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Chapter 6: Problem 21

For a normal distribution, a. Show that a total probability of 0.01 falls more than \(z=2.58\) standard deviations from the mean. b. Find the \(z\) -score for which the two-tail probability that falls more than that many standard deviations from the mean in either direction equals (a) \(0.05,\) (b) 0.10 . Sketch the two cases on a single graph.

Short Answer

Expert verified
(a) P = 0.01 at z = 2.58 confirms; (b) z = 1.96 for P = 0.05, z = 1.645 for P = 0.10. Graph shows z-scores around the mean.

Step by step solution

01

Understanding the Normal Distribution

A normal distribution is symmetric about the mean. The area under the curve of a normal distribution equals 1, representing the total probability of all outcomes. The standard deviation measures how dispersed the data is with respect to the mean.
02

Identifying Areas for Given Probabilities

In the normal distribution, extreme tails beyond a certain z-score contain a specific area representing a given probability. We look for these z-scores that correlate with certain probability thresholds above the tails of the distribution.
03

Use of Z-table for Probability 0.01

To show that a total probability of 0.01 falls outside the range of 7z, where uz=2.58, check the z-table for the cumulative probability corresponding to z=2.58. Notice that symmetric tails divide the "0.01" equally, meaning each tail contains 0.005 probability.
04

Verifying with Probability Table

From the z-table, for z=2.58, the probability in the right tail is about 0.495. Therefore, for two tails to sum to 0.01, each side is 0.005, which verifies that the right-side tail probability (greater than z=2.58) causes half of total 0.01.
05

Finding Z-scores for Two-Tail Probabilities

Determine the z-score needed when our two-tail areas equal 0.05 and 0.10. Divide these probabilities equally (0.025 and 0.025, respectively) in the tails.
06

Z-score for 0.05 Total Two-Tail Probability

Locate the cumulative probability of 0.025 in the tail (for 0.05), corresponding to a z-score. Using a z-table or inverse function, z is approximately 1.96.
07

Z-score for 0.10 Total Two-Tail Probability

For a 0.10 probability, divide equally into the tails as 0.05 each. The cumulative probability in the left tail (0.05) indicates z is approximately 1.645.
08

Sketching the Distribution

Plot a normal distribution curve, marking the mean at the center. Show z=1.96 and z=1.645 on both sides of the mean to represent probabilities of 0.05 and 0.10, respectively.

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Key Concepts

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

Standard Deviation
The concept of standard deviation is central to understanding the distribution of data around the mean. In simple terms, it measures the average distance each data point lies from the mean. This statistic helps us grasp how "spread out" the data is.
  • If the standard deviation is small, data points tend to be close to the mean, indicating less variability.
  • Conversely, a large standard deviation indicates that data points are spread out over a wider range of values.
In normal distribution, the standard deviation is especially useful because it tells us about the data's concentration around the mean. A key property of normal distributions is that about 68% of data falls within one standard deviation, 95% within two, and 99.7% within three. This is known as the empirical rule.
Z-score
The Z-score is a numerical measurement that describes a value's position relative to the mean of a group of values. It is expressed in terms of standard deviations from the mean. A Z-score can tell you how unusual a data point is within the set.
When you calculate a Z-score, you're essentially translating a data point into standard deviations:
  • A Z-score of 0 indicates the data point is exactly at the mean.
  • A positive Z-score shows the data point is above the mean, while a negative indicates it's below.
Z-scores are particularly useful in identifying outliers and for comparing results from different data sets or scale units by standardizing them. For a point to be unusual, or to observe a low probability outcome, its Z-score would typically be far from zero, indicating it's several standard deviations away from the mean.
Probability Distribution
Probability distribution describes how the probabilities of a dataset are distributed over its values. It is essentially a function that provides the probabilities of occurrence of different possible outcomes.
The normal distribution, a type of probability distribution, is bell-shaped and symmetric about the mean:
  • It shows that data near the mean are more frequent in occurrence as compared to data far from the mean.
  • The total area under the curve equals 1, representing the entirety of all possible outcomes.
Understanding probability distributions helps in determining the likelihood of different outcomes in experiments and is a foundation for statistical inference. In practical applications, they allow statisticians to calculate confidence intervals, conduct hypothesis testing, and model the real-world phenomena.
Bell Curve
The Bell Curve, formally known as the Gaussian distribution or normal distribution, is characterized by its distinct symmetrical shape. This curve represents how data values are distributed across a range of values:
  • It has a single peak, indicating the most frequent occurrence of data, which aligns at the mean.
  • As you move further from the mean, the frequency of data points falls off, resulting in the characteristic bell shape.
A Bell Curve is vital in statistics due to its unique properties:
  • Many natural phenomena, like heights or test scores, tend to follow a bell-shaped distribution.
  • It also serves as a key tool in quality control and measurement comparisons.
In practical applications, recognizing the bell curve in data helps with predictions and determining the probability of various outcomes. It is an integral part of processes involving data-driven decision-making.

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Most popular questions from this chapter

Movies sample Five of the 20 movies running in movie theatres this week are comedies. A selection of four movies are picked at random. Does \(X=\) the number of movies in the sample which are comedies have the binomial distribution with \(n=4\) and \(p=0.25 ?\) Explain why or why not.

For each of the following situations, explain whether the binomial distribution applies for \(X\). a. You are bidding on four items available on eBay. You think that you will win the first bid with probability \(25 \%\) and the second through fourth bids with probability \(30 \%\). Let \(X\) denote the number of winning bids out of the four items you bid on. b. You are bidding on four items available on eBay. Each bid is for \(\$ 70\), and you think there is a \(25 \%\) chance of winning a bid, with bids being independent events. Let \(X\) be the total amount of money you pay for your winning bids.

SAT math scores follow a normal distribution with an approximate \(\mu=500\) and \(\sigma=100\). Also ACT math scores follow a normal distrubution with an approximate \(\mu=21\) and \(\sigma=4.7\). You are an admissions officer at a university and have room to admit one more student for the upcoming year. Joe scored 600 on the SAT math exam, and Kate scored 25 on the \(\mathrm{ACT}\) math exam. If you were going to base your decision solely on their performances on the exams, which student should you admit? Explain.

The mean and standard deviation of the grades of a statistics course and an English course are \((\mu=80, \sigma=4.5)\) and \((\mu=85, \sigma=4.0),\) respectively. A student attends both the courses and scores 85 in statistics and 95 in English. Which grade is relatively better? Explain why.

San Francisco Giants hitting The table shows the probability distribution of the number of bases for a randomly selected time at bat for a San Francisco Giants player in 2010 (excluding times when the player got on base because of a walk or being hit by a pitch). In \(74.29 \%\) of the at-bats the player was out, \(17.04 \%\) of the time the player got a single (one base), \(5.17 \%\) of the time the player got a double (two bases), \(0.55 \%\) of the time the player got a triple, and \(2.95 \%\) of the time the player got a home run. a. Verify that the probabilities give a legitimate probability distribution. b. Find the mean of this probability distribution. c. Interpret the mean, explaining why it does not have to be a whole number, even though each possible value for the number of bases is a whole number. $$ \begin{array}{cc} \hline {\text { San Francisco Giants Hitting }} \\ \hline \text { Number of Bases } & \text { Probability } \\ \hline 0 & 0.7429 \\ 1 & 0.1704 \\ 2 & 0.0517 \\ 3 & 0.0055 \\ 4 & 0.0295 \\ \hline \end{array} $$
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