/*! 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 21 Use a table or technology to ans... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a table or technology to answer each question. Include an appropriately labeled sketch of the Normal curve for each part. Shade the appropriate region. a. Find the probability that a z-score will be \(1.76\) or less. b. Find the probability that a z-score will be \(1.76\) or more. c. Find the probability that a \(z\) -score will be between \(-1.3\) and \(-1.03\).

Short Answer

Expert verified
The probabilities that a z-score will be \(1.76\) or less, \(1.76\) or more, and between \(-1.3\) and \(-1.03\) can be found using a z-table or similar technology. The exact probabilities would depend on the values provided by the specific z-table or technology used.

Step by step solution

01

Using z-table for finding probabilities

To solve part a, look up \(1.76\) in the z-table. This table gives the cumulative probability to the left of the given z-score. So for \(1.76\), this will give the probability that a z-score is \(1.76\) or less.
02

Finding complement of the probability

For part b, remember that the total probability under the normal curve is 1. So the probability that a z-score will be \(1.76\) or more is simply the complement of the probability found in step 1, i.e., \(1 - \text{probability from step 1}\).
03

Calculating probability between two z-scores

For part c, to find the probability that a z-score will be between \(-1.3\) and \(-1.03\), we need to subtract the cumulative probability of \(-1.03\) from the cumulative probability of \(-1.3\). This is done by looking up both values in the z-table and performing the subtraction.

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.

Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a symmetric, bell-shaped curve that represents the distribution of many types of natural and social phenomena. It's defined mathematically by its mean (average) and standard deviation (a measure of spread). Think of it like a real-world data set where most of the values are clustered around the mean, creating a peak, and fewer values are found as you move away from the center, creating the tails.
Many everyday datasets approximately follow a normal distribution, including heights, test scores, and measurement errors. When the mean is at the center of the curve, the distribution is said to be 'standardized', leading to a very useful form known as the standard normal distribution, which has a mean of 0 and a standard deviation of 1. It's the reference for finding out how unusual or usual a piece of data is compared to the expected pattern - that's where the z-score comes in.
Cumulative Probability
Imagine you're filling up a glass with water, and you're interested in knowing how much water is in there at different points, cumulatively, as you pour. Cumulative probability in statistics works in a similar way but instead of water, we're filling a graph with probability as we move left to right across the scores. It's the chance that a variable will take a value less than or equal to a specific value.
For a normal distribution, this is illustrated as the area under the curve to the left of a given z-score, which tells us the proportion of the data that falls below that score. This concept is crucial - it allows us to quantify the 'usualness' of a score within a dataset, hence helping us make decisions or predictions based on statistical likelihood.
Z-Table

Demystifying the Z-Table

Using a z-table can seem like reading a map of a very complex cityscape at first. But once you understand it, it becomes an invaluable tool for navigating the world of statistics. The z-table contains cumulative probabilities for a standard normal distribution, with z-scores along one axis and associated probabilities along the other.
When using a z-table, you are essentially 'matching' a z-score to its cumulative probability. For a z-score of 1.76, the table lists a probability that corresponds to the percentage of data points lying to the left of this z-score in a normal distribution. This is like finding out how many people are shorter than a certain height on a graph where height is plotted on the horizontal axis, and the number of people is on the vertical axis.
Complement of Probability
Once you understand cumulative probability, the complement of probability is like looking at the other side of the coin. If cumulative probability is concerned with 'this much or less', the complement is all about 'this much or more'. In essence, it's the probability of the event not occurring, and for normal distributions, it refers to the area under the curve to the right of a given z-score.
The total area under a normal curve represents 100% of the possible outcomes, or a probability of 1. So, if you want to find the probability that a z-score is greater than 1.76 (as opposed to less than or equal to), you simply subtract the cumulative left-tail probability (found from the z-table for 1.76) from 1. This is an important concept in hypothesis testing, where researchers are often interested in the likelihood of results being due to chance, and they look to the tail ends of the distribution for answers.

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

A study of human body temperatures using healthy women showed a mean of \(98.4^{\circ} \mathrm{F}\) and a standard deviation of about \(0.70^{\circ} \mathrm{F}\). Assume the temperatures are approximately Normally distributed. a. Find the percentage of healthy women with temperatures below \(98.6^{\circ} \mathrm{F}\) (this temperature was considered typical for many decades). b. What temperature does a healthy woman have if her temperature is at the 76 th percentile?

Assume a standard Normal distribution. Draw a separate, well-labeled Normal curve for each part. a. Find the \(z\) -score that gives a left area of \(0.8577\). b. Find the \(z\) -score that gives a left area of \(0.0146\).

Extreme Negative z-Scores For each question, find the area to the right of the given z-score in a standard Normal distribution. In this question, round your answers to the nearest \(0.000 .\) Include an appropriately labeled sketch of the \(N(0,1)\) curve. a. \(z=-4.00\) b. \(z=-8.00\) c. \(z=-30.00\) d. If you had the exact probability for these right proportions, which would be the largest and which would be the smallest? e. Which is equal to the area in part b: the area below (to the left of) \(z=8.00\) or the area above (to the right of) \(z=8.00 ?\)

A study of U.S. births published on the website Medscape from WebMD reported that the average birth length of babies was \(20.5\) inches and the standard deviation was about \(0.90\) inch. Assume the distribution is approximately Normal. Find the percentage of babies who have lengths of 19 inches or less at birth.

College women have heights with the following distribution (inches): \(N(65,2.5)\). a. Find the height at the 75 th percentile. b. Find the height at the 25 th percentile. c. Find the interquartile range for heights. d. Is the interquartile range larger or smaller than the standard deviation? Explain.

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.