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An insurance office buys paper by the ream, 500 sheets, for use in the copier, fax, and printer. Each ream lasts an average of 4 days, with standard deviation 1 day. The distribution is normal, independent of previous reams. a. Find the probability that the next ream outlasts the present one by more than 2 days. b. How many reams must be purchased if they are to last at least 60 days with probability at least \(80 \%\) ?

Short Answer

Expert verified
a. 0.079. b. 17 reams.

Step by step solution

01

Understanding the Question

We're asked to calculate probabilities and determine quantities based on the Normal distribution of ream durations. Given an average duration of 4 days and a standard deviation of 1 day, we want to find the likelihood of one ream lasting longer than another by more than 2 days and then calculate how many reams are needed to cover 60 days with at least 80% probability.
02

Part A: Express the Problem in Terms of a Normal Distribution

We need to find the probability that the next ream lasts more than 2 days longer than the current one, i.e., the difference in durations exceeds 2 days. Let's denote the lifespan of the present ream by the random variable \(X\) and the lifespan of the next ream by \(Y\), both normally distributed with mean 4 and standard deviation 1. The difference \( Z = Y - X \) follows a Normal distribution with mean \( \mu_Z = 0 \) and standard deviation \( \sigma_Z = \sqrt{1^2 + 1^2} = \sqrt{2} \).
03

Calculate the Probability for Part A

We're looking for \( P(Y - X > 2) = P(Z > 2) \). Standardize \(Z\) using \( Z^* = \frac{Z - 0}{\sqrt{2}} \). We need \( P(Z^* > \frac{2}{\sqrt{2}}) = P(Z^* > \sqrt{2}) \). Use the standard Normal distribution table to find \( P(Z^* > \sqrt{2}) = 1 - P(Z^* \leq \sqrt{2}) = 1 - 0.921 = 0.079 \).
04

Part B: Determine Number of Reams for 60 Days

We want the total usage of several reams, say \(n\), to last at least 60 days with at least 80% probability. Define \(T\) as the total duration of \(n\) reams, which is normally distributed with mean \(4n\) and standard deviation \(\sqrt{n} \). We need \( P(T \geq 60) \), which translates to \( P\left( \frac{T - 4n}{\sqrt{n}} \geq \frac{60 - 4n}{\sqrt{n}} \right) \geq 0.8 \). From the standard Normal table, \(P(Z \geq k) = 0.20\) leads to \(k = -0.84\). So, \( \frac{60 - 4n}{\sqrt{n}} = -0.84 \).
05

Solve the Inequality for Part B

We solve \(60 - 4n = -0.84\sqrt{n}\). Square both sides: \((60 - 4n)^2 = (0.84)^2n\). This leads to solving the quadratic equation \(n^2 - 30n + 225 = 0\), which gives roots using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Calculating, we find \(n \approx 16.23\). Since only an integer number of reams can be bought, round up to 17.

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Key Concepts

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

Probability Calculation
Probability calculations help us determine how likely it is for an event to occur based on given data. In the context of the exercise, we're interested in finding the probability of a ream of paper outlasting another by a certain number of days. To do this, we first define the lifespans as random variables, which are quantities that can vary. Each ream lasts an average of 4 days with a standard deviation of 1 day, following a Normal distribution.

To calculate the probability of the next ream lasting more than the current one by over 2 days, we need to find the probability that the difference between the durations is more than 2 days. We denote this difference by a new variable, say, "Z", which also follows the Normal distribution. By computing the standard deviation of this difference and using statistical tables, we can find that the probability is approximately 0.079, or 7.9%. This means there is a low chance for the next ream to last significantly longer than the current one.
Standard Deviation
Standard deviation is a measure of how spread out the values in a data set are. In statistical terms, it shows the average distance of each data point from the mean. Here, each ream is expected to last 4 days on average, but not all reams will last exactly this duration.

In our problem, the lifespan of each ream has a standard deviation of 1 day. This implies that the actual number of days a ream lasts will fluctuate around the mean of 4 days, most likely within a range of 3 to 5 days. Standard deviation is pivotal for understanding the variation in any normally distributed data, allowing you to compute probabilities of outcomes within certain ranges.

Understanding standard deviation assists in visualizing how far the durations can vary, making it easier to calculate probabilities, as we see with the variance of these paper reams.
Statistical Problem Solving
Statistical problem solving involves meticulous steps to address a question using statistical means. This exercise is centered around predicting the use of paper reams. To solve such statistical problems, it's important to:
  • Identify the random variable and its distribution.
  • Define the problem in statistical terms, such as calculating probabilities or determining needed quantities.
  • Use statistical formulas and tables to find solutions to these defined problems.
For instance, step 4 of our solution required establishing the conditions under which a set of reams would last 60 days with an 80% probability. This involves modeling the problem using the Normal distribution and standardizing it to find a cutoff point using statistical tables. Solving the resulting equations provided the necessary number of reams, prioritizing statistical thinking and methodical calculations.
Random Variable
A random variable is a fundamental concept in statistics that represents a quantity with potential variability. It acts as the mechanism by which randomness is quantified and analyzed. This exercise features random variables "X" and "Y", representing the lifespans of consecutive reams.

Understanding random variables involves recognizing the distribution they follow—in this case, the Normal distribution. This allows predictions and calculations based on known properties like the mean (average duration) and standard deviation (variation from the average).

The random variable is crucial in determining probabilities and expectations. For example, when looking at ream durations, knowing that these values follow a Normal distribution, helps in describing the range of likely outcomes and the likelihood of these occurring, forming the basis for questions a and b in the exercise.

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