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Chapter 6: Problem 45

Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2. About __________\(\%\) of the area is between \(z=-1\) and \(z=1\) (or within 1 standard deviation of the mean).

Short Answer

Expert verified
About 68.26% of the area is between z = -1 and z = 1.

Step by step solution

01

- Understanding the Problem

To find the area between the values of the standard normal distribution from z = -1 to z = 1, which is equivalent to finding the area under the curve between these z-scores.
02

- Using the Z-table

Look up the values of the standard normal distribution for z = -1 and z = 1 on the Z-table.
03

- Finding the Cumulative Area

The cumulative area from z = -鈭� to z = 1 is approximately 0.8413, and the cumulative area from z = -鈭� to z = -1 is approximately 0.1587.
04

- Calculating the Area Between Z = -1 and Z = 1

Subtract the cumulative area at z = -1 from the cumulative area at z = 1: 0.8413 - 0.1587 = 0.6826.
05

- Converting to a Percentage

Convert the decimal area to a percentage by multiplying by 100: 0.6826 x 100 = 68.26%.
06

- Filling in the Blank

Insert the percentage into the blank.

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Key Concepts

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

z-scores
In statistics, a z-score represents the number of standard deviations a data point is from the mean. To calculate a z-score, use the formula: \( z = \frac{{(X - \mu)}}{{\sigma}} \), where X is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. This transformation allows different datasets to be compared using a common scale. For example, z = -1 indicates that the value is one standard deviation below the mean, while z = 1 indicates it is one standard deviation above the mean.
area under the curve
The area under the curve in a normal distribution graph represents the probability of a value falling within a specific range. For the standard normal distribution, the total area under the curve is 1. This area can be divided into segments based on z-scores. For instance, the area between z = -1 and z = 1 contains approximately 68.26% of the data. Understanding this area helps in probability calculations and statistical predictions, as it shows the proportion of values that lie within certain ranges.
cumulative area
The cumulative area refers to the total area under the curve from the leftmost boundary up to a specific z-score. This cumulative area can be found using Z-tables. For instance, the cumulative area from z = -鈭� to z = 1 is about 0.8413. By subtracting the cumulative area at one z-score from another, we can find the area between those two scores. This concept is crucial for understanding probabilities and distributions in statistical analysis.
percentage conversion
To convert the cumulative area into a percentage, multiply the decimal value by 100. For instance, an area of 0.6826 between z = -1 and z = 1 converts to 68.26%. This conversion is helpful in expressing probabilities in a more intuitive manner. For example, saying that 68.26% of values fall within one standard deviation of the mean is more immediately understandable than quoting a decimal figure. This step makes the concept more accessible, especially in contexts like the empirical rule.
range rule of thumb
The range rule of thumb provides a quick estimate of the spread of a dataset. It states that approximately 95% of the data in a normal distribution lies within two standard deviations from the mean. However, for practical purposes, many use the approximation that about 70% of data lies within one standard deviation (as we've seen this to be about 68.26%). This rule is especially useful for making quick, rough estimates about the distribution of data without detailed calculations.

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Most popular questions from this chapter

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Find the probability that a male has a back-to-knee length between \(22.0\) in. and \(24.0\) in.

Assume that \(29.2 \%\) of people have sleepwalked (based on "Prevalence and Comorbidity of Nocturnal Wandering in the U.S. Adult General Population," by Ohayon et al., Neurology, Vol. 78, No. 20). Assume that in a random sample of 1480 adults, 455 have sleepwalked. a. Assuming that the rate of \(29.2 \%\) is correct, find the probability that 455 or more of the 1480 adults have sleepwalked. b. Is that result of 455 or more significantly high? c. What does the result suggest about the rate of \(29.2 \%\) ?

Use these parameters (based on Data Set 1 "Body Data" in Appendix B): \- Men's heights are normally distributed with mean \(68.6 \mathrm{in.}\) and standard deviation \(2.8 \mathrm{in} .\) \- Women's heights are normally distributed with mean 63.7 in. and standard deviation \(2.9 \mathrm{in}\). The Gulfstream 100 is an executive jet that seats six, and it has a doorway height of \(51.6 \mathrm{in}\). a. What percentage of adult men can fit through the door without bending? b. Does the door design with a height of \(51.6\) in. appear to be adequate? Why didn't the engineers design a larger door? c. What doorway height would allow \(40 \%\) of men to fit without bending?

Assume that cans of Coke are filled so that the actual amounts are normally distributed with a mean of \(12.00 \mathrm{oz}\) and a standard deviation of \(0.11\) oz. a. Find the probability that a single can of Coke has at least \(12.19 \mathrm{oz}\). b. The 36 cans of Coke in Data Set 26 "Cola Weights and Volumes" in Appendix \(\mathrm{B}\) have a mean of \(12.19\) oz. Find the probability that 36 random cans of Coke have a mean of at least \(12.19 \mathrm{oz}\) c. Given the result from part (b), is it reasonable to believe that the cans are actually filled with a mean equal to \(12.00\) oz? If the mean is not equal to \(12.00 \mathrm{oz}\), are consumers being cheated?

When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 929 peas, with 705 of them having red flowers. If we assume, as Mendel did, that under these circumstances, there is a \(3 / 4\) probability that a pea will have a red flower, we would expect that \(696.75\) (or about 697 ) of the peas would have red flowers, so the result of 705 peas with red flowers is more than expected. a. If Mendel's assumed probability is correct, find the probability of getting 705 or more peas with red flowers. b. Is 705 peas with red flowers significantly high? c. What do these results suggest about Mendel's assumption that \(3 / 4\) of peas will have red flowers?
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