/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Sketch the areas under the stand... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. To the left of \(z=-1.32\)

Short Answer

Expert verified
The area to the left of z = -1.32 is approximately 0.0934.

Step by step solution

01

Understand the Problem

We need to find the area to the left of the given z-score on the standard normal curve. The z-score tells us how many standard deviations away from the mean a value is in a standard normal distribution.
02

Use Z-Score Table

Locate the z-score of -1.32 in the standard normal distribution table (Z-table). This table shows the area to the left of a given z-score.
03

Read the Z-Score Table Value

From the Z-table, find the value corresponding to a z-score of -1.32. Typically, this value will be listed directly in the table and represents the cumulative probability.
04

Interpret the Result

The value obtained from the Z-table is the area under the standard normal curve to the left of z = -1.32. This value represents the probability of a data point being below z = -1.32.

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Key Concepts

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

Z-score
In the realm of statistics, the **Z-score** is an essential concept that helps us understand where a data point lies concerning the mean of a distribution. It is essentially a measure of how many standard deviations an element is from the mean. When working with the standard normal distribution, the mean is always zero, and the standard deviation is one. Thus, the Z-score allows us to standardize different data points, making them comparable even if they originate from different datasets.
  • A positive Z-score indicates the data point is above the mean.
  • A negative Z-score signifies that the point is below the mean.
  • A Z-score of zero reflects that the data point is precisely at the mean.
In practical terms, if a Z-score is -1.32, it means the data point is 1.32 standard deviations below the mean. This concept is vital in determining probabilities and comparing statistical data.
Cumulative Probability
**Cumulative Probability** refers to the probability that a random variable is less than or equal to a particular value. In the context of the standard normal distribution, it represents the area under the curve to the left of a specific Z-score. When you look at a Z-table, the numbers indicate the cumulative probability for each Z-score value. Essentially, it tells you the likelihood of a value being below a certain point in a standard normal distribution.
  • For instance, if you have a Z-score of -1.32, you would use the Z-table to find the cumulative probability, which is the area under the standard normal curve to the left of -1.32.
  • This probability can be directly interpreted as the chance of a data point being less than or equal to the Z-score in question.
These cumulative probabilities are helpful in various fields like finance, psychology, and social sciences, where understanding the likelihood of events within a range is crucial.
Standard Deviation
In statistics, **Standard Deviation** is a fundamental measure that quantifies the amount of variation or dispersion in a set of data values. For the standard normal distribution, the standard deviation has a fixed value of one. This distribution is known for its characteristic bell-shaped curve.
  • The smaller the standard deviation, the closer the data points are to the mean, showing less spread.
  • A larger standard deviation indicates more spread out data points away from the mean.
Understanding standard deviation is crucial when interpreting a Z-score because: - The Z-score is calculated by taking the difference between an individual data point and the mean, and then dividing by the standard deviation. - This process standardizes the data, making it easier to compare. For example, calculating Z-scores in different datasets allows for comparison without being affected by different scales. Grasping how standard deviation works in standard normal distribution enables you to better understand statistical variations and how they affect the probability distributions.

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Most popular questions from this chapter

Attendance at large exhibition shows in Denver averages about 8000 people per day, with standard deviation of about \(500 .\) Assume that the daily attendance figures follow a normal distribution. (a) What is the probability that the daily attendance will be fewer than 7200 people? (b) What is the probability that the daily attendance will be more than 8900 people? (c) What is the probability that the daily attendance will be between 7200 and 8900 people?

Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let \(x\) be a random variable that represents the number of milligrams of porphyrin per deciliter of blood. In healthy adults, \(x\) is approximately normally distributed with mean \(\mu=38\) and standard deviation \(\sigma=12\) (see reference in Problem 25 ). What is the probability that : (a) \(x\) is less than \(60 ?\) (b) \(x\) is greater than \(16 ?\) (c) \(x\) is between 16 and \(60 ?\) (d) \(x\) is more than \(60 ?\) (This may indicate an infection, anemia, or another type lness.)

Basic Computation: \(\hat{p}\) Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose \(n=100\) and \(p=0.23 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why? Compute \(\mu_{j}\) and \(\sigma_{\tilde{p}}\) (b) Suppose \(n=20\) and \(p=0.23 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why or why not?

Let \(x\) be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in Mesa Verde National Park. Then \(x\) has a distribution that is approximately normal, with mean \(\mu=63.0 \mathrm{kg}\) and standard deviation \(\sigma=7.1 \mathrm{kg}\) (Source: The Mule Deer of Mesa Verde National Park, by G. W. Micrau and J. L. Schmidt, Mesa Verde Museum Association). Suppose a doe that weighs less than \(54 \mathrm{kg}\) is considered undernourished. (a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (b) If the park has about 2200 does, what number do you expect to be undernourished in December? (c) Interpretation To estimate the health of the December doe population, park rangers use the rule that the average weight of \(n=50\) does should be more than \(60 \mathrm{kg}\). If the average weight is less than \(60 \mathrm{kg}\), it is thought that the entire population of does might be undernourished. What is the probability that the average weight \(\bar{x}\) for a random sample of 50 does is less than \(60 \mathrm{kg}\) (assume a healthy population)? (d) Interpretation Compute the probability that \(\bar{x}<64.2 \mathrm{kg}\) for 50 does (assume a healthy population). Suppose park rangers captured, weighed, and released 50 does in December, and the average weight was \(\bar{x}=64.2 \mathrm{kg} .\) Do you think the doe population is undernourished or not? Explain.

Supermarkets: Free Samples Do you take the free samples offered in supermarkets? About \(60 \%\) of all customers will take free samples. Furthermore, of those who take the free samples, about \(37 \%\) will buy what they have sampled. (See reference in Problem 8.) Suppose you set up a counter in a supermarket offering free samples of a new product. The day you are offering free samples, 317 customers pass by your counter. (a) What is the probability that more than 180 take your free sample? (b) What is the probability that fewer than 200 take your free sample? (c) What is the probability that a customer takes a free sample and buys the product? Hint: Use the multiplication rule for dependent events. Notice that we are given the conditional probability \(P(\text { buy } | \text { sample })=0.37\) while \(P(\text { sample })=0.60.\) (d) What is the probability that between 60 and 80 customers will take the free sample and buy the product? Hint: Use the probability of success calculated in part (c).

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