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Chapter 17: Problem 7

A company producing and packaging sugar for home consumption put labels on their sugar bags noting the weight to be \(500 \mathrm{g}\). Their machines are known to fill the bags with weights that are normally distributed with a standard deviation of \(5.7 \mathrm{g} .\) A bag that contains less than \(500 \mathrm{g}\) is considered to be underweight and is not appreciated by consumers. a) If the company decides to set their machines to fill the bags with a mean of \(512 \mathrm{g},\) what fraction will be underweight? b) If they wish the percentage of underweight bags to be at most \(4 \%,\) what mean setting must they have? c) If they do not want to set the mean as high as \(512 \mathrm{g},\) but instead at \(510 \mathrm{g}\) what standard deviation gives them at most \(4 \%\) underweight bags?

Short Answer

Expert verified
a) 1.74%, b) 510 g, c) 5.71 g.

Step by step solution

01

Understanding the Normal Distribution

The problem involves a normal distribution where the machine fills the bags with a mean weight. We will use the Z-score formula to transform our normal distribution into a standard normal distribution to calculate probabilities. For any normal distribution with mean \( \mu \) and standard deviation \( \sigma \), the Z-score for a value \( x \) is given by \( Z = \frac{x - \mu}{\sigma} \).
02

Calculating the Fraction of Underweight Bags for \( \mu = 512 \) g

To find the fraction of underweight bags with \( \mu = 512 \) g and \( \sigma = 5.7 \) g, we first calculate the Z-score for \( 500 \) g:\[ Z = \frac{500 - 512}{5.7} = \frac{-12}{5.7} \approx -2.11 \]Using standard normal distribution tables or tools, find the probability associated with \( Z = -2.11 \). This probability approximates to \( P(Z < -2.11) \approx 0.0174 \), indicating 1.74% of the bags are underweight.
03

Setting Mean for 4% Underweight Bags

To find the mean that results in 4% of bags being underweight, we find the Z-score associated with a cumulative probability of 0.04. The Z-score for 4% is approximately \( Z = -1.75 \). Using the Z-score formula solve for \( \mu \):\[ -1.75 = \frac{500 - \mu}{5.7} \]\[ \mu = 500 + 1.75 \times 5.7 \approx 510 \text{ grams} \].
04

Determining Standard Deviation for \( \mu = 510 \) g

With a mean of \( 510 \) g, solve the Z-score equation for \( \sigma \) to ensure at most 4% are underweight. Use \( Z = -1.75 \):\[ -1.75 = \frac{500 - 510}{\sigma} \]\[ \sigma = \frac{10}{1.75} \approx 5.71 \text{ grams} \].

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Key Concepts

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

Z-score
The Z-score is a crucial concept when dealing with normal distributions. It is a measure that indicates how many standard deviations an element is from the mean. If you have a value and you want to understand its position within a distribution, you'd convert it to a Z-score using the formula:\[ Z = \frac{x - \mu}{\sigma} \]where:
  • \( x \) is the value you are examining,
  • \( \mu \) is the mean of the distribution,
  • \( \sigma \) is the standard deviation.
Converting to a Z-score allows comparing elements from different distributions. It's a way to standardize different data points so they can be compared universally. This makes it simpler to use statistical tables to find probabilities linked with a normal distribution.
For example, in the sugar bag problem, evaluating Z-scores helps determine the probability of bags being under or overweight.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation indicates high variability.
In the sugar bag exercise, the standard deviation is given as \(5.7 \, \mathrm{g}\). This means the weights of the bags typically deviate from the mean by this value.
Understanding this concept is important because it affects the calculation of the Z-score, and helps determine the proportion of data points (such as underweight sugar bags) in specific segments of the distribution. Adjusting the standard deviation can majorly influence the precision of the distributions we study.
  • A greater \(\sigma\) spreads out the distribution, possibly increasing the count of underweight bags.
  • A smaller \(\sigma\) concentrates values close to the mean and may help meet quality requirements, thereby reducing the underweight bags.
Using standard deviation effectively lets manufacturers control how tightly the product weights are regulated around the desired mean.
Probability
Probability within the context of normal distributions indicates the likelihood an observed event, like a sugar bag being underweight, occurs. Probability values are determined using the Z-score in a standard normal distribution table, which provides the probability that a value lies to the left of a given Z-score (often denoting less than or equal probabilities).
The main aim in our exercise is to calculate the probability that bags are underweight, which is essentially determining how often consumers will receive an unsatisfactory product.
Using our calculated Z-score, we employed the normal distribution table to discover this probability. In this scenario, if a bag's probability to be underweight was calculated as \(0.0174\), that translates to \(1.74\%\) of bags, meaning only a small fraction is expected to fall below satisfactory weight.
  • A probability of \(0.04\) or \(4\%\) conveys managing fewer underweight bags, and it becomes a target for manufacturing settings to adjust mean or \(\sigma\scriptsize'\) depending on production flexibility.
Understanding probability in this way is crucial for making data-driven decisions to minimize issues and maintain product consistency.

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Most popular questions from this chapter

A machine is set to produce bags of salt, whose weights are distributed normally, with a mean of \(110 \mathrm{g}\) and standard deviation of \(1.142 \mathrm{g}\). If the weight of a bag of salt is less than \(108 \mathrm{g}\), the bag is rejected. With these settings, \(4 \%\) of the bags are rejected. The settings of the machine are altered and it is found that \(7 \%\) of the bags are rejected. a) (i) If the mean has not changed, find the new standard deviation, correct to three decimal places. The machine is adjusted to operate with this new value of the standard deviation. (ii) Find the value, correct to two decimal places, at which the mean should be set so that only \(4 \%\) of the bags are rejected. b) With the new settings from part a), it is found that \(80 \%\) of the bags of salt have a weight which lies between \(A \mathrm{g}\) and \(\mathrm{Bg}\), where \(A\) and \(B\) are symmetric about the mean. Find the values of \(A\) and \(B\), giving your answers correct to two decimal places.

Let \(X\) denote a random variable that has a Poisson distribution with mean \(\mu=3\) Find the following probabilities, both manually and with a GDC: a) \(P(x=5)\) b) \(P(x<5)\) c) \(P(x \geqslant 5)\) d) \(P(x \geqslant 5 | x \geq 3)\)

Medical research has shown that a certain type of chemotherapy is successful \(70 \%\) of the time when used to treat skin cancer. In a study to check the validity of such a claim, researchers chose different treatment centres and chose five of their patients at random. Here is the probability distribution of the number of successful treatments for groups of five: $$\begin{array}{|l|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(x) & 0.002 & 0.029 & 0.132 & 0.309 & 0.360 & 0.168 \\ \hline \end{array}$$ a) Find the probability that at least two patients would benefit from the treatment. b) Find the probability that the majority of the group does not benefit from the treatment. c) Find \(E(X)\) and interpret the result. d) Show that \(\sigma(x)=1.02\) e) Graph \(P(x)\). Locate \(\mu, \mu \pm \sigma\) and \(\mu \pm 2 \sigma\) on the graph. Use the empirical rule to approximate the probability that \(x\) falls in this interval. Compare this with the actual probability.

Two children, Alan and Belle, each throw two fair cubical dice simultaneously. The score for each child is the sum of the two numbers shown on their respective dice. a) (i) Calculate the probability that Alan obtains a score of 9 (ii) Calculate the probability that Alan and Belle both obtain a score of \(9 .\) b) (i) Calculate the probability that Alan and Belle obtain the same score. (ii) Deduce the probability that Alan's score exceeds Belle's score. c) Let \(X\) denote the largest number shown on the four dice. (i) Show that for \(P(X \leqslant x)=\left(\frac{x}{6}\right)^{4},\) for \(x=1,2, \ldots, 6\) (ii) Copy and complete the following probability distribution table. $$\begin{array}{|l|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X=x) & \frac{1}{1296} & \frac{15}{1296} & & & & \frac{671}{1296} \\ \hline \end{array}$$ (iii) Calculate \(E(X)\)

\(X \sim N\left(\mu, \sigma^{2}\right) . P(x>19.6)=0.16\) and \(P(x<17.6)=0.012 .\) Find \(\mu\) and \(\sigma\)
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