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Chapter 13: Problem 59

Suppose that the actual amount of cement that a filling machine puts into 'six-kilogram' bags is a normal random variable with \(\sigma=0.05 \mathrm{~kg}\). If only \(3 \%\) of bags are to contain less than \(6 \mathrm{~kg}\), what must be the mean fill of the bags?

Short Answer

Expert verified
The mean fill of the bags must be approximately 6.094 kg.

Step by step solution

01

Understanding the Given Data

We are given a normally distributed random variable representing the weight of cement bags. It has a standard deviation \( \sigma = 0.05 \) kg. Additionally, we know that only \(3\%\) of the bags should weigh less than \(6\) kg. Our goal is to find the mean \( \mu \) that achieves this condition.
02

Using the Z-Score Definition

The Z-score describes how many standard deviations a value is from the mean. For a normal distribution, \(3\%\) of the data lies below a certain point, corresponding to a Z-score that often needs to be determined using statistical tables or a normal distribution calculator. Since 3% corresponds to the lower tail, we look for the Z-score that satisfies \( P(Z < a) = 0.03 \).
03

Finding the Z-Score

Using standard normal distribution tables, we find that \( P(Z < a) = 0.03 \) corresponds to a Z-score \( a \approx -1.88 \). Statistical software or calculators can confirm this value.
04

Setting Up the Equation

The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \), where \(X\) is the value we are interested in (\(6\) kg in this problem), \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Using our Z-score, \(-1.88 = \frac{6 - \mu}{0.05} \).
05

Solving for the Mean

Rearrange the equation \(-1.88 = \frac{6 - \mu}{0.05}\) to solve for \(\mu\):Multiply both sides by \(0.05\): \(-1.88 \times 0.05 = 6 - \mu\)Simplify the left side:\(-0.094 = 6 - \mu\)Add \(\mu\) to both sides and add \(0.094\) to both sides:\(\mu = 6 + 0.094 = 6.094\).

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Key Concepts

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

Standard Deviation
In statistics, the standard deviation is a measure of the amount of variation or dispersion in a set of values. It indicates how much individual data points differ from the mean of the sample.
In this exercise, the standard deviation is given as \( \sigma = 0.05 \) kg.
This number tells us that the weights of the cement bags tend to deviate by about 0.05 kg from their average value.
  • A small standard deviation like 0.05 indicates that the data points are close to the mean.
  • A larger standard deviation would imply more spread out data.
Understanding the standard deviation is crucial for calculating probabilities in a normal distribution and helps assess how consistent the cement filling process is.
Z-score
The Z-score is a statistic that tells us how many standard deviations an element is from the mean. It's a standard normal variable, offering a way to compare different data points to the mean more easily.
To find the Z-score for a specific value, use the formula:\[Z = \frac{X - \mu}{\sigma}\]where:
  • \( X \) is the value of interest (6 kg in our problem),
  • \( \mu \) is the mean we are trying to find,
  • \( \sigma \) is the standard deviation (0.05 kg).
In the context of the exercise: we found the Z-score corresponding to the lower 3% of our data using statistical tables, which is approximately -1.88.
This means 6 kg is 1.88 standard deviations below the desired mean.
Mean Calculation
Calculating the mean involves determining the central value around which data points are distributed. In this exercise, we need to find the mean that results in only 3% of bags containing less than 6 kg of cement.
Using the Z-score formula:\[Z = \frac{X - \mu}{\sigma}\]We can rearrange to solve for the mean (\( \mu \)):
  • Substitute the given values: \( -1.88 = \frac{6 - \mu}{0.05} \)
  • Multiply by 0.05: \(-1.88 \times 0.05 = 6 - \mu\)
  • Solve for \( \mu \) by adding 6 on both sides: \(-0.094 + 6 = \mu\)
  • Thus, \( \mu = 6.094 \) kg.
This mean ensures that only 3% of the bags fall below the 6 kg mark.
Statistical Tables
Statistical tables, particularly Z-tables, are essential tools for finding the probabilities and Z-scores in a normal distribution. Z-tables show the percentage of values expected to be below (or above) a particular Z-score.
In this exercise, these tables were used to find the Z-score that corresponds to the lower 3% of the distribution.
  • Typically, these tables are organized to show cumulative probabilities from the left tail of the distribution.
  • By looking up the value 0.03 in the Z-table, we determine the Z-score to be approximately -1.88.
Such statistical tables are crucial in both education and applied statistics to make sense of raw data in the context of theoretical models like the normal distribution.

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Most popular questions from this chapter

A manufacturer has agreed to dispatch small servomechanisms in cartons of 100 to a distributor. The distributor requires that \(90 \%\) of cartons contain at most one defective servomechanism. Assuming the Poisson approximation to the binomial distribution, write down an equation for the Poisson parameter \(\lambda\) such that the distributor's requirements are just satisfied. Solve by trial and error (approximate solution \(0.5\) ), and hence find the required proportion of manufactured servomechanisms that must be satisfactory.

A major airline operates 350 flights a day throughout the world. The probability that a flight will be delayed for more than one hour, for any reason, is \(0.7 \%\). If more than four flights suffer such delays in any one day, the implications for route organization and crewing become serious. Call such a day a 'flap-day'. Using approximations as appropriate, find the probabilities that (a) any particular day is a flap-day; (b) two flap-days (not more) occur in one week; (c) more than fifty flap-days occur in a year of 365 days.

The distribution of the daily number of malfunctions of a certain computer is given by the following table: $$ \begin{array}{l|ccccccc} \hline \begin{array}{l} \text { Number of } \\ \text { malfunctions } \end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Probability } & 0.17 & 0.29 & 0.27 & 0.16 & 0.07 & 0.03 & 0.01 \\ \hline \end{array} $$Find the mean, the median and the standard deviation of this distribution.

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Assume that (a) an aircraft can land safely if at least half of its engines are working, (b) the probability of an engine failing is \(0.1\), and (c) engine failures are independent. Which is safer, a four-engine aircraft or a twoengine aircraft?
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