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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 1 day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?

Short Answer

Expert verified
Probabilities are approximately 0.8686 and 0.8897 for each day.

Step by step solution

01

Define the Problem

We are given that the time taken to fill out the form follows a normal distribution with mean \( \mu = 10 \) minutes and standard deviation \( \sigma = 2 \) minutes. We need to find the probability that the average time taken by samples of five people on one day and six people on another day is at most 11 minutes.
02

Identify the Sampling Distribution

The sample mean from a normally distributed population is also normally distributed. If we have a sample of size \( n \), the mean of the sample means is \( \mu = 10 \) and the standard deviation of the sample means is \( \sigma/\sqrt{n} \). For the first sample (five people): \( \sigma_1 = 2/\sqrt{5} \); For the second sample (six people): \( \sigma_2 = 2/\sqrt{6} \).
03

Standardize the Sample Averages

To use the standard normal distribution, convert the sample average to a \( Z \)-score for both samples. For five people: \[ Z_1 = \frac{11 - 10}{2/\sqrt{5}} \]. For six people: \[ Z_2 = \frac{11 - 10}{2/\sqrt{6}} \]. This simplifies to: \( Z_1 = \frac{1}{2/\sqrt{5}} = \frac{\sqrt{5}}{2} \) and \( Z_2 = \frac{1}{2/\sqrt{6}} = \frac{\sqrt{6}}{2} \).
04

Calculate Z-scores

Calculate \( Z_1 \) and \( Z_2 \): \[ Z_1 = \frac{\sqrt{5}}{2} \approx 1.118 \] and \[ Z_2 = \frac{\sqrt{6}}{2} \approx 1.225 \].
05

Find Probabilities from Z-table

Use a standard normal distribution table or calculator to find \( P(Z \leq Z_1) \) and \( P(Z \leq Z_2) \). \( P(Z \leq 1.118) \approx 0.8686 \) and \( P(Z \leq 1.225) \approx 0.8897 \).
06

Interpret Results

The probability that the average time is at most 11 minutes for the first sample is approximately 0.8686, and for the second sample, it is approximately 0.8897.

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Key Concepts

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

Normal Distribution
Imagine if you charted the times taken by applicants to fill out a mortgage form. The time they take is usually around an average, but there's some randomness in it. Often, this randomness looks like a bell-shaped curve when charted — this is the normal distribution. The peak of this curve represents the average time, here given as 10 minutes.
This distribution is symmetric, meaning it is equally likely for an applicant to take either slightly less or slightly more time than the mean. The normal distribution is crucial because it allows for the calculation of probabilities over an interval or range of values using the parameters of the mean and standard deviation.
  • Mean (\( \mu \) )— the center of the distribution, here 10 mins.
  • The Standard Deviation (\( \sigma\) ) — measures how spread out the times are, here 2 mins.
Understanding how the data is spread helps predict how likely a particular data point is — useful when determining probabilities like in this problem.
Z-score
When comparing different data points within a normal distribution, we use something called the Z-score. It tells us how far a data point is from the mean in terms of standard deviations. This is key to understanding probabilities in the normal distribution context.
To find a Z-score:
  • Subtract the mean from your data point,
  • Divide by the standard deviation of the sampling distribution.
In our exercise, the fixed amount of time we're interested in comparing is 11 minutes. We want to know how far this is from the average of 10 minutes in terms of our sampling distributions. Calculating Z-scores enables us to standardize these comparisons, making it easier to use statistical tables to find the probabilities associated with them.
Once calculated, these Z-scores can be looked up in a Z-table to find the probability of observing a value less than or equal to 11 minutes.
Sampling Distribution
In a real-world scenario, it's not always practical to look at the entire population, so we often take samples. When we calculate means from all possible samples of a given size from a population, these means form a distribution — the sampling distribution.
The distribution of the sample mean will follow a normal distribution even if the source population is not, provided the sample size is large enough. However, with our data being normally distributed even with smaller sample sizes, the sample means will still be normally distributed.
  • The mean of this sampling distribution remains the same as the population mean.
  • The standard deviation of the sampling distribution is called the standard error, calculated as \( \sigma/\sqrt{n}\).
Sampling distributions help ensure data reliability by giving statistical evidence through probability, which is what we've established in the exercise with five and six applications.
Standard Deviation
Standard deviation is a critical concept to grasp when dealing with variability in your data. It tells you how much your data points typically differ from the mean.
In the context of our exercise, it's important because it indicates how much the time taken by mortgage applicants varies around the mean of 10 minutes.
  • A small standard deviation indicates that most applicants take almost the same time.
  • A large standard deviation suggests more variety in filling times.
In the sample means' context, the standard deviation of the sample means is what we refer to as the standard error. It decreases with increasing sample size because the average of a larger sample tends to be closer to the population mean.
Understanding standard deviation and standard error helps in assessing the reliability of our sample mean estimates in representing the population properly.

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Most popular questions from this chapter

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