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Chapter 6: Problem 13

Find the following areas under the standard normal curve. a. To the right of \(z=3.18, P(z>3.18)\) b. To the right of \(z=1.84, P(z>1.84)\) c. To the right of \(z=0.75, P(z>0.75)\)

Short Answer

Expert verified
The probabilities for the given z-scores are \(P(z > 3.18) = 0.0008\), \(P(z > 1.84) = 0.0336\), and \(P(z > 0.75) = 0.2266\).

Step by step solution

01

Understand the Problem

In this exercise, we need to find out the probabilities \(P(z > 3.18)\), \(P(z > 1.84)\), and \(P(z > 0.75)\). These are probabilities for three different cases in a standard normal distribution.
02

Calculate Probability for \(z = 3.18\)

Probability values for a standard normal distribution can be found in a Z-table or using calculator functions. Assume till now, you have calculated it by looking into the Z-table or in calculator for \(z = 3.18\), where you get \(P(z < 3.18) = 0.9992\). Since this is a right-tail probability, subtract it from 1 to get \(P(z > 3.18) = 1 - P(z < 3.18) = 1 - 0.9992 = 0.0008\).
03

Calculate Probability for \(z = 1.84\)

Doing the similar calculations for \(z = 1.84\), and finding the value in Z-table or calculator, you get \(P(z < 1.84) = 0.9664\). Since we are looking for a right-tail probability, calculate \(P(z > 1.84) = 1 - P(z < 1.84) = 1 - 0.9664 = 0.0336\).
04

Calculate Probability for \(z = 0.75\)

For \(z = 0.75\), you get \(P(z < 0.75) = 0.7734\), using Z-table or calculator. For right-tail probability, calculate \(P(z > 0.75) = 1 - P(z < 0.75) = 1 - 0.7734 = 0.2266\).

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Key Concepts

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

Z-Table
The Z-Table is a crucial tool when working with standard normal distributions. It helps us to find the probability that a given value falls below a particular standard score, or "z-score". A z-score indicates how many standard deviations an element is from the mean of the distribution.
To use a Z-Table, locate the row that corresponds to the first two digits of your z-score and the column for the third decimal place. This intersection provides the cumulative probability of that z-score.
  • For a positive z-score, the table gives the area to the left under the standard normal curve.
  • To find the probability of being to the right of a z-score, as in the given problems, subtract the table value from 1. This is because the total area under the curve equals 1.
By understanding how to navigate the Z-Table, you can efficiently calculate probabilities for different z-scores.
Right-Tail Probability
Right-tail probability refers to the area under the normal curve that falls to the right of a specified z-score.
This is essentially the probability that a standard normal random variable will take on a value greater than a specific z-score.

This concept is pivotal when you need to find how likely an observation more extreme than your calculated z-score is to occur. To determine this, simply use the Z-Table to find the cumulative probability of being less than your z-score and subtract this number from 1.
  • This technique applies to each part of the exercise: calculate the left-tail probability and subtract from 1 to find the right-tail probability.
  • This helps in situations where high z-scores imply extreme data points relative to the mean.
Understanding right-tail probabilities assists in evaluating outliers and assessing the distribution of data.
Normal Curve Areas
The normal curve, or the Gaussian distribution, is the bell-shaped curve that represents the distribution of many types of data. Each segment of the curve corresponds to a particular probability or area.
The total area under the curve is always equal to 1, representing the entire probability space.

When we work with normal curve areas, we're often interested in certain sections like the left-tail, right-tail, or even specific intervals.
  • In the standard normal distribution, finding areas involves calculating probabilities that a random variable fall into these sections.
  • The curve is symmetric, so often calculations for the left or right side can mirror each other.
Whether determining probability for a statistic, or testing hypotheses, understanding normal curve areas is essential for statistics.

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Most popular questions from this chapter

There is a new working class with money to burn, according to the USA Today March 1, 2005, article "New "gold-collar' young workers gain clout." "Gold- collar" is a subset of blue-collar workers, defined by researchers as those working in fast food and retail jobs, or as security guards, office workers, or hairdressers. These 18- to 25-yearold "gold-collar" workers are spending an average of \(\$ 729\) a month on themselves (versus \(\$ 267\) for college students and \(\$ 609\) for blue-collar workers). Assuming this spending is normally distributed with a standard deviation of \(\$ 92.00\) what percentage of gold-collar workers spend: a. between \(\$ 600\) and \(\$ 900\) a month on themselves? b. between \(\$ 400\) and \(\$ 1000\) a month on themselves? c. more than \(\$ 1050\) a month on themselves? d. less than \(\$ 500\) a month on themselves?

According to the Federal Highway Administration's 2006 highway statistics, the distribution of ages for licensed drivers has a mean of 47.5 years and a standard deviation of 16.6 years [www.fhwa.dot.gov]. Assuming the distribution of ages is normally distributed, what percentage of the drivers are: a. between the ages of 17 and \(22 ?\) b. younger than 25 years of age? c. older than 21 years of age? d. between the ages of 48 and \(68 ?\) e. older than 75 years of age?

a. Find the standard \(z\)-score such that \(80 \%\) of the distribution is below (to the left of) this value. b. Find the standard \(z\)-score such that the area to the right of this value is 0.15. c. Find the two \(z\)-scores that bound the middle \(50 \%\) of a normal distribution.

A machine is programmed to fill 10-oz containers with a cleanser. However, the variability inherent in any machine causes the actual amounts of fill to vary. The distribution is normal with a standard deviation of 0.02 oz. What must the mean amount \(\mu\) be so that only \(5 \%\) of the containers receive less than 10 oz?

Given that \(z\) is the standard normal variable, find the value of \(c\) such that: a. \(P(|z|>c)=0.0384\). b. \(P(|z|< c)=0.8740\).
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