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Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 6, Problem 6.4.7a

Using the Central Limit Theorem. In Exercises 5–8, assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of 1.2 kg and a standard deviation of 4.9 kg (based on Data Set 13 “Freshman 15” in Appendix B).
a. If 1 male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year.

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Step 1: Identify the given parameters. The problem states that the weight gain is normally distributed with a mean (μ) of 1.2 kg and a standard deviation (σ) of 4.9 kg. We are tasked with finding the probability that a randomly selected male college student gains between 0 kg and 3 kg.
Step 2: Standardize the values of 0 kg and 3 kg using the z-score formula: z = (X - μ) / σ. For X = 0 kg, calculate z₁ = (0 - 1.2) / 4.9. For X = 3 kg, calculate z₂ = (3 - 1.2) / 4.9.
Step 3: Use the z-scores calculated in Step 2 to find the corresponding probabilities from the standard normal distribution table (or use statistical software). Let P(z₁) represent the cumulative probability for z₁ and P(z₂) represent the cumulative probability for z₂.
Step 4: To find the probability that the weight gain is between 0 kg and 3 kg, subtract the cumulative probability for z₁ from the cumulative probability for z₂. Mathematically, this is expressed as P(0 ≤ X ≤ 3) = P(z₂) - P(z₁).
Step 5: Interpret the result. The value obtained in Step 4 represents the probability that a randomly selected male college student gains between 0 kg and 3 kg during their freshman year.

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Key Concepts

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

Central Limit Theorem

The Central Limit Theorem (CLT) states that the distribution of the sample means will approach a normal distribution as the sample size increases, regardless of the original distribution of the population. This theorem is crucial for making inferences about population parameters based on sample statistics, especially when dealing with large samples.
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Calculating the Mean

Normal Distribution

A normal distribution is a continuous probability distribution characterized by its bell-shaped curve, defined by its mean and standard deviation. In this context, the weights gained by male college students are normally distributed, which allows us to use properties of the normal distribution to calculate probabilities related to weight gain.
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Finding Standard Normal Probabilities using z-Table

Probability Calculation

Probability calculation involves determining the likelihood of a specific outcome occurring within a defined range. In this scenario, we need to calculate the probability that a randomly selected male college student gains between 0 kg and 3 kg, which requires using the properties of the normal distribution and potentially standardizing the values using the z-score formula.
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Related Practice
Textbook Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.


Aircraft Cockpit The overhead panel in an aircraft cockpit typically includes controls for such features as landing lights, fuel booster pumps, and oxygen. It is important for pilots to be able to reach those overhead controls while sitting. Seated adult males have overhead grip reaches that are normally distributed with a mean of 51.6 in. and a standard deviation of 2.2 in.


a. If an aircraft is designed for pilots with an overhead grip reach of 53 in., what percentage of adult males would not be able to reach the overhead controls? Is that percentage too high?

Textbook Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.


Safe Loading of Elevators The elevator in the car rental building at San Francisco International Airport has a placard stating that the maximum capacity is “4000 lb—27 passengers.” Because 4000/27=148, this converts to a mean passenger weight of 148 lb when the elevator is full. We will assume a worst-case scenario in which the elevator is filled with 27 adult males. Based on Data Set 1 “Body Data” in Appendix B, assume that adult males have weights that are normally distributed with a mean of 189 lb and a standard deviation of 39 lb.


a. Find the probability that 1 randomly selected adult male has a weight greater than 148 lb.

Textbook Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.


Redesign of Ejection Seats When women were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ACES-II ejection seats were designed for men weighing between 140 lb and 211 lb. Weights of women are now normally distributed with a mean of 171 lb and a standard deviation of 46 lb (based on Data Set 1 “Body Data” in Appendix B).


a. If 1 woman is randomly selected, find the probability that her weight is between 140 lb and 211 lb.

Textbook Question

Transformations The heights (in inches) of women listed in Data Set 1 “Body Data” in Appendix B have a distribution that is approximately normal, so it appears that those heights are from a normally distributed population.


a. If 2 inches is added to each height, are the new heights also normally distributed?

Textbook Question

Continuity Correction In testing the assumption that the probability of a baby boy is 0.512, a geneticist obtains a random sample of 1000 births and finds that 502 of them are boys. Using the continuity correction, describe the area under the graph of a normal distribution corresponding to the following. (For example, the area corresponding to “the probability of at least 502 boys” is this: the area to the right of 501.5.)


a. The probability of 502 or fewer boys

Textbook Question

Mendelian Genetics 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.