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 female has a back-to-knee length greater than \(24.0 \mathrm{in}\).

Step by step solution

01

Understand the provided data

The mean back-to-knee length for females is 22.7 inches, and the standard deviation is 1.0 inch. The distribution is normal.

02

Identify the target measurement

We need to find the probability that a female has a back-to-knee length greater than 24.0 inches.

03

Standardize the target measurement

Convert the target measurement to a standard score (z-score) using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where: X = 24.0 inches, \( \mu \) = 22.7 inches, \( \sigma \) = 1.0 inch.

04

Calculate the z-score

Substitute the values into the formula: \[ Z = \frac{24.0 - 22.7}{1.0} = 1.3 \]

05

Use the z-score to find the probability

Using a standard normal distribution table or a calculator, find the probability that corresponds to the z-score of 1.3. This gives the probability of a female having a back-to-knee length less than 24.0 inches.

06

Calculate the complement probability

Since the problem asks for the probability of having a back-to-knee length greater than 24.0 inches, we take the complement of the result from Step 5. \[ P(X > 24.0) = 1 - P(Z < 1.3) \]

07

Obtain the final probability

From the standard normal distribution table, \( P(Z < 1.3) \approx 0.9032 \). Thus, \[ P(X > 24.0) = 1 - 0.9032 = 0.0968 \]

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Key Concepts

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

Z-Score Calculation

In statistics, a z-score helps us understand how far a specific data point is from the mean of the data set. The z-score is calculated using this formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Here, X is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For our exercise, X is 24.0 inches, the mean (\( \mu \)) is 22.7 inches, and the standard deviation (\( \sigma \)) is 1.0 inch. Plugging these numbers in, we get:
\[ Z = \frac{24.0 - 22.7}{1.0} = 1.3 \]
The z-score of 1.3 tells us that 24.0 inches is 1.3 standard deviations above the mean of 22.7 inches.

Probability Determination

Once we have the z-score, we can find the probability associated with this z-score. This gives us the proportion of observations below our data point in a normal distribution.
We look up the z-score in a standard normal distribution table or use a calculator that provides cumulative probabilities for z-scores. For a z-score of 1.3, the table or calculator tells us that the probability of a back-to-knee length being less than 24.0 inches is approximately 0.9032.
So, if 90.32% of females have a back-to-knee length less than 24.0 inches, we conclude that the remaining females (the ones greater than 24.0 inches) make up the rest of the 100%.

Standard Normal Distribution Table

The standard normal distribution table, also known as the z-table, shows the cumulative probability of a z-score in a standard normal distribution. The values in the table represent the area under the curve to the left of the z-score.
To find the probability associated with our z-score of 1.3, we look it up in the table:

• The intersection of the row for 1.3 and the column for 0.00 gives us the value 0.9032.
• This means the area to the left of z=1.3 is 0.9032, or 90.32%.

Using the standard normal distribution table is a crucial step in determining probabilities from z-scores.

Complement Rule

Finally, we use the complement rule to find the probability of a female having a back-to-knee length greater than 24.0 inches.
The complement rule states:
\[ P(A') = 1 - P(A) \]
Here, P(A') is the probability of the complement of event A. In our case, event A is having a back-to-knee length less than 24.0 inches, and P(A) is 0.9032. Therefore,
\[ P(X > 24.0) = 1 - P(Z < 1.3) = 1 - 0.9032 = 0.0968 \]
Hence, the probability of a female having a back-to-knee length greater than 24.0 inches is approximately 9.68%.