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Chapter 6: Problem 44

The weight (in pounds) of "medium-size" watermelons is normally distributed with mean 15 and variance 4. A packing container for several melons has a nominal capacity of 140 pounds. What is the maximum number of melons that should be placed in a single packing container if the nominal weight limit is to be exceeded only \(5 \%\) of the time? Give reasons for your answer.

Short Answer

Expert verified
The maximum number of watermelons is 8.

Step by step solution

01

Understand the Problem

We need to determine the maximum number of watermelons that can be placed in a container such that the combined weight exceeds the container's nominal capacity of 140 pounds only 5% of the time.
02

Calculate Mean and Variance for n Watermelons

The mean weight of one watermelon is 15 pounds, and the variance is 4. If we have \(n\) watermelons, the mean of the total weight will be \(15n\) and the variance of the total weight will be \(4n\), since variances add up for independent random variables.
03

Find the Normal Distribution Parameters

The total weight of \(n\) watermelons is normally distributed with mean \(15n\) and standard deviation \(\sqrt{4n} = 2\sqrt{n}\). Thus, the weight distribution is \(N(15n, (2\sqrt{n})^2)\).
04

Determine the Z-Score

We seek a Z-score corresponding to the upper 5% tail of the normal distribution. The Z-score that corresponds to the 95th percentile (0.05 tail) is approximately 1.645.
05

Set up the Inequality

We want the total weight mean and standard deviation to satisfy: \[ 15n + 1.645 \, \cdot\, 2\sqrt{n} = 140 \] Simplify and solve for \(n\).
06

Solve for n

Rewriting the inequality gives: \( 1.645 \, \cdot\, 2\sqrt{n} = 140 - 15n \). Solve for \( n \):\[ 3.29\sqrt{n} = 140 - 15n \] Calculate to find \( n \) approximately as integer satisfying \( n \approx 8 \).
07

Verification

For \( n = 8 \), the calculation gives \( 15*8 = 120 \) as mean, with \( 3.29\sqrt{8} \approx 9.32 \) making 129.32. Check for the weight exceeding only 5% case. Continue to verify and round if necessary.

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Key Concepts

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

Normal Distribution
Normal distribution is a fundamental concept in statistics, often depicted as a bell-shaped curve. This distribution is symmetric around the mean, meaning most data points cluster around a central point, with fewer points as you move away. The normal distribution is characterized by two parameters:
  • Mean (\( \mu \) variable): It represents the average of all data points.
  • Variance (\( \sigma^2 \) variable): It indicates how spread out the data points are around the mean.
In our watermelon example, the weights follow a normal distribution with a mean (\( \mu = 15 \) pounds) and a variance (\( \sigma^2 = 4 \) pounds2). This ensures that while most watermelons weigh about 15 pounds, some can weigh a bit more or less.
Mean and Variance
The mean and variance are crucial in understanding the distribution of data. In the watermelon problem, the mean tells us the average or expected weight of a single watermelon, which is 15 pounds. The variance of 4 implies how much the weights of these watermelons deviate from the mean, helping us determine the spread or dispersion.
When we consider multiple watermelons, the combined weight has a mean of \( 15n \), where \( n \) is the number of watermelons, and variance \( 4n \). This shows the additivity of variances: the variance of a sum equals the sum of the variances, assuming the weights are independent. The standard deviation, a measure of dispersion closely related to variance, is \( 2\sqrt{n} \), derived from the square root of variance.
Z-score
A Z-score is a statistical measure that indicates how many standard deviations a data point is from the mean. It allows us to compare data from different normal distributions.
For example, in the given problem, we want to find out the weight at which only the top 5% of watermelon totals exceed the nominal capacity. Using a Z-score can help with this, as it transforms the problem into querying standard normal tables.
The Z-score for the 95th percentile (or the top 5% tail, as we want weight exceeding only 5% of the time) is approximately 1.645. This means weights beyond 1.645 standard deviations above the mean fall in the top 5%.
Percentile Calculation
Percentile calculation is used to understand the relative standing of a value within a dataset. It tells us the percentage of data points below a certain value.
In our scenario, we're interested in the 95th percentile. This is where 95% of watermelon weight totals will fall below, and only 5% will exceed. By finding this threshold, we address the problem of not exceeding the given weight limit 95% of the time.
  • We first identify the mean total weight (\(15n\)) and the threshold capacity (140 pounds).
  • Applying the Z-score (1.645), which corresponds to the 95th percentile, we set up the inequality: \( 15n + 1.645 \times 2\sqrt{n} = 140 \).
Solving this equation allows us to find the maximum number of watermelons that satisfy the packing constraint.

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Most popular questions from this chapter

The length of time necessary to tune up a car is exponentially distributed with a mean of .5 hour. If two cars are waiting for a tune-up and the service times are independent, what is the probability that the total time for the two tune-ups will exceed 1.5 hours? [Hint: Recall the result of Example 6.12.]

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If \(Y_{1}\) and \(Y_{2}\) are independent and identically distributed normal random variables with mean \(\mu\) and variance \(\sigma^{2}\), find the probability density function for \(U=(1 / 2)\left(Y_{1}-3 Y_{2}\right)\)

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