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Chapter 5: Problem 12

Suppose the distribution of the time \(X\) (in hours) spent by students at a certain university on a particular project is gamma with parameters \(\alpha=50\) and \(\beta=2\). Because \(\alpha\) is large, it can be shown that \(X\) has approximately a normal distribution. Use this fact to compute the approximate probability that a randomly selected student spends at most 125 hours on the project.

Short Answer

Expert verified
The probability is approximately 0.9616.

Step by step solution

01

Identify the Parameters for Normal Approximation

The gamma distribution with parameters \(\alpha = 50\) and \(\beta = 2\) can be approximated by a normal distribution. Recall that for a gamma distribution, the mean \(\mu\) is given by \(\mu = \alpha \cdot \beta\) and the variance \(\sigma^2\) is \(\alpha \cdot \beta^2\). We have \(\mu = 50 \cdot 2 = 100\) and \(\sigma^2 = 50 \cdot (2)^2 = 200\). Therefore, the standard deviation \(\sigma\) is \(\sqrt{200} \approx 14.14\).
02

Define the Problem within the Context of a Normal Distribution

We need to find the probability that \(X \leq 125\) using the normal approximation. This corresponds to finding the probability that a normally distributed random variable with mean \(\mu = 100\) and standard deviation \(\sigma = 14.14\) is less than or equal to 125.
03

Standardize the Normal Distribution Variable

We calculate the z-score, which is the number of standard deviations the value 125 is away from the mean. The z-score is given by \(z = \frac{125 - \mu}{\sigma} = \frac{125 - 100}{14.14} \approx 1.77\).
04

Use the Z-score to Find the Probability

Using standard normal distribution tables or a calculator, find the probability that the z-value is less than or equal to 1.77. This corresponds to the cumulative distribution function (CDF) for the standard normal distribution, denoted as \(P(Z \leq 1.77)\). From tables or a calculator, \(P(Z \leq 1.77) \approx 0.9616\).

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Key Concepts

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

Gamma Distribution
The gamma distribution is a continuous probability distribution that is widely used in statistics. It's defined by two parameters: shape parameter \( \alpha \) and scale parameter \( \beta \). This distribution is particularly useful when modeling waiting times or lifespans of objects.
  • Mean: For a gamma distribution, the mean is calculated as \( \mu = \alpha \cdot \beta \). In our example, the mean is 100 hours.
  • Variance: The variance is found using the formula \( \sigma^2 = \alpha \cdot \beta^2 \). Here, it鈥檚 200 hours squared.
As the shape parameter \( \alpha \) becomes larger, the gamma distribution starts to resemble a normal distribution due to the Central Limit Theorem. This is why, with a large \( \alpha \), we can approximate the gamma distribution with a normal distribution.
Z-score
The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It's measured in terms of standard deviations from the mean.
  • Calculation: The formula to calculate the z-score is \( z = \frac{X - \mu}{\sigma} \), where \( X \) is the value being evaluated, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • Purpose: Z-scores allow us to compare different data points from different sets or distributions by leveling them on the same scale.
In our problem, to find out how far 125 hours is from the mean of 100 hours in terms of the standard deviation of 14.14 hours, we calculated a z-score of 1.77.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. All normal distributions, regardless of their specific parameters, can be transformed into a standard normal distribution through z-score calculation.
  • Mean and Variance: Always centered at 0 with a variance of 1.
  • Use: Once data is in the form of a standard normal distribution, it becomes easy to calculate probabilities using standard normal tables or calculators.
In practical applications, when you convert the original data to a standard normal distribution using z-scores, you simplify the process of determining probabilities for any interval or cumulative distribution query.
Cumulative Distribution Function
The cumulative distribution function (CDF) describes the probability that a random variable takes on a value less than or equal to a specified number. It effectively accumulates the probabilities from the lower bound of the distribution up to that number.
  • Purpose: The CDF maps each value of a random variable to the probability that the variable is less than or equal to that value. Hence, it provides complete information about the distribution.
  • Relation to Z-score: In a standard normal distribution, the CDF is used to find the probability associated with a z-score.
For a z-score of 1.77, corresponding to the case where a student spends at most 125 hours on a project, the CDF equals about 0.9616. This value means there is a 96.16% chance that a student will spend 125 hours or fewer on the project.

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Most popular questions from this chapter

Two airplanes are flying in the same direction in adjacent parallel corridors. At time \(t=0\), the first airplane is \(10 \mathrm{~km}\) ahead of the second one. Suppose the speed of the first plane \((\mathrm{km} / \mathrm{hr})\) is normally distributed with mean 520 and standard deviation 10 and the second plane's speed is also normally distributed with mean and standard deviation 500 and 10 , respectively. a. What is the probability that after 2 hr of flying, the second plane has not caught up to the first plane? b. Determine the probability that the planes are separated by at most \(10 \mathrm{~km}\) after \(2 \mathrm{hr}\).

The National Health Statistics Reports dated Oct. 22, 2008, stated that for a sample size of 277 18-year-old American males, the sample mean waist circumference was \(86.3 \mathrm{~cm}\). A somewhat complicated method was used to estimate various population percentiles, resulting in the following values: \(\begin{array}{lllllll}5^{\text {th }} & 10^{\text {th }} & 25^{\text {th }} & 50^{\text {th }} & 75^{\text {th }} & 90^{\text {th }} & 95^{\text {th }} \\\ 69.6 & 70.9 & 75.2 & 81.3 & 95.4 & 107.1 & 116.4\end{array}\) a. Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning. If your answer is no, conjecture the shape of the population distribution. b. Suppose that the population mean waist size is \(85 \mathrm{~cm}\) and that the population standard deviation is \(15 \mathrm{~cm}\). How likely is it that a random sample of 277 individuals will result in a sample mean waist size of at least \(86.3 \mathrm{~cm}\) ? c. Referring back to (b), suppose now that the population mean waist size in \(82 \mathrm{~cm}\). Now what is the (approximate) probability that the sample mean will be at least \(86.3 \mathrm{~cm}\) ? In light of this calculation, do you think that \(82 \mathrm{~cm}\) is a reasonable value

Suppose that when the \(\mathrm{pH}\) of a certain chemical compound is \(5.00\), the \(\mathrm{pH}\) measured by a randomly selected beginning chemistry student is a random variable with mean \(5.00\) and standard deviation .2. A large batch of the compound is subdivided and a sample given to each student in a morning lab and each student in an afternoon lab. Let \(\bar{X}=\) the average \(\mathrm{pH}\) as determined by the morning students and \(\bar{Y}=\) the average \(\mathrm{pH}\) as determined by the afternoon students. a. If \(\mathrm{pH}\) is a normal variable and there are 25 students in each lab, compute \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\). [Hint: \(\bar{X}-\bar{Y}\) is a linear combination of normal variables, so is normally distributed. Compute \(\mu_{\bar{x}-\bar{Y}}\) and \(\sigma_{\bar{x}-\bar{y}}\). b. If there are 36 students in each lab, but \(\mathrm{pH}\) determinations are not assumed normal, calculate (approximately) \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\).

Let \(X\) be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of \(X\) is as follows: \begin{tabular}{l|rrrr} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.4\) & \(.3\) & \(.2\) & \(.1\) \end{tabular} a. Consider a random sample of size \(n=2\) (two customers), and let \(\bar{X}\) be the sample mean number of packages shipped. Obtain the probability distribution of \(\bar{X}\). b. Refer to part (a) and calculate \(P(\bar{X} \leq 2.5)\). c. Again consider a random sample of size \(n=2\), but now focus on the statistic \(R=\) the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of \(R\). [Hint: Calculate the value of \(R\) for each outcome and use the probabilities from part (a).] d. If a random sample of size \(n=4\) is selected, what is \(P(\bar{X} \leq 1.5)\) ? [Hint: You should not have to list all possible outcomes, only those for which \(\bar{x} \leq 1.5\).]

Consider 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. Disregarding the signs of the observations, rank them from smallest to largest in absolute value, and 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 \(15, W\) will be called Wilcoxon's signed-rank statistic. W can be represented as follows: a. Determine \(E(Y)\) and then \(E(W)\) using the equation for \(W\). [Hint: The first \(n\) positive integers sum to \(n(n+1) / 2 .]\) b. Determine \(V\left(Y_{i}\right)\) and then \(V(W)\). [Hint: The sum of the squares of the first \(n\) positive integers can be expressed as \(n(n+1)(2 n+1) / 6\).] $$ \begin{aligned} W &=1 \cdot Y_{1}+2 \cdot Y_{2}+3 \cdot Y_{3}+\cdots+n \cdot Y_{n} \\ &=\sum_{i=1}^{n} i \cdot Y_{i} \end{aligned} $$ where the \(Y_{i}\) 's are independent Bernoulli rv's, each with \(p=.5\left(Y_{i}=1\right.\) corresponds to the observation with rank \(i\) being positive).
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