/*! 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 43 Let \(z\) be a random variable w... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(z\) be a random variable with a standard normal distribution. Find the indicated probability, and shade the corresponding area under the standard normal curve. $$ P(-2.18 \leq z \leq-0.42) $$

Short Answer

Expert verified
The probability is approximately 0.3226.

Step by step solution

01

Understanding the Problem

We need to find the probability that the standard normal variable \( z \) lies between -2.18 and -0.42. This corresponds to finding the area under the standard normal curve between these two points.
02

Locate Z-Scores

First, identify the z-scores for the given values. We are looking for \( z = -2.18 \) and \( z = -0.42 \). These z-scores are based on the standard normal distribution, which has a mean of 0 and a standard deviation of 1.
03

Use Standard Normal Distribution Table

Look up the values of \( z \) in the standard normal distribution table (Z-table). Find the probability corresponding to \( z = -2.18 \) which is approximately 0.0146, and for \( z = -0.42 \) which is approximately 0.3372.
04

Calculate the Probability

Calculate the probability \( P(-2.18 \leq z \leq -0.42) \) by subtracting the lower probability from the higher probability: \( 0.3372 - 0.0146 = 0.3226 \).
05

Shade the Corresponding Area

On a standard normal curve, shade the area between \( z = -2.18 \) and \( z = -0.42 \). This shaded region represents the probability we calculated, which is 0.3226.

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Key Concepts

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

Z-scores
Z-scores are an essential part of understanding the standard normal distribution. They represent how many standard deviations an element is from the mean of the distribution. In a standard normal distribution, which is a special type of normal distribution with a mean of 0 and a standard deviation of 1, any data point can be converted to a z-score. This is important because it allows for the standardization of results, aiding in the comparison across different datasets.

For our exercise, the z-scores given are -2.18 and -0.42. These scores are less than the mean since they are negative. A negative z-score indicates that the data point is below the mean of the dataset. The specific values tell us precisely how far below the mean they are. For instance, a z-score of -2.18 means that the value is 2.18 standard deviations below the mean, while a z-score of -0.42 is 0.42 standard deviations below the mean.
Probability Calculation
Calculating the probability in a standard normal distribution involves finding the area under the curve between two z-score values. This is typically done using a Z-table, which lists z-score values alongside their corresponding probabilities. The probabilities represent the cumulative probability from the left end of the distribution up to a given z-score.

In our example, we need to find the probability that the random variable falls between -2.18 and -0.42. This requires us to look up each z-score in the Z-table. For -2.18, the probability is approximately 0.0146, and for -0.42, it is approximately 0.3372. To find the probability of falling between these two z-scores, we subtract the smaller probability from the larger one:
  • Find the cumulative probability for each z-score.
  • Subtract the lower cumulative probability from the higher one.
  • Thus, the probability for our range is 0.3372 - 0.0146 = 0.3226.
This calculation tells us that there is a 32.26% chance of the variable falling within this range.
Normal Curve Shading
Shading the area under the normal curve helps in visually understanding probability. The standard normal curve is symmetric around the mean, which is 0. To represent probabilities on this curve, we often shade regions between specific z-score values, illustrating the concept of probability as an area under the curve.

In the given problem, the task is to shade the area between z = -2.18 and z = -0.42. This shaded area visually represents the probability we calculated, which is 0.3226. When shading this on a graph:
  • Start at z = -2.18 and shade continuously to z = -0.42.
  • The shaded area gives a visual impression of the probability and helps in reinforcing the concept of distribution spread.
  • Understanding the shaded area offers insight into how likely a random variable will fall within that range under the standard normal distribution.
By matching the shaded area with the numerical probability, students can bridge the gap between mathematical theory and visual representation, strengthening their grasp of the concept.

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Most popular questions from this chapter

Let \(z\) be a random variable with a standard normal distribution. Find the indicated probability, and shade the corresponding area under the standard normal curve. $$ P(z \geq-1.50) $$

Let \(z\) be a random variable with a standard normal distribution. Find the indicated probability, and shade the corresponding area under the standard normal curve. $$ P(0 \leq z \leq 0.54) $$

Consider two \(\bar{x}\) distributions corresponding to the same \(x\) distribution. The first \(\bar{x}\) distribution is based on samples of size \(n=100\) and the second is based on samples of size \(n=225\). Which \(\bar{x}\) distribution has the smaller standard error? Explain.

Let \(x\) be a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the \(x\) distribution is about \(\mu=2.7\) minutes, with standard deviation \(\sigma=0.6\) minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of \(x\) values is more or less symmetrical and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let \(x_{i}\) (for \(\left.i=1,2,3, \ldots, 30\right)\) represent the checkout time for each customer. For example, \(x_{1}\) is the checkout time for the first customer, \(x_{2}\) is the checkout time for the second customer, and so forth. Each \(x_{i}\) has mean \(\mu=2.7\) minutes and standard deviation \(\sigma=0.6\) minute. Let \(w=x_{1}+x_{2}+\cdots+x_{30}\) Explain why the problem is asking us to compute the probability that \(w\) is less than 90 . (b) Use a little algebra and explain why \(w<90\) is mathematically equivalent to \(w / 30<3 .\) Since \(w\) is the total of the \(30 x\) values, then \(w / 30=\bar{x}\). Therefore, the statement \(\bar{x}<3\) is equivalent to the statement \(w<90\). From this we conclude that the probabilities \(P(\bar{x}<3)\) and \(P(w<90)\) are equal. (c) What does the central limit theorem say about the probability distribution of \(\bar{x} ?\) Is it approximately normal? What are the mean and standard deviation of the \(\bar{x}\) distribution? (d) Use the result of part (c) to compute \(P(\bar{x}<3)\). What does this result tell you about \(P(w<90)\) ?

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