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Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? a. Yes b. No c. Unable to determine

Short Answer

Expert verified
a. Yes, it would be surprising.

Step by step solution

01

Understanding the Normal Distribution

The problem states that the time to find a parking space is normally distributed with a mean (12 9.7 6.321 u, measured in minutes) of 5 minutes and a standard deviation (12 2.15 5.6 7.3 sigma) of 2 minutes. We are interested in the probability that the parking search takes less than 1 minute.
02

Standardizing the Time

To find this probability, we need to convert the time to a standard normal variable (Z). The formula to convert a normal variable X to a standard normal variable Z is: \[ Z = \frac{X - \mu}{\sigma} \]Substitute the values for X = 1 (since we want to find for less than 1 minute), \( \mu = 5 \), and \( \sigma = 2 \). So, \[ Z = \frac{1 - 5}{2} = -2 \]
03

Finding the Probability Utilizing Z-tables

We use the Z-table to find the probability of Z being less than -2. Looking up Z = -2 in the Z-table gives a probability of 0.0228. This means that there is a 2.28% chance the time to find a parking space is less than 1 minute.
04

Interpreting the Probability

Since the probability of finding a parking spot in less than 1 minute is very low (just 2.28%), it would be considered surprising if it took less than one minute to find a spot.

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a key concept in statistics that helps handle data that follows a pattern. It takes any normal distribution and converts it into a special normal distribution with a mean of 0 and a standard deviation of 1.
This is mainly done using the Z-score. Understanding the characteristics of the standard normal distribution can make statistical calculations a lot easier. The bell curve shape of the distribution helps visualize how data clusters around the mean. With most data falling within one standard deviation from the mean, it reflects how respective probabilities are spread over the values.
By converting any normal distribution to a standard normal, you can use a Z-table, simplifying the process of looking up probabilities without needing to recalculate them every time. In our scenario, the time it takes to find a parking space at 9 A.M. is normally distributed with given mean and standard deviation. By converting this distribution to a standard normal one, we can easily calculate the required probabilities.
Probability Calculation
Calculating probabilities in a normal distribution requires an understanding of how likely certain outcomes are. With a normal distribution, these probabilities often concern questions like the likelihood of a random variable being less than or greater than a particular value. Probabilities in a normal distribution are determined with the help of the Z-table. Once the Z-score for a particular point in the distribution is calculated, you can reference the Z-table to find the probability that a value is below the Z-score. This allows you to determine how big or small the likelihood is for a specific event.
In probability terms, the area under the curve of the distribution graph tells us about these probabilities. A larger area signifies a higher probability and vice versa. In the exercise, we're finding the probability of the time taken to find a parking spot being less than one minute. The calculated probability from the Z-table turned out to be 0.0228, indicating a rare occurrence. This low probability confirms our surprise if such an event occurs.
Z-Score Conversion
Z-Score conversion is a critical process in standardizing normal distributions. This conversion helps translate the original data point in a distribution to one that fits the standard normal distribution. By doing so, it quantifies how far and in what direction a data point deviates from the mean.The formula to compute the Z-score is: \[ Z = \frac{X - \mu}{\sigma} \]Here, X represents the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. The Z-score tells you how many standard deviations an element is from the mean. Positive Z-scores indicate values above the mean, while negative values suggest those below.In our solution, we calculated the Z-score for finding a parking spot in less than 1 minute, resulting in a Z-score of -2. This score translates to the point being 2 standard deviations below the mean. Once calculated, it became straightforward to use the Z-table to find the associated probability.

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Most popular questions from this chapter

In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. a. In words, define the random variable X. b. X ~ _____(_____,_____) c. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement. d. What percent of the children spend over ten hours per day unsupervised? e. Seventy percent of the children spend at least how long per day unsupervised?

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. a. If X = distance in feet for a fly ball, then X ~ _____(_____,_____) b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean ? = 125 and standard deviation ? = 14. Systolic blood pressure for males follows a normal distribution. a. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters. b. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him?

In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean.

\(X \sim N(-2,1)\) \(\mu=\)_____.

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