/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Use the data in the table below ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 less than 21 in.

Short Answer

Expert verified
The probability is approximately 0.0116 or 1.16%.

Step by step solution

01

- Understand the Given Data

We are provided with a mean and standard deviation for the sitting back-to-knee length of adult males. The mean is 23.5 inches and the standard deviation is 1.1 inches. The distribution is normal.
02

- Standardize the Value

To find the probability that a male has a back-to-knee length less than 21 inches, we need to convert this value to a z-score using the formula: \[ Z = \frac{X - \text{Mean}}{\text{Standard Deviation}} \]Substitute the given values: \[ Z = \frac{21 - 23.5}{1.1} \]
03

- Calculate the Z-Score

Calculate the z-score:\[ Z = \frac{21 - 23.5}{1.1} = \frac{-2.5}{1.1} \ Z = -2.27 \]
04

- Look Up the Z-Score in a Standard Normal Distribution Table

Using a standard normal distribution table, find the probability corresponding to a z-score of -2.27. This value is approximately 0.0116.
05

- Interpret the Probability

The probability that a male has a back-to-knee length less than 21 inches is 0.0116 or 1.16%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

z-score calculation
Understanding z-scores is crucial when working with normal distributions. A **z-score** tells us how many standard deviations an element is from the mean.
To calculate it, use the formula: .
In our example, we're trying to find the probability that a male has a back-to-knee length < 21 inches. First, we standardize this value by calculating its z-score.
Here’s the formula:

.
  • Mean (μ): 23.5 inches
  • Standard Deviation (σ): 1.1 inches
  • Measured Value (X): 21 inches

Apply the given values: .Calculate the result step-by-step:

The final z-score is -2.27. This means that 21 inches is 2.27 standard deviations below the mean.
probability
Now that you've calculated the z-score, the next step is to find the corresponding **probability**. This tells us how likely a given event is to occur.
For a normal distribution, we use the **z-score** to find this probability. First, understand that probabilities in a normal distribution curve range from 0 to 1.
If we're looking up the probability for a z-score of -2.27, it means we need the area under the normal distribution curve to the left of this z-score.
This area represents the percentage of values that fall below 21 inches. We can find this using the **Standard Normal Distribution Table* or a z-score table, which is a statistical tool.
Each value in the table corresponds to a particular probability. For any z-score, simply locate it in the table to find the associated probability.
In our case, the table shows that a **z-score** of -2.27 corresponds to a probability of approximately 0.0116.
Therefore, the probability that a male's back-to-knee length is less than 21 inches is approximately **0.0116** or **1.16%**.
standard normal distribution table
The **Standard Normal Distribution Table** is vital for finding probabilities in a normal distribution. This table gives the area (or probability) to the left of a given z-score in a standard normal distribution.
Here's a quick guide on how to use it:
  • Find the row corresponding to the first two digits of your z-score
  • Locate the column corresponding to the second decimal place
Let's say our z-score is -2.27:
  • First two digits: -2.2
  • Second decimal place: 0.07

Combining these two finds the probability value in the table: 0.0116.
Keep in mind that this table only provides positive z-scores, because the normal distribution is symmetric. For negative z-scores like -2.27, the probabilities look the same as for positive values, just on the opposite side of the mean.
Use the table by following general rules:
  • If your z-score is positive, look it up directly.
  • If your z-score is negative, also look up the positive counterpart and infer the result (since tables show cumulative areas).
Being comfortable with this table greatly empowers your understanding of probabilities in a normal distribution.
normal distribution graph
The **normal distribution graph** is one of the most useful tools for visualizing data. It is characterized by its bell-shaped curve and is symmetrical about the mean.
  • Mean (μ): Central peak
  • Standard Deviation (σ): Measures the spread or width
For our exercise, the mean back-to-knee length for males is 23.5 inches, and the standard deviation is 1.1.
The graph shows how data is distributed around the mean. The highest point of the bell curve corresponds to the mean. As you move away from the mean, the curve decreases, indicating fewer occurrences of extreme values.
You can visualize the probability we calculated earlier by shading the area of the curve to the left of z = -2.27. This shaded area represents 1.16% of the total area under the curve, depicting the proportion of males having a sitting back-to-knee length less than 21 inches.
Understanding the normal distribution graph helps you grasp the concept of probabilities in normally distributed data. Not all datasets will form a perfect bell curve, but this graph provides a strong foundation for analyzing and interpreting data distributions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(29.2 \%\) ?

Computers are commonly used to randomly generate digits of telephone numbers to be called when conducting a survey. Can the methods of this section be used to find the probability that when one digit is randomly generated, it is less than 3 ? Why or why not? What is the probability of getting a digit less than 3 ?

In a study of babies born with very low birth weights, 275 children were given IQ tests at age 8 , and their scores approximated a normal distribution with \(\mu=95.5\) and \(\sigma=16.0\) (based on data from "Neurobehavioral Outcomes of School-age Children Born Extremely Low Birth Weight or Very Preterm," by Anderson et al., Journal of the American Medical Association, Vol. 289, No. 24 ). Fifty of those children are to be randomly selected without replacement for a follow-up study. a. When considering the distribution of the mean IQ scores for samples of 50 children, should \(\sigma_{-}\) be corrected by using the finite population correction factor? Why or why not? What is the value of \(\sigma_{\bar{x}}\) ? b. Find the probability that the mean IQ score of the follow-up sample is between 95 and 105 .

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?

Use the given data values to identify the corresponding z scores that are used for a normal quantile plot, then identify the coordinates of each point in the normal quantile plot. Construct the normal quantile plot, then determine whether the data appear to be from a population with a normal distribution. A sample of depths (km) of earthquakes is obtained from Data Set 21 "Earthquakes" in Appendix B: 17.3, 7.0,7.0,7.0,8.1,6.8.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.