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The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value \(10 \mathrm{~min}\) and standard deviation \(2 \mathrm{~min}\). If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most \(11 \mathrm{~min}\) ?

Short Answer

Expert verified
The probability is approximately 0.7722.

Step by step solution

01

Understand the Problem

We need to determine the probability that the average time for filling out forms on two different days is at most 11 minutes. The form-filling time is normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.
02

Determine the Sampling Distribution

For five individuals, the average time \(\bar{X}_1\) follows a normal distribution with mean \(10\) and standard deviation\(\frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}}\). Similarly, for six individuals, the average \(\bar{X}_2\) follows a normal distribution with mean \(10\) and standard deviation \(\frac{2}{\sqrt{6}}\).
03

Calculate the Probability for Each Day

We calculate the probability for day one: \[ P(\bar{X}_1 \leq 11) = P\left(Z \leq \frac{11 - 10}{\frac{2}{\sqrt{5}}}\right) = P\left(Z \leq \frac{1}{0.894}\right) = P(Z \leq 1.118)\] Using the standard normal distribution table, we find approximately 0.8686.For day two: \[ P(\bar{X}_2 \leq 11) = P\left(Z \leq \frac{11 - 10}{\frac{2}{\sqrt{6}}}\right) = P\left(Z \leq \frac{1}{0.8165}\right) = P(Z \leq 1.225)\] Using the standard normal distribution table, we find approximately 0.8897.
04

Combine the Probabilities

Since the events are independent, the probability that the average on both days is at most 11 minutes is the product of individual probabilities:\[ P(\bar{X}_1 \leq 11, \bar{X}_2 \leq 11) = 0.8686 \times 0.8897 \approx 0.7722.\]

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Key Concepts

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

Normal Distribution
The normal distribution is a key concept in statistics that you encounter frequently in probability exercises. It is a continuous probability distribution characterized by its bell-shaped curve, known as the Gaussian curve, which is symmetric around its mean. In the exercise with mortgage applications, the time taken to fill out a form follows a normal distribution with a mean of 10 minutes and a standard deviation of 2 minutes.

A few important properties of normal distribution include:
  • The mean (\(\mu\)) determines the center of the curve.
  • The standard deviation (\(\sigma\)) measures the spread or width of the distribution.
  • Approximately 68% of the data falls within one standard deviation of the mean, while about 95% falls within two.

These properties make the normal distribution useful for modeling various natural phenomena and for making predictions about outcomes, such as the time it takes to complete a task.
Sampling Distribution
In statistics, when we take a sample from a population, we are often interested in understanding the distribution of a sample statistic, like the sample mean. This brings us to the concept of sampling distribution, which is the probability distribution of the sample statistic.

In the given exercise, the sample means for the times taken by five individuals and six individuals to fill out the forms are considered. Each of these sample means has its own distribution, which is centered around the same population mean (10 minutes) but has a smaller spread due to averaging. The standard deviations of these sample means are computed as \(\frac{2}{\sqrt{5}}\) and \(\frac{2}{\sqrt{6}}\) respectively. This reduction in standard deviation as the sample size increases is known as standard error, and it indicates that larger samples provide more reliable estimates of the population mean.

Sampling distributions allow us to make inferences about the population from samples, which is critical in statistical analysis.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. This transformation is very useful because it allows us to use standard normal tables to find probabilities for any normal distribution.

In the mortgage form exercise, to find the probability that the sample average on a certain day is at most 11 minutes, we convert the sample mean to a standard normal variable (\(Z\)) using the transformation: \[Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}\]where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. By converting to a standard normal distribution, we can easily find the probability using standard normal tables or software applications. This makes calculating and understanding normal probabilities much more straightforward.
Sample Average
The sample average, also known as the sample mean, is the sum of a set of observations divided by the number of observations. It serves as an estimate of the population mean and is denoted by \(\bar{X}\).

In the problem given, the sample average is the average time taken by individuals from samples of five and six people to fill out a form. It is a crucial measure because it provides a point estimate of the population mean, which is not always directly accessible.

Key points about the sample average:
  • It is affected by sample size—more observations generally lead to a more accurate estimate of the true population mean.
  • It reduces variability in the estimation process since it's influenced by the law of large numbers, meaning the larger the sample, the closer the sample average is to the population mean.
Understanding the sample average and its relation to the population mean is foundational in statistics, especially when estimating parameters or testing hypotheses.

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Most popular questions from this chapter

Garbage trucks entering a particular waste-management facility are weighed prior to offloading their contents. Let \(X=\) the total processing time for a randomly selected truck at this facility (waiting, weighing, and offloading). The article "Estimating Waste Transfer Station Delays Using GPS" (Waste Mgmt., 2008: 1742-1750) suggests the plausibility of a normal distribution with mean \(13 \mathrm{~min}\) and standard deviation \(4 \mathrm{~min}\) for \(X\). Assume that this is in fact the correct distribution. a. What is the probability that a single truck's processing time is between 12 and 15 min? b. Consider a random sample of 16 trucks. What is the probability that the sample mean processing time is between 12 and \(15 \mathrm{~min}\) ? c. Why is the probability in (b) much larger than the probability in (a)? d. What is the probability that the sample mean processing time for a random sample of 16 trucks will be at least \(20 \mathrm{~min}\) ?

Suppose a randomly chosen individual's verbal score \(X\) and quantitative score \(Y\) on a nationally administered aptitude examination have a joint pdf $$ f(x, y)=\left\\{\begin{array}{cl} \frac{2}{5}(2 x+3 y) & 0 \leq x \leq 1,0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ You are asked to provide a prediction \(t\) of the individual's total score \(X+Y\). The error of prediction is the mean squared error \(E\left[(X+Y-t)^{2}\right]\). What value of \(t\) minimizes the error of prediction?

A binary communication channel transmits a sequence of "bits" (0s and 1s). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000 -bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

Two different professors have just submitted final exams for duplication. Let \(X\) denote the number of typographical errors on the first professor's exam and \(Y\) denote the number of such errors on the second exam. Suppose \(X\) has a Poisson distribution with parameter \(\mu_{1}, Y\) has a Poisson distribution with parameter \(\mu_{2}\), and \(X\) and \(Y\) are independent. a. What is the joint pmf of \(X\) and \(Y\) ? b. What is the probability that at most one error is made on both exams combined? c. Obtain a general expression for the probability that the total number of errors in the two exams is \(m\) (where \(m\) is a nonnegative integer). [Hint: \(A=\) \(\\{(x, y): x+y=m\\}=\\{(m, 0),(m-1,1), \ldots\), \((1, m-1),(0, m)\\}\). Now sum the joint pmf over \((x, y) \in A\) and use the binomial theorem, which says that $$ \sum_{k=0}^{m}\left(\begin{array}{c} m \\ k \end{array}\right) a^{k} b^{m-k}=(a+b)^{m} $$ for any \(a, b .]\)

Annie and Alvie have agreed to meet for lunch between noon (0:00 P.M.) and 1:00 P.M. Denote Annie's arrival time by \(X\), Alvie's by \(Y\), and suppose \(X\) and \(Y\) are independent with pdf's $$ \begin{aligned} &f_{X}(x)=\left\\{\begin{array}{cc} 3 x^{2} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \\ &f_{X}(y)=\left\\{\begin{array}{cl} 2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: \(h(X, Y)=|X-Y| .]\)

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