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Chapter 8: Problem 6

Find the indicated probabilities. $$ P(-1.71 \leq Z \leq 0.23) $$

Short Answer

Expert verified
Using the Z-table and given Z-scores of -1.71 and 0.23, we can calculate the probability of a standard normal distribution variable falling within this range: \(P(-1.71 \leq Z \leq 0.23) = P(Z \leq 0.23) - P(Z \leq -1.71)\) \(P(-1.71 \leq Z \leq 0.23) = 0.5893 - 0.0436 = 0.5457\) Therefore, the probability that a standard normal distribution variable falls between -1.71 and 0.23 is 0.5457 or 54.57%.

Step by step solution

01

Understand the problem

We are given a Z-score range of -1.71 to 0.23, and we need to find the probability of a standard normal distribution variable falling within this range. The standard normal distribution has a mean of 0 and a standard deviation of 1.
02

Consult the Z-table

To solve this problem, we need to use a Z-table, which shows the probabilities associated with various Z-scores. Locate the probabilities associated with Z-scores of -1.71 and 0.23 in the table.
03

Find the probability associated with Z = -1.71

Look for the row labeled -1.7 and the column labeled 0.01 (as -1.71 has 71 in the decimals) in the Z-table. The intersection of this row and column will show the probability associated with Z = -1.71. This value is 0.0436.
04

Find the probability associated with Z = 0.23

Look for the row labeled 0.2 and the column labeled 0.03 (as 0.23 has 23 in the decimals) in the Z-table. The intersection of this row and column will show the probability associated with Z = 0.23. This value is 0.5893.
05

Calculate the probability between the given Z-scores

To find the probability of Z falling between -1.71 and 0.23, subtract the probability of Z = -1.71 from the probability of Z = 0.23: \(P(-1.71 \leq Z \leq 0.23) = P(Z \leq 0.23) - P(Z \leq -1.71)\) \(P(-1.71 \leq Z \leq 0.23) = 0.5893 - 0.0436\) \(P(-1.71 \leq Z \leq 0.23) = 0.5457\) The probability that a standard normal distribution variable falls between -1.71 and 0.23 is 0.5457 or 54.57%.

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Key Concepts

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

Z-score
A Z-score, also known as a standard score, measures how many standard deviations an element is from the mean of a distribution. In a standard normal distribution, the mean is 0 and the standard deviation is 1. Calculating a Z-score can help you determine the relative position of a data point within a data set.

When you encounter a Z-score:
  • If it's positive, the data point is above the mean.
  • If it's negative, the data point is below the mean.
The Z-score provides insight into whether a score is typical for a data set or if it's an outlier. For example, a Z-score of 0 means the data point is exactly at the mean.

This concept is key when performing probability calculations within the context of normal distribution, as it standardizes various elements for easy comparisons.
Probability Calculation
Probability calculation within a standard normal distribution involves finding the likelihood of a variable falling between, above, or below certain Z-scores. This is essential for interpreting and predicting patterns within data sets.

Here's how you can think about probability calculations:
  • Identify the Z-scores that define your range of interest. For the interval between Z = -1.71 and Z = 0.23, you'll want to calculate the probability of a variable falling within this range.

  • Use the Z-table to determine the areas (probabilities) for each Z-score involved.

  • Subtract the smaller probability from the larger to find the probability of falling between two Z-scores. Like in our example, finding that 54.57% is the probability of a variable lying between -1.71 and 0.23.

This method of calculation provides a quantifiable way to understand behaviors and outcomes in data analysis. Probability tells us how likely it is for something to occur based on historical patterns, which is helpful in many fields, such as economics, management, and social sciences.
Z-table
A Z-table, also known as a standard normal distribution table, is a vital tool used alongside Z-scores to determine probabilities. It contains different probabilities for Z-scores and helps to calculate the area under the standard normal distribution curve.

When you use a Z-table:
  • Find the row matching the first decimal of your Z-score and locate the column for the second decimal. For instance, for a Z-score of -1.71, find -1.7 in the row and 0.01 in the column.

  • The intersection shows the cumulative probability up to that Z-score. Hence, for Z = -1.71, the probability is 0.0436, indicating the area under the curve to the left of this Z-score.

  • Repeat this process for other Z-scores in your problem, like 0.23, which gives a probability of 0.5893.

Understanding how to navigate a Z-table will allow you to make accurate probability predictions quickly. This tool helps translate the abstract math of normal distribution into practical, actionable information you can apply in real-world situations.

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Most popular questions from this chapter

In one Finite Math class, the average grade was 75 and the standard deviation of the grades was \(5 .\) In another Finite Math class, the average grade was 65 and the standard deviation of the grades was \(20 .\) What conclusions can you draw about the distributions of the grades in each class?

If we model after-tax household income with a normal distribution, then the figures of a 1995 study imply the information in the following table, which should be used for Exercises \(35-40^{55}\) Assume that the distribution of incomes in each country is normal, and round all percentages to the nearest whole number: Which country has a higher proportion of very poor families (income \(\$ 12,000\) or less): Canada or Switzerland?

In a certain set of scores, the median occurs more often than any other score. It follows that (A) The median and mean are equal. (B) The mean and mode are equal. (C) The mode and median are equal. (D) The mean, mode, and median are all equal.

Your grade in a recent midterm was \(80 \%\), but the class median was \(100 \%\). Your score was lower than the average score, right?

The new computer your business bought lists a mean time between failures of 1 year, with a standard deviation of 2 months. Ten months after a repair, it breaks down again. Is this surprising? (Assume that the times between failures are normally distributed.)
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