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A pizza company advertises that it puts \(0.5 \mathrm{lb}\) of real mozzarella cheese on its medium pizzas. In fact, the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of \(0.5 \mathrm{lb}\) and \(\mathrm{a}\) standard deviation of \(0.025 \mathrm{lb}\). a. What is the probability that the amount of cheese on a medium pizza is between \(0.525\) and \(0.550 \mathrm{lb}\) ? b. What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? c. What is the probability that three randomly selected medium pizzas all have at least \(0.475 \mathrm{lb}\) of cheese?

Short Answer

Expert verified
a. The probability that the amount of cheese on a medium pizza is between \(0.525\) and \(0.550 \) lb is the area under the standard normal curve between \(1\) and \(2\). b. The probability that the amount of cheese on a medium pizza exceeds the mean value by more than \(2\) standard deviations is the area under the standard normal curve to the right of \(2\). c. The probability that three randomly selected medium pizzas all have at least \(0.475 \) lb of cheese is \((P(X \geq 0.475))^3\).

Step by step solution

01

Compute Z-score for part a

Firstly, the Z-scores for \(0.525 \) lb and \(0.550 \) lb are calculated using the formula mentioned. Subtract the mean \(0.5 \) from each value and then divide the result by the standard deviation \(0.025 \). The calculated Z-scores are \(1\) and \(2\) respectively.
02

Calculate probability for part a

Using the standard normal distribution (mean \(0\) standard deviation \(1\)), the probability that Z is between \(1\) and \(2\) is found using the Z-table or using statistical software. This represents the probability that the amount of cheese is between \(0.525\) and \(0.550 \) lb.
03

Compute probability for part b

For part b, the Z-score for the point \(0.5 + 2*0.025\) (mean plus two standard deviations) is computed as \(2\). Using the standard normal distribution, the probability that Z exceeds \(2\) is then calculated. This is the same as finding the area under the curve to the right of the point \(2\).
04

Compute probability for part c

For part c, firstly the Z-score for the point \(0.475 \) lb is computed. Then \(P(X \geq 0.475)\) (probability that one pizza has at least \(0.475\) lb of cheese) is found using the standard normal distribution. As the pizzas are selected randomly, assume they are independent. The probability that all three have at least \(0.475 \) lb of cheese is \((P(X \geq 0.475))^3\).

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Key Concepts

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

Z-score
The concept of a Z-score is a statistical way to measure how far a particular data point is from the mean of a data set. It is expressed in terms of standard deviations. Suppose you have an observed value from your data. In that case, the Z-score tells you how many standard deviations it is from the mean. This makes it easier to compare different data points from different data sets or to the same data set.

To calculate a Z-score, use the formula:
  • Subtract the mean of the data set from your observed value.
  • Then, divide the result by the standard deviation of the data set.
Mathematically, it can be expressed as:\[ Z = \frac{(X - \mu)}{\sigma} \]where:
  • \(X\) is the observed value.
  • \(\mu\) is the mean of the data.
  • \(\sigma\) is the standard deviation of the data.
By calculating the Z-score, we can tell if the observed value is typical or unusual compared to the mean. It's a useful way to understand where exactly a particular observation stands within a distribution.
Standard Deviation
Standard deviation is a measure of the variation or spread in a set of values. It shows how much the individual observations of a data set typically vary from the mean. A smaller standard deviation means that the values tend to be close to the mean, while a larger standard deviation indicates the values are more spread out.

In the context of normal distribution, the standard deviation helps in describing the shape of the distribution curve. It's important because:
  • It plays a vital role in calculating Z-scores.
  • It defines the width of the bell-shaped curve in a normal distribution graph.
  • Most data points in a normal distribution lie within three standard deviations from the mean.
For example, if you have a very small standard deviation, it means the data points are clustered very closely around the mean. Conversely, a large standard deviation means that the data points are spread over a wider range. Understanding standard deviation helps in predicting the likelihood of an observation occurring within a particular range of the mean.
Probability Calculation
Probability calculation in the context of normal distribution often involves using Z-scores to find the likelihood of a particular range of outcomes. To calculate these probabilities, a Z-table, statistical software, or a calculator with statistical functions is typically used.

There are three steps commonly involved when calculating probabilities related to a normal distribution:
  • Compute the Z-score for the observed value(s) you're interested in. This helps in standardizing your data point concerning the mean and standard deviation.
  • Use the Z-score to find the probability from the standard normal distribution. This can be done using a Z-table or statistical software.
  • If dealing with independent events, use the rule of multiplication for probabilities to find the chance of all events occurring.
For instance, you might be interested in finding the probability of cheese on a pizza weighing between two values. You would compute the Z-scores for both values, find the corresponding probabilities in the Z-table, and calculate the difference between these probabilities. This would tell you how likely it is to find a value within that specific range. Probability calculations like these enable analysts and researchers to make informed predictions based on statistical data.

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