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Chapter 6: Problem 18

Let \(X\) represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the mean value and standard deviation of \(X\) are \(11.5\) and \(4.0\), respectively. a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons? b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons. c. What is the approximate value of the 95 th percentile for the total amount purchased by 50 randomly selected customers.

Short Answer

Expert verified
a. Probability is 0.1881. b. Probability is 0.8944. c. 95th percentile is 607.9 gallons.

Step by step solution

01

Identify the distribution of the sample mean

Since the amount of gasoline purchased, \(X\), has a mean of 11.5 and a standard deviation of 4.0, and we are taking a sample of 50 customers, the sampling distribution of the sample mean \( \bar{X} \) will be approximately normal due to the Central Limit Theorem. The mean of \( \bar{X} \) is 11.5, and the standard deviation (standard error) is \( \frac{4.0}{\sqrt{50}} \approx 0.5657 \).
02

Calculate probability for part a

To find the probability that the sample mean is at least 12 gallons, we use the standard normal distribution. Compute the z-score: \[ z = \frac{12 - 11.5}{0.5657} \approx 0.8844 \]Using a standard normal distribution table or calculator, find \( P(Z \geq 0.8844) \). This corresponds to approximately 0.1881.
03

Identify the distribution for the sample total

The total amount of gasoline purchased by the 50 customers, \( T \), is the sample sum and will also follow a normal distribution. The mean of \( T \) is \( 50 \times 11.5 = 575 \), and the standard deviation is \( 50 \times 4.0 = 20 \).
04

Calculate probability for part b

To find the probability that the total amount is at most 600 gallons, compute the z-score:\[ z = \frac{600 - 575}{20} = 1.25 \]Using a standard normal distribution table or calculator, find \( P(Z \leq 1.25) \). This probability is approximately 0.8944.
05

Calculate 95th percentile for part c

To find the 95th percentile for the total amount purchased, identify the z-score that corresponds to the 0.95 cumulative probability, which is approximately 1.645. Use the mean and standard deviation of the total to find the percentile:\[ T_{0.95} = 575 + 1.645 \times 20 \approx 607.9 \]

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Key Concepts

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

Probability distributions
Probability distributions describe how the values of a random variable are spread or distributed. They provide a model that shows the likelihood of different outcomes. In the case of the gasoline purchase, the random variable \(X\) represents the amount of gasoline bought by a customer. Based on the information given, \(X\) follows a normal distribution.
  • The mean (\(\mu\)) is 11.5 gallons, indicating the average purchase amount.
  • The standard deviation (\(\sigma\)) is 4.0 gallons, showing the variability from the mean.
This model helps in understanding how much gas customers typically buy and the extent of deviation from this average. Knowing this allows us to make predictions using the properties of the normal distribution.
Sampling distribution
Sampling distribution refers to the probability distribution of a statistic (like mean or sum) based on a random sample. When dealing with the sample of gasoline purchases, we look at the sample mean \( \bar{X} \) and sample sum \(T\). The Central Limit Theorem comes into play here, as it states that with a large enough sample size, the distribution of the sample mean approaches a normal distribution, irrespective of the original distribution.
  • The mean of the sampling distribution of \( \bar{X} \) is the same as \(X\), which is 11.5.
  • The standard error (standard deviation of the sample mean) is calculated as \( \frac{4.0}{\sqrt{50}} \approx 0.5657 \).
These calculations allow predictions about the sample mean, such as in part a of the original exercise.
Z-scores
Z-scores are standardized scores that indicate how many standard deviations an element is from the mean. In probability distributions, we often use z-scores to find the probability of a given outcome.
In the context of the gasoline purchase, the calculation of z-scores helps determine the likelihood of different quantities being purchased. For instance:
  • For the sample mean to be at least 12 gallons, the z-score is calculated as \( z = \frac{12 - 11.5}{0.5657} \approx 0.8844 \).
  • This z-score corresponds to the probability \( P(Z \geq 0.8844) \).
Similarly, for the total purchase, we find z-scores to determine related probabilities, such as \( P(Z \leq 1.25) \) when checking for a maximum of 600 gallons purchased.
Percentiles in statistics
Percentiles are values that divide a dataset into 100 equal parts, providing insights into the distribution of the data. They indicate the relative standing of a value within a dataset.
In the exercise, to find the 95th percentile, we looked at the total amount of gasoline purchased. The 95th percentile is the value below which 95% of the data falls.
  • Using the z-score for the 95th percentile (\(1.645\)), the calculation for the total amount was \( T_{0.95} = 575 + 1.645 \times 20 \approx 607.9 \).
  • This means that 95% of the time, the total gasoline purchased by 50 customers will be 607.9 gallons or less.
Understanding percentiles helps in determining thresholds or limits for data, providing a valuable perspective on data positioning.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}\), and \(X_{3}\) represent the times necessary to perform three successive repair tasks at a service facility. Suppose they are independent, normal rv's with expected values \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\sigma_{3}^{2}\), respectively. a. If \(\mu_{1}=\mu_{2}=\mu_{3}=60\) and \(\sigma_{1}^{2}=\sigma_{2}^{2}=\) \(\sigma_{3}^{2}=15\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 200\right)\) What is \(P\left(150 \leq X_{1}+X_{2}+X_{3} \leq 200\right) ?\) b. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate \(P(55 \leq \bar{X})\) and \(P(58 \leq \bar{X} \leq 62)\). c. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate \(P\left(-10 \leq X_{1}-.5 X_{2}-.5 X_{3} \leq 5\right)\). d. If \(\mu_{1}=40, \quad \mu_{2}=50, \quad \mu_{3}=60, \quad \sigma_{1}^{2}=10\), \(\sigma_{2}^{2}=12\), and \(\sigma_{3}^{2}=14\), calculate \(P\left(X_{1}+\right.\) \(X_{2}+X_{3} \leq 160\) ) and \(P\left(X_{1}+X_{2} \geq 2 X_{3}\right)\).

Suppose the amount of liquid dispensed by a machine is uniformly distributed with lower limit \(A=8 \mathrm{oz}\) and upper limit \(B=10 \mathrm{oz}\). Describe how you would carry out simulation experiments to compare the sampling distribution of the (sample) fourth spread for sample sizes \(n=5\), 10,20 , and 30 .

Apply the Law of Large Numbers to show that \(\chi_{v}^{2} / v\) approaches 1 as \(v\) becomes large.

A box contains ten sealed envelopes numbered 1 , \(\ldots, 10\). The first five contain no money, the next three each contain \(\$ 5\), and there is a \(\$ 10\) bill in each of the last two. A sample of size 3 is selected with replacement (so we have a random sample), and you get the largest amount in any of the envelopes selected. If \(X_{1}, X_{2}\), and \(X_{3}\) denote the amounts in the selected envelopes, the statistic of interest is \(M=\) the maximum of \(X_{1}, X_{2}\), and \(X_{3}\). a. Obtain the probability distribution of this statistic. b. Describe how you would carry out a simulation experiment to compare the distributions of \(M\) for various sample sizes. How would you guess the distribution would change as \(n\) increases?

In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let \(X=\) the number of trees planted in sandy soil that survive 1 year and \(Y=\) the number of trees planted in clay soil that survive I year. If the probability that a tree planted in sandy soil will survive l year is. 7 and the probability of 1 -year survival in clay soil is \(.6\), compute \(P(-5 \leq X-Y \leq 5)\) (use an approximation, but do not bother with the continuity correction).
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