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Chapter 6: Problem 31

A concert has three pieces of music to be played before intermission. The time taken to play each piece has a normal distribution. Assume that the three times are independent of each other. The mean times are 15,30 , and \(20 \mathrm{~min}\), respectively, and the standard deviations are 1,2, and \(1.5 \mathrm{~min}\), respectively. What is the probability that this part of the concert takes at most \(1 \mathrm{~h}\) ? Are there reasons to question the independence assumption? Explain.

Short Answer

Expert verified
The probability that this part of the concert takes at most 1 hour is approximately 0.0314. The independence assumption should be questioned if there are factors like performer fatigue influencing the performances.

Step by step solution

01

Define the Problem

We need to calculate the probability that the total time taken by three independent musical pieces is at most 60 minutes, given their means and standard deviations.
02

Calculate Total Mean and Variance

For independent normal distributions, the total mean is the sum of the individual means and the total variance is the sum of the individual variances. Therefore, the total mean is \[ \mu_{total} = 15 + 30 + 20 = 65 \]And the variances are calculated by squaring the standard deviations and summing them: \[ \sigma_{total}^2 = 1^2 + 2^2 + 1.5^2 = 1 + 4 + 2.25 = 7.25 \]Thus, the total standard deviation is \[ \sigma_{total} = \sqrt{7.25} \approx 2.69 \].
03

Standardize the Total Time

To find the probability, we need to standardize the total time to fit a standard normal distribution. The corresponding standardized value is calculated as: \[ Z = \frac{X - \mu_{total}}{\sigma_{total}} = \frac{60 - 65}{2.69} \approx -1.86 \].
04

Find the Probability Using Standard Normal Distribution

Using the standard normal distribution table or a calculator, find the probability that \( Z < -1.86 \). From the Z-table, \( P(Z < -1.86) \approx 0.0314 \). This is the probability that the total time taken is at most 60 minutes.
05

Question Independence Assumption

The independence assumption is crucial because if the pieces are somehow dependent (e.g., musicians may tire as the performance progresses), the calculation doesn't hold. We must evaluate the practical situation to ensure that such dependencies do not exist; otherwise, the calculated probability might not be accurate.

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Key Concepts

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

Normal Distribution
In probability theory, the normal distribution is a foundational concept. It describes how values of a variable are distributed. Often referred to as the bell curve due to its shape, the normal distribution is symmetric around its mean.
  • The mean, median, and mode are all equal and located at the center.
  • The curve is defined by two parameters: mean (average) and standard deviation (spread).
  • Most of the data falls within three standard deviations of the mean.
In the original exercise, each piece of music has play times that follow this distribution, with different means and standard deviations. This is typical in real-world scenarios where performance times can vary slightly.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out the numbers are from the mean.
  • A small standard deviation means the values are close to the mean.
  • A large standard deviation means the values are more spread out.
In this exercise, each piece of music has a different standard deviation: 1, 2, and 1.5 minutes, respectively. To find the total standard deviation when adding independent normal variables, like music pieces, we first calculate variance (square of standard deviation), sum them up, and then take the square root. This combined value, approximately 2.69, represents the total standard deviation of all pieces played together.
Independence Assumption
The independence assumption is vital in many statistical calculations. It means that the occurrence of one event does not affect the occurrence of another.
  • This assumption simplifies calculations as events can be considered separately.
In this music scenario, independence would mean that the duration of playing one piece doesn't affect the others. However, in real life, this might not always hold true. For instance, if musicians tire over time, the playtime of subsequent pieces might be extended slightly. Not taking potential dependencies into account can lead to inaccurate probability calculations.
Z-Score Calculation
A Z-score is a statistical measurement that describes a score's relationship to the mean of a group of scores. It's expressed in terms of standard deviations from the mean.
  • A Z-score of 0 indicates the score is identical to the mean.
  • A positive Z-score shows the score is above the mean.
  • A negative Z-score shows it's below the mean.
For the concert timing problem, we calculate the Z-score to determine how probable it is that the total playing time is at most 60 minutes. The Z-score formula is \[ Z = \frac{X - \mu_{total}}{\sigma_{total}} \] where \( X \) is the desired time, \( \mu_{total} \) the total mean time, and \( \sigma_{total} \) the total standard deviation. A calculated Z of approximately -1.86 suggests the total time is less than average, and using Z tables, we see there's only a 3.14% chance of finishing within 60 minutes.

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Most popular questions from this chapter

A company maintains three offices in a region, each staffed by two employees. Information concerning yearly salaries (1000's of dollars) is as follows: $$ \begin{array}{lcccccc} \text { Office } & 1 & 1 & 2 & 2 & 3 & 3 \\ \text { Employee } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Salary } & 29.7 & 33.6 & 30.2 & 33.6 & 25.8 & 29.7 \end{array} $$ a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary \(\bar{X}\). b. Suppose one of the three offices is randomly selected. Let \(X_{1}\) and \(X_{2}\) denote the salaries of the two employees. Determine the sampling distribution of \(X\). c. How does \(E(X)\) from parts (a) and (b) compare to the population mean salary \(\mu\) ?

The lifetime of a type of battery is normally distributed with mean value \(10 \mathrm{~h}\) and standard deviation \(1 \mathrm{~h}\). There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \(5 \%\) of all packages?

Use moment generating functions to show that if \(X_{3}=X_{1}+X_{2}\), with \(X_{1} \sim \chi_{v_{1}}^{2}, X_{3} \sim \chi_{v_{5}}^{2}, v_{3}>v_{1}\), and \(X_{1}\) and \(X_{2}\) are independent, then \(X_{2} \sim \chi_{v_{3}-v_{1}}^{2}\).

Knowing that \((n-1) S^{2} / \sigma^{2} \sim \gamma_{n-1}^{2}\) for a normal random sample, a. Show that \(E\left(S^{2}\right)=\sigma^{2}\) b. Show that \(V\left(S^{2}\right)=2 \sigma^{4} /(n-1)\). What happens to this variance as \(n\) gets large? c. Apply Equation (6.12) to show that $$ E(S)=\sigma \frac{\sqrt{2} \Gamma(n / 2)}{\sqrt{n-1 \Gamma}[(n-1) / 2]} $$ Then show that \(E(S)=\sigma \sqrt{2 / \pi}\) if \(n=2\). Is it true that \(E(S)=\sigma\) for normal data?

Suppose we take a random sample of size \(n\) from a continuous distribution having median 0 so that the probability of any one observation being positive is .5. We now disregard the signs of the observations, rank them from smallest to largest in absolute value, and then let \(W=\) the sum of the ranks of the observations having positive signs. For example, if the observations are \(-.3,+.7\), \(+2.1\), and \(-2.5\), then the ranks of positive observations are 2 and 3, so \(W=5\). In Chapter \(14, W\) will be called Wilcoxon's signed-rank statistic. W can be represented as follows: $$ \begin{aligned} W &=1 \cdot Y_{1}+2 \cdot Y_{2}+3 \cdot Y_{3}+\cdots+n \cdot Y_{n} \\ &=\sum_{i=1}^{n} i \cdot Y_{i} \end{aligned} $$ where the \(Y_{i}^{\prime}\) s are independent Bernoulli rv's, each with \(p=.5\left(Y_{i}=1\right.\) corresponds to the observation with rank \(i\) being positive). Compute the following: a. \(E\left(Y_{i}\right)\) and then \(E(W)\) using the equation for \(W\) [Hint: The first \(n\) positive integers sum to \(n(n+1) / 2 .]\) b. \(V\left(Y_{i}\right)\) and then \(V(W)\) [Hint: The sum of the squares of the first \(n\) positive integers is \(n(n+1)(2 n+1) / 6 .]\)
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