Question

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

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

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.

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.

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 \(2\).

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 \((P(X \geq 0.475))^3\).