/*! 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 33 Find the indicated probability, ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(z \leq-0.13)$$

Short Answer

Expert verified
The probability is 0.4483.

Step by step solution

01

- Understand the Problem

We need to find the probability that the standard normal variable \( z \) is less than or equal to \(-0.13\). This is denoted as \( P(z \leq -0.13) \).
02

- Use the Standard Normal Distribution Table

Consult the standard normal distribution table which gives the probability that a standard normal variable is less than a given value of \( z \). Find \( z = -0.13 \) in the table.
03

- Find the Corresponding Probability

Using the standard normal distribution table, locate the value of \( -0.1 \) on the left column and \( 0.03 \) on the top row. The intersection of this row and column provides the probability. In most standard normal distribution tables, the probability \( P(z \leq -0.13) \) is approximately 0.4483.
04

- Interpret and Shade the Area

The probability \( P(z \leq -0.13) = 0.4483 \) tells us the area under the z-curve to the left of \( -0.13 \). Shade this area on the standard normal curve, which begins at \( -\infty \) and ends at \( -0.13 \).

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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 special type of normal distribution that has a mean of 0 and a standard deviation of 1. This distribution is symmetrical, meaning it looks like a bell-shaped curve. It is used frequently in statistics because it simplifies the process of finding probabilities for standard scores, or z-scores. The shape of the curve allows us to understand probabilities over continuous ranges of values. When we talk about the area under the curve, we are referring to probabilities of a variable falling inside a certain range. More than just numbers, these probabilities can give us insights into how likely certain outcomes are, based purely on the distribution's characteristics.
Z-Score
A z-score is a way of expressing a number from a dataset in terms of its position relative to the mean and standard deviation of the dataset. Essentially, a z-score tells us how many standard deviations away a particular value is from the mean.To calculate a z-score, you use the following formula: \[ z = \frac{X - \mu}{\sigma} \]where:- \( X \) is the value in the dataset,- \( \mu \) is the mean of the dataset, and- \( \sigma \) is the standard deviation.Z-scores are very useful because they allow comparison of scores from different distributions. They also enable the use of the standard normal distribution table to find probabilities.
Probability Tables
Probability tables, specifically the standard normal distribution tables, are tools that help us find probabilities associated with standard normal variables, or z-scores.These tables list cumulative probabilities for values of \( z \). Each entry in the table represents the area under the standard normal curve to the left of a specific \( z \)-value. This is equivalent to knowing what portion of data lies below a certain point in a standard normal distribution.To use a probability table:- Look down the left-hand side to find the row corresponding to the first digit and the first decimal of the z-score.- Across the top, find the column corresponding to the second decimal place of your z-score.- The number where your row and column intersect is your probability.Understanding how to read these tables is critical for tasks such as identifying probabilities like \( P(z \leq -0.13) \). Mastery over probability tables ensures we can interpret and leverage distributions effectively in real-world data analysis scenarios.

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Most popular questions from this chapter

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.1 millimeters (mm) and a standard deviation of \(0.9 \mathrm{mm}\) (Source: Homol'ovi II: Archaeology of an Ancestral Hopi Village, Arizona, edited by E. C. Adams and K. A. Hays, University of Arizona Press). For a randomly found shard, what is the probability that the thickness is (a) less than \(3.0 \mathrm{mm} ?\) (b) more than \(7.0 \mathrm{mm} ?\) (c) between \(3.0 \mathrm{mm}\) and \(7.0 \mathrm{mm} ?\)

Measurement errors from instruments are often modeled using the uniform distribution (see Problem 16 ). To determine the range of a large public address system, acoustical engineers use a method of triangulation to measure the shock waves sent out by the speakers. The time at which the waves arrive at the sensors must be measured accurately. In this context, a negative error means the signal arrived too early. A positive error means the signal arrived too late. Measurement errors in reading these times have a uniform distribution from -0.05 to +0.05 microseconds. (Reference: J. Perruzzi and E. Hilliard, "Modeling Time Delay Measurement Errors," Journal of the Acoustical Society of America, Vol. 75, No. 1, pp. 197-201.) What is the probability that such measurements will be in error by (a) less than \(+0.03 \text { microsecond (i.e., }-0.05 \leq x<0.03) ?\) (b) more than -0.02 microsecond? (c) between -0.04 and +0.01 microsecond? (d) Find the mean and standard deviation of measurement errors. Measurements from an instrument are called unbiased if the mean of the measurement errors is zero. Would you say the measurements for these acoustical sensors are unbiased? Explain.

Let \(\alpha\) and \(\beta\) be any two constants such that \(\alpha<\beta .\) Suppose we choose a point \(x\) at random in the interval from \(\alpha\) to \(\beta .\) In this context the phrase at random is taken to mean that the point \(x\) is as likely to be chosen from one particular part of the interval as any other part. Consider the rectangle. The base of the rectangle has length \(\beta-\alpha\) and the height of the rectangle is \(1 /(\beta-\alpha),\) so the area of the rectangle is \(1 .\) As such, this rectangle's top can be thought of as part of a probability density curve. since we specify that \(x\) must lie between \(\alpha\) and \(\beta,\) the probability of a point occurring outside the interval \([\alpha, \beta]\) is, by definition, \(0 .\) From a geometric point of view, \(x\) chosen at random from \(\alpha\) to \(\beta\) means we are equally likely to land anywhere in the interval from \(\alpha\) to \(\beta .\) For this reason, the top of the (rectangle's) density curve is flat or uniform. Now suppose that \(a\) and \(b\) are numbers such that \(\alpha \leq a

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. To the left of \(z=0.72\)

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