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The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of \(7.7\) minutes and a standard deviation of \(2.1\) minutes. Find the probability that the mean delivery time for a random sample of 16 such orders at this restaurant is a, between 7 and 8 minutes b. within 1 minute of the population mean c. less than the population mean by 1 minute or more

Short Answer

Expert verified
The probabilities that the mean delivery time for a random sample of 16 such orders at this restaurant is (a) between 7 and 8 minutes is 0.6243, (b) within 1 minute of the population mean is 0.6874, (c) less than the population mean by 1 minute or more is 0.0287.

Step by step solution

01

Calculation of Standard Error

The standard Error (SE) is calculated with the formula: \(SE = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation and \(n\) is the sample size. Here, \(\sigma = 2.1\) and \(n = 16\). So, \(SE = \frac{2.1}{\sqrt{16}} = 0.525\).
02

Calculate Z-scores for part (a)

The formula for Z-score is \(Z = \frac{x - \mu}{SE}\). For 7 minutes, \(Z_1 = \frac{7 - 7.7}{0.525} = -1.33\) and for 8 minutes, \(Z_2 = \frac{8 - 7.7}{0.525} = 0.57\).
03

Find Probability for part (a)

The probability that the mean lies between two values is the area under the curve between these points that can be obtained from the Z-table. For part (a), the required probability is \(P(Z_1 < Z < Z_2) = P(-1.33 < Z < 0.57)\). From the standard normal distribution table, \(P(Z < -1.33) = 0.0918\) and \(P(Z < 0.57) = 0.7161\). Therefore, \(P(-1.33 < Z < 0.57) = P(Z < 0.57) - P(Z < -1.33) = 0.7161 - 0.0918 = 0.6243\).
04

Calculate Z-scores for part (b)

For part (b), the z-score corresponding to the mean +1 minute is \(Z_3 = \frac{(7.7 + 1) - 7.7}{0.525} = 0.57\) and for mean -1 minute, \(Z_4 = \frac{(7.7 - 1) - 7.7}{0.525} = -1.905\).
05

Find Probability for part (b)

Similarly, using the Z-table, for part (b), \(P(-1.905 < Z < 0.57) = P(Z < 0.57) - P(Z < -1.905) = 0.7161 - 0.0287 = 0.6874\).
06

Calculate Z-score for part (c)

For part (c), \(Z_5 = \frac{(7.7 - 1) - 7.7}{0.525} = -1.905\).
07

Find Probability for part (c)

Using the Z-table, the probability for part (c), \(P(Z < -1.905) = 0.0287\).

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Key Concepts

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

Normal Distribution
Understanding the normal distribution is key for solving problems related to probabilities of continuous data. The normal distribution is a bell-shaped curve that shows how data is spread symmetrically around the mean. In this distribution, most of the data points are close to the mean, creating the peak of the curve, and fewer data points are found as you move away from the mean.

Why is it important? Many real-world variables, like delivery times or people's heights, follow a normal distribution. This makes the normal distribution incredibly useful in statistics to predict outcomes and understand variance in data.

  • The mean, median, and mode are all equal in a perfectly normal distribution.
  • About 68% of data lies within one standard deviation from the mean, 95% within two, and 99.7% within three.
Understanding this concept helps us make predictions about the likelihood of certain outcomes.
Standard Error
The concept of standard error (SE) is crucial when dealing with sample means. SE helps us measure how much the sample mean of our data is expected to fluctuate from the actual population mean. In simpler terms, it tells us how 'off' our sample's mean could be from the actual mean.

SE is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
  • A smaller SE indicates that the sample mean is more likely to be close to the population mean.
  • Increasing the sample size (\( n \)) typically reduces the SE, leading to more precise estimates of the population mean.
Understanding standard error enables you to gauge the reliability of your sample data and how tightly it clusters around the population mean.
Z-scores
Z-scores are a statistical measurement that describe a value's relationship to the mean of a group of values. They tell us how many standard deviations away a data point is from the average. Z-scores help standardize data, giving a clear picture of how unusual or typical a specific data point is within a distribution.

The formula to find a z-score is: \[ Z = \frac{x - \mu}{SE} \]where \( x \) is the data point, \( \mu \) is the mean, and \( SE \) is the standard error.
  • A z-score of 0 means the data point is exactly at the mean.
  • Positive z-scores are above the mean, while negative ones are below.
  • Z-scores are indispensable in calculating probabilities in a normal distribution using z-tables.
Grasping z-scores allows you to pinpoint where your data stands in comparison to the average, making it essential for probability calculations.
Sample Mean
The sample mean is the average of a sample, which is a subset of a population. It provides a quick and useful estimate of the population mean. Calculating the sample mean is often more practical than determining the population mean because it involves less data, making it more manageable and cost-effective.

Here’s how you calculate the sample mean: \[ \bar{x} = \frac{\sum x}{n} \]where \( \sum x \) is the sum of all sample values and \( n \) is the size of the sample.
  • The sample mean is a single-value estimate of the population mean.
  • It is used to make inferences about the population from which the sample was drawn.
In statistics, the sample mean is essential as it provides an approximation of what the population mean could be, helping to make informed decisions based on smaller datasets.

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Most popular questions from this chapter

A 2009 nonscientific poll on the Web site of the Daily Gazette of Schenectady, New York, asked readers the following question: "Are you less inclined to buy a General Motors or Chrysler vehicle now that they have filed for bankruptcy?" Of the readers who responded, \(56.1 \%\) answered Yes (http://www. dailygazette.com/polls/2009/jun/Bankruptcy/). Assume that this result is true for the current population of vehicle owners in the United States. Let \(\hat{p}\) be the proportion in a random sample of \(340 \mathrm{U.S}\), vehicle owners who are less inclined to buy a General Motors or Chrysler vehicle after these corporations filed for bankruptcy. Find the mean and standard deviation of the sampling distribution of \(\hat{p}\), and describe its shape.

The amounts of electricity bills for all houscholds in a particular city have an approximately normal distribution with a mean of \(\$ 140\) and a standard deviation of \(\$ 30 .\) Let \(\bar{x}\) be the mean amount of electricity bills for a random sample of 25 households selected from this city. Find the mean and standard deviation of \(\bar{x}\), and comment on the shape of its sampling distribution.

Explain briefly the meaning of nonsampling errors. Give an example. Do such errors occur only in a sample survey, or can they occur in both a sample survey and a census?

Annual per capita (average per person) chewing gum consumption in the United States is 200 pieces (http://www.iplcricketlive.com/). Suppose that the standard deviation of per capita consumption is 145 pieces per year. Let \(\bar{x}\) be the average annual chewing gum consumption of 84 randomly selected Americans. Find the mean and the standard deviation of the sampling distribution of \(\bar{x}\). What is the shape of the sampling distribution of \(\bar{x}\) ? Do you need to know the shape of the population distribution to make this conclusion? Explain why or why not.

The amounts of electricity bills for all households in a city have a skewed probability distribution with a mean of \(\$ 140\) and a standard deviation of \(\$ 30\). Find the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be \(\mathrm{a}_{2}\) between \(\$ 132\) and \(\$ 136\) b. within \(\$ 6\) of the population mean \(c\), more than the population mean by at least \(\$ 4\)

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