referring to Exercise \(3.5\) (page 84 ) for the measurement instructions. What is your arm length in centimeters? What is your percentile?

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