Ch. 6 - Normal Probability Distributions

Triola - Elementary Statistics 14th Edition

Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook

Triola 14th Edition

Ch. 6 - Normal Probability Distributions

Problem 6.q.7

Chapter 6, Problem 6.q.7

Seat Designs. In Exercises 7–9, assume that when seated, adult males have back-to-knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. (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.

Find the probability that a male has a back-to-knee length greater than 25.0 in.

Verified step by step guidance

Step 1: Identify the key parameters of the normal distribution. The problem states that the back-to-knee lengths are normally distributed with a mean (μ) of 23.5 inches and a standard deviation (σ) of 1.1 inches.

Step 2: Define the event of interest. We are tasked with finding the probability that a male has a back-to-knee length greater than 25.0 inches. This corresponds to P(X > 25.0), where X represents the back-to-knee length.

Step 3: Standardize the value of 25.0 inches using the z-score formula: z = (X - μ) / σ. Substitute the values: X = 25.0, μ = 23.5, and σ = 1.1. This will give the z-score corresponding to 25.0 inches.

Step 4: Use the z-score obtained in Step 3 to find the cumulative probability from a standard normal distribution table or using statistical software. The cumulative probability corresponds to P(Z ≤ z), where Z is the standard normal variable.

Step 5: Since we are interested in P(X > 25.0), calculate the complement of the cumulative probability found in Step 4. This is given by P(X > 25.0) = 1 - P(Z ≤ z).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean. It is characterized by its bell-shaped curve, defined by its mean and standard deviation. In this context, the back-to-knee lengths of adult males follow a normal distribution, which allows for the calculation of probabilities related to specific length values.

Mean and Standard Deviation

The mean is the average of a set of values, while the standard deviation measures the amount of variation or dispersion in a set of values. In this scenario, the mean back-to-knee length is 23.5 inches, and the standard deviation is 1.1 inches. These statistics are crucial for understanding the distribution of lengths and for calculating probabilities of lengths exceeding a certain value.

Z-Score

A Z-score indicates how many standard deviations an element is from the mean. It is calculated by subtracting the mean from the value of interest and then dividing by the standard deviation. In this case, to find the probability that a male has a back-to-knee length greater than 25.0 inches, one would first calculate the Z-score for 25.0 inches, which can then be used to find the corresponding probability from the standard normal distribution table.

Recommended video:

Guided course

06:31

Z-Scores From Given Probability - TI-84 (CE) Calculator

Related Practice

Textbook Question

Seat Designs. In Exercises 7–9, assume that when seated, adult males have back-to-knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. (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.

Find the probability that nine males have back-to-knee lengths with a mean greater than 23.0 in.

Textbook Question

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

c. For a randomly selected subject, find the probability of a bone density test score between -0.67 and 1.29.

Textbook Question

a. For a randomly selected subject, find the probability of a bone density test score greater than -1.37.

Textbook Question

Bone Density Test. In Exercises 1–4, assume that scores on a bone mineral density test are normally distributed with a mean of 0 and a standard deviation of 1.

Bone Density For a randomly selected subject, find the probability of a bone density score between and 2.00.

Textbook Question

Bone Density Test. In Exercises 1–4, assume that scores on a bone mineral density test are normally distributed with a mean of 0 and a standard deviation of 1.

Bone Density Find the bone density score that is the 90th percentile, which is the score separating the lowest 90% from the top 10%.

Textbook Question

Normal Distribution Using a larger data set than the one given for the preceding exercises, assume that cell phone radiation amounts are normally distributed with a mean of 1.17 W/kg and a standard deviation of 0.29 W/kg.

b. Find the value of Q3, the cell phone radiation amount that is the third quartile.