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91Ó°ÊÓ

Sketch the areas under the standard normal curve over the indicated intervals, and find the specified areas. Between \(z=1.42\) and \(z=2.17\)

Short Answer

Expert verified
The area between z=1.42 and z=2.17 is approximately 0.0628.

Step by step solution

01

Understanding the Standard Normal Curve

The standard normal distribution is a bell-shaped curve that is symmetric around the mean, which is 0. The standard deviation is 1. The area under this curve represents probability, with the total area equaling 1.
02

Identify the Interval

We are interested in finding the area under the curve between two z-scores: 1.42 and 2.17. These z-scores correspond to points along the horizontal axis of the standard normal distribution.
03

Use Z-tables or Calculators

To find the area between two z-scores, we can use a standard normal distribution table (Z-table) or a calculator that has functions for calculating normal distribution probabilities. Look up the z-score of 1.42, which gives an area of approximately 0.9222, and look up the z-score of 2.17, which gives an area of approximately 0.9850.
04

Calculate the Area Between Z-scores

Subtract the area corresponding to z=1.42 from the area corresponding to z=2.17. This gives us the area between these two z-scores. \[ Area = P(Z < 2.17) - P(Z < 1.42) = 0.9850 - 0.9222 = 0.0628 \]
05

Sketching the Area

Draw the standard normal distribution curve. On the horizontal axis, mark the z-scores 1.42 and 2.17. Shade the region under the curve between these two points to represent the area you calculated.

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Key Concepts

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

Z-scores in Standard Normal Distribution
A Z-score is a numeric value that represents how many standard deviations a data point is from the mean. In the context of the standard normal distribution, the mean is 0 and the standard deviation is 1. So, if a Z-score is 1, it means the data point is 1 standard deviation away from the mean. Z-scores are useful because they allow us to compare different data points within a distribution. They help normalize different data sets, making them easily comparable.
In our exercise, we're considering Z-scores of 1.42 and 2.17. This tells us where these points lie on the standard normal distribution curve. We can think of Z-scores as specific markers on a number line that lies below the bell-shaped curve, helping us to locate specific intervals.
Understanding Probability in Normal Distribution
Probability in the standard normal distribution is found under the curve. The total area under this curve is equal to 1, representing all possible outcomes. Probability thus ranges from 0 to 1, where 0 means an impossible event and 1 means a certain event. For each interval on the normal distribution curve, the area represents the probability of a random variable falling within that interval.
In our scenario, we calculate the probability that a random variable falls between Z-scores of 1.42 and 2.17. Using Z-tables makes it easy to find these probabilities, as they show the cumulative probability from the far left of the curve up to any given Z-score. Thus, finding the probability between two Z-scores involves simple subtraction of these areas. This results in a practical way to measure likelihoods using the normal distribution.
Role of Standard Deviation in the Distribution
Standard deviation measures how spread out the numbers in a data set are. In a standard normal distribution, this spread is fixed at 1. It acts as a consistent measure that quantifies the amount of variation or dispersion in a set of data points.
A large standard deviation indicates that the data points are spread out over a wider range of values. Conversely, a small standard deviation means that the data points tend to be close to the mean. The concept of standard deviation is essential because it directly impacts the shape and spread of the normal distribution.
For example, knowing the standard deviation in the context of Z-scores helps us understand how far a particular score is from the average in a way that is comparable regardless of other factors, making data analysis more reliable and insightful.
The Beauty of the Normal Distribution Curve
The normal distribution curve, commonly known as the bell curve, is a graphical representation where the mean, median, and mode of the data are all located at the peak. The curve displays a symmetrical distribution of data, with the left half mirroring the right. The curve's tails approach but never quite touch the horizontal axis, which theoretically extends to infinity in both directions. This curve is important in statistics because it models many natural phenomena and serves as the basis for inferential statistics.
In the example exercise, the normal distribution curve helps visualize and calculate the probability between the Z-scores of 1.42 and 2.17. A key property is that approximately 68% of data lies within 1 standard deviation of the mean, approximately 95% within 2 standard deviations, and about 99.7% falls within 3 standard deviations. This distribution ensures that we can predict how data behaves consistently and efficiently.

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Most popular questions from this chapter

Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and the time a patrol car radios that it has arrived at the scene (based on information from The Denver Post). Over a long period of time, it has been determined that the police response time has a normal distribution with a mean of \(8.4\) minutes and a standard deviation of \(1.7\) minutes. For a randomly received emergency call, what is the probability that the response time will be (a) between 5 and 10 minutes? (b) less than 5 minutes? (c) more than 10 minutes?

Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(40 \leq x \leq 47) ; \mu=50 ; \sigma=15 $$

Suppose we have a binomial experiment with \(n=40\) trials and a probability of success \(p=0.50\) . (a) Is it appropriate to use a normal approximation to this binomial distribution? Why? (b) Compute \(\mu\) and \(\sigma\) of the approximating normal distribution. (c) Use a continuity correction factor to convert the statement \(r \geq 23\) successes to a statement about the corresponding normal variable \(x\). (d) Estimate \(P(r \geq 23)\). (e) Is it unusual for a binomial experiment with 40 trials and probability of success \(0.50\) to have 23 or more successes? Explain.

Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(x \geq 120) ; \mu=100 ; \sigma=15 $$

Let \(x\) represent the dollar amount spent on supermarket impulse buying in a 10 -minute (unplanned) shopping interval. Based on a Denver Post article, the mean of the \(x\) distribution is about \(\$ 20\) and the estimated standard deviation is about \(\$ 7\). (a) Consider a random sample of \(n=100\) customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of \(\bar{x}\), the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the \(\bar{x}\) distribution? Is it necessary to make any assumption about the \(x\) distribution? Explain. (b) What is the probability that \(\bar{x}\) is between \(\$ 18\) and \(\$ 22\) ? (c) Let us assume that \(x\) has a distribution that is approximately normal. What is the probability that \(x\) is between \(\$ 18\) and \(\$ 22 ?\) (d) In part (b), we used \(\bar{x}\), the average amount spent, computed for 100 customers. In part (c), we used \(x\), the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen? In this example, \(\bar{x}\) is a much more predictable or reliable statistic than \(x\). Consider that almost all marketing strategies and sales pitches are designed for the average customer and not the individual customer. How does the central limit theorem tell us that the average customer is much more predictable than the individual customer?

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