/*! 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 46 Find the indicated probability, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(0 \leq z \leq 0.54)$$

Short Answer

Expert verified
The probability \( P(0 \leq z \leq 0.54) \) is approximately 0.2054.

Step by step solution

01

Understand the Problem

We are asked to find the probability of a standard normal variable, often denoted as \( z \), falling between 0 and 0.54. This involves determining the area under the standard normal curve, which is a Gaussian distribution with mean 0 and standard deviation 1, between \( z = 0 \) and \( z = 0.54 \).
02

Reference the Standard Normal Table

Use the standard normal distribution table or a calculator with statistical functions to find the probability that \( z \) takes a value less than 0.54. These tables give the area to the left of a specific \( z \) value.
03

Look Up \( z = 0.54 \) in the Table

Search for \( z = 0.54 \) in the standard normal distribution table. The table indicates that the area to the left of \( z = 0.54 \) is approximately 0.7054. This means \( P(z \leq 0.54) = 0.7054 \).
04

Determine the Probability for \( z = 0 \)

For the standard normal distribution, the probability of \( z \leq 0 \) is 0.5 since it is the midpoint of the distribution. Thus, \( P(z \leq 0) = 0.5 \).
05

Calculate the Desired Probability

The probability we need is for \( 0 \leq z \leq 0.54 \). This can be found by subtracting \( P(z \leq 0) \) from \( P(z \leq 0.54) \): \[ P(0 \leq z \leq 0.54) = P(z \leq 0.54) - P(z \leq 0) = 0.7054 - 0.5 = 0.2054 \]
06

Shade the Area on the Curve

To visually represent this probability on the standard normal curve, shade the region between \( z = 0 \) and \( z = 0.54 \). This is the portion of the curve corresponding to the probability \( 0.2054 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 often used in probability and statistics. This distribution is characterized by its bell-shaped curve, known as the Gaussian curve. What makes it "standard" is that it has a mean (average) of 0 and a standard deviation of 1.
This simplifies a lot of calculations because it provides a standardized way to understand how data is spread. Each point on this distribution relates to a specific "z" value.
  • The mean of 0 means that it is perfectly centered around the y-axis.
  • A standard deviation of 1 means that approximately 68% of the data falls within one unit of standard deviation from the mean.
The area under the curve represents the probability of random variables falling within a certain range. So for any two points on the horizontal axis, the area under the curve between those points will tell us the probability of a value falling within that range.
Z-score
The concept of a Z-score is a way to quantify how far away a particular data point is from the mean of a data set, measured in terms of standard deviations. A Z-score allows us to compare different data points from different normal distributions.
  • A positive Z-score indicates the data point is above the mean.
  • A negative Z-score means it is below the mean.
  • A Z-score of 0 means the data point is exactly at the mean.
To find a Z-score, use the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the data point you are analyzing, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation. In the standard normal distribution, since \( \mu = 0 \) and \( \sigma = 1 \), the Z-score is simply the "z" value itself, making calculations straightforward.
Gaussian distribution
The Gaussian distribution, named after the mathematician Carl Friedrich Gauss, is another term for the normal distribution. It appears naturally in many physical, biological, and social measurement scenarios, which is why it is often used in statistical analysis.
  • It is symmetrically bell-shaped, with its peak at the mean.
  • The tails of the curve extend infinitely in both directions and never touch the horizontal axis.
The Gaussian distribution is important because it provides a common framework for understanding variability and estimating probabilities. This distribution is defined by its mean and standard deviation. Changing these parameters will shift the center of the curve and affect how spread out the data is.
In practical applications, many techniques exist to determine the probability of an event within a Gaussian distribution, such as Z-scores or transforming data to fit the standard normal distribution. Its wide applicability is a testament to its mathematical robustness and real-world relevance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(-2.37 \leq z \leq 0)$$

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.

Inverse Normal Distribution Most exhibition shows open in the morning and close in the late evening. A study of Saturday arrival times showed that the average arrival time was 3 hours and 48 minutes after the doors opened, and the standard deviation was estimated at about 52 minutes. Assume that the arrival times follow a normal distribution. (a) At what time after the doors open will \(90 \%\) of the people who are coming to the Saturday show have arrived? (b) At what time after the doors open will only \(15 \%\) of the people who are coming to the Saturday show have arrived? (c) Do you think the probability distribution of arrival times for Friday might be different from the distribution of arrival times for Saturday? Explain.

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(z \geq 1.35)$$

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. To the left of \(z=0\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.