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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 must be 6.094 kg.

Step by step solution

01

Understand the Problem

We are given a normal distribution representing the amount of cement in bags, with a standard deviation \(\sigma = 0.05\) kg. We need to find the mean so that only 3% of bags contain less than 6 kg.
02

Use Z-score formula

To solve this, we can use the Z-score formula: \( Z = \frac{X - \mu}{\sigma} \), where \(X = 6\) kg, \(\mu\) is the mean, and \( \sigma = 0.05 \) kg.
03

Determine the Z-score for 3%

Consult a standard normal distribution table or use a calculator to find that a probability of 3% corresponds to a Z-score of approximately \(-1.88\).
04

Rearrange Z-score formula to solve for mean

We rearrange the Z-score formula to solve for \(\mu\): \( \mu = X - Z \cdot \sigma \). Substituting the known values gives \( \mu = 6 - (-1.88) \cdot 0.05 \).
05

Calculate the mean

Calculate \( \mu = 6 - (-1.88) \cdot 0.05 = 6 + 0.094 = 6.094 \) kg.

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Key Concepts

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

Standard Deviation
Standard deviation is a key concept in statistics that measures how spread out the numbers are in a data set. In the context of a normal distribution, it indicates how much the individual data points differ from the mean. It's denoted by the Greek letter \( \sigma \). Let's consider our cement bag example:
  • The standard deviation \( \sigma \) is given as 0.05 kg.
  • This means that most of the bags' weights are within 0.05 kg of the mean weight.
A smaller standard deviation implies that the data points are closer to the mean, while a larger standard deviation indicates that they are more spread out. In our problem, understanding the standard deviation helps us determine how much the mean can vary while keeping the specified percentage of weights under 6 kg.
Z-score
The Z-score is a measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. The Z-score tells us exactly how many standard deviations away a certain point is from the mean. It's crucial for interpreting data in a normal distribution like the one in our bag example.
In the given exercise:
  • We know that a 3% chance of a bag weighing less means we are looking at values on the lower tail of the distribution.
  • The associated Z-score for a 3% probability is approximately -1.88.
  • A negative Z-score indicates that the value is below the mean.
Using Z-scores allows us to pinpoint where specific values fall within the entire distribution, thereby helping us calculate the required mean that ensures only 3% of the bags fall below the 6 kg mark.
Mean Calculation
The mean is the average value in a set of numbers and is central to describing a normal distribution. In our context, when we determine a mean weight for cement bags, we establish a central tendency around which the bags' weights are distributed.
To solve for the mean in this exercise:
  • We rearranged the Z-score formula: \( Z = \frac{X - \mu}{\sigma} \) to solve for \( \mu \).
  • We had an observed value \( X = 6 \, \text{kg} \), and found that a Z-score of -1.88 aligns with a 3% probability.
  • Substitute these into the rearranged formula: \( \mu = X - Z \cdot \sigma \).
  • The calculation \( \mu = 6 - (-1.88) \times 0.05 \) gives us \( \mu = 6.094 \, \text{kg} \).
This mean ensures that the tipping point, representing bags below 6 kg, is in the 3% lowest range of weights.
Probability
Probability is the measure of the likelihood that an event will occur. It is expressed as a number between 0 and 1, where 0 means the event will not happen and 1 means it will certainly happen. Let's see how it plays out in our exercise:
  • We wanted the probability of a bag containing less than 6 kg to be just 3% or 0.03.
  • This places these bags in the lower extreme of the distribution, which corresponds to that small slice of the area under the curve of a normal distribution.
  • Using standard scales and Z-scores, we translate these probability percentages into numerical scores.
Understanding probability in this context enables us to quantify how likely it is to place, fill, or adjust the mean, allowing us to meet specified conditions like ensuring only 3% of bags have less than 6 kg of cement.

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