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Ch. 5 - Normal Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 5, Problem 5.3.39

Bags of Baby Carrots The weights of bags of baby carrots are normally distributed, with a mean of 32 ounces and a standard deviation of 0.36 ounce. Bags in the upper 4.5% are too heavy and must be repackaged. What is the most a bag of baby carrots can weigh and not need to be repackaged?

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Identify the key parameters of the normal distribution: the mean (μ = 32 ounces) and the standard deviation (σ = 0.36 ounces). The problem asks for the maximum weight a bag can have and not be in the upper 4.5% of the distribution.
Determine the z-score corresponding to the upper 4.5% of the normal distribution. Since the upper 4.5% corresponds to the top tail of the distribution, find the z-score for the cumulative probability of 1 - 0.045 = 0.955 using a z-table or statistical software.
Use the z-score formula to relate the z-score to the weight (X): z = (X - μ) / σ. Rearrange this formula to solve for X: X = μ + z * σ.
Substitute the known values into the formula: μ = 32, σ = 0.36, and the z-score obtained in the previous step. This will give the maximum weight a bag can have without being in the upper 4.5%.
Interpret the result: The calculated weight represents the threshold above which bags are considered too heavy and must be repackaged.

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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, showing 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 weights of bags of baby carrots follow a normal distribution, which allows us to use statistical methods to determine probabilities and thresholds.
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Z-Score

A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is calculated by subtracting the mean from the value and then dividing by the standard deviation. In this scenario, the Z-score will help identify the weight threshold above which the top 4.5% of bags fall, indicating the maximum weight a bag can have without needing repackaging.
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Percentiles

Percentiles are measures that indicate the value below which a given percentage of observations in a group of observations falls. For example, the 95th percentile is the value below which 95% of the data points lie. In this question, we need to find the weight corresponding to the 95th percentile to determine the maximum weight a bag can have before it is considered too heavy.
Related Practice
Textbook Question

Finding Area

In Exercises 23–36, find the indicated area under the standard normal curve. If convenient, use technology to find the area.


To the left of z=1.365

Textbook Question

Approximating a Binomial Distribution In Exercises 17 and 18, a binomial experiment is given. Determine whether you can use a normal distribution to approximate the binomial distribution. If you can, find the mean and standard deviation. If you cannot, explain why.


Bachelor’s Degrees Twenty-two percent of adults over 18 years of age have a bachelor’s degree. You randomly select 20 adults over 18 years of age and ask whether they have a bachelor’s degree.

Textbook Question

Finding a z-Score Given an Area In Exercises 23–30, find the indicated z-score.


Find the z-score that has 78.5% of the distribution’s area to its left.

Textbook Question

Finding Probability In Exercises 47–56, find the indicated probability using the standard normal distribution. If convenient, use technology to find the probability.


P(z < - 1.11)

Textbook Question

Interpreting the Central Limit Theorem In Exercises 19–26, find the mean and standard deviation of the indicated sampling distribution of sample means. Then sketch a graph of the sampling distribution.


Renewable Energy The zloty is the official currency of Poland. During a recent period of two years, the day-ahead prices for renewable energy in Poland (in zlotys per mega-watt hour) have a mean of 158.51 and a standard deviation of 33.424. Random samples of size 100 are drawn from this population, and the mean of each sample is determined. (Adapted from Multidisciplinary Digital Publishing Institute)

Textbook Question

True or False? In Exercises 5–8, determine whether the statement is true or false. If it is false, rewrite it as a true statement.


As the sample size increases, the standard deviation of the distribution of sample means increases.