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Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1 .\) Determine each of the following probabilities: a. \(P(z < 0.10)\) b. \(P(z < -0.10)\) c. \(P(0.40 < z < 0.85)\) d. \(P(-0.85 < z < -0.40)\) e. \(P(-0.40 < z < 0.85)\) f. \(P(z > \- 1.25)\) g. \(P(z < -1.50\) or \(z > 2.50)\)

Short Answer

Expert verified
Use standard normal distribution table to find the values corresponding to Z-scores. Be vigilant about the negative Z-scores, use the symmetric property of normal distribution where necessary.

Step by step solution

01

Understanding Z-Scores and Standard Normal Tables

A Z-Score is a measurement of how many standard deviations a particular data point is from the mean. It is calculated by subtracting the mean from a particular data point (in this case already given as Z-scores), and dividing this by the standard deviation. Standard Normal Tables (also known as Z-tables) is a mathematical table that allows us to know the percentage of values below (to the left of) a z-score in a standard normal distribution. Here, since we are dealing with a normal distribution where \(\mu = 0\) and \(\sigma = 1\), our Z-score is equal to the value of z itself.
02

Calculate Probability for Each Case

We will refer to the standard normal distribution table to find the probabilities. a. \(P(z < 0.10)\): Find the value from the standard normal table that corresponds to 0.10. That value is the probability. b. \(P(z < -0.10)\): Here, since it's a negative value, we use symmetry properties of the normal distribution which states probability of being below -0.10 is same as the probability of being above 0.10; it equals \(1 - P(z < 0.10)\). c. \(P(0.40 < z < 0.85)\): To find the probability between two values, we find the probability of lower bound and upper bound separately from the standard normal distribution table and subtract the former from the latter i.e., \(P(z < 0.85) - P(z < 0.40)\). d. \(P(-0.85 < z < -0.40)\): Similar to the previous case, we subtract the lower bound from the upper bound but as these are negative, we use the symmetry property which states probability of being below -0.85 is same as probability of being above 0.85 and likewise for -0.40. Emerges as \(P(z > 0.40) - P(z > 0.85)\). e. \(P(-0.40 < z < 0.85)\): This is calculated as \(P(z < 0.85) - P(z > 0.40)\). f. \(P(z > -1.25)\): Probability of being above -1.25 is same as probability of being below 1.25 as per symmetric property, calculated as \(P(z < 1.25)\). g. \(P(z < -1.50\) or \(z > 2.50)\): Since these are disjoint events we add the corresponding probabilities i.e., \(P(z > -1.50) + P(z < 2.50)\).

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Key Concepts

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

Z-Scores
Understanding Z-Scores is crucial when working with normal distribution probability. A Z-Score essentially tells us how far from the mean, in standard deviations, a particular score lies. It's calculated by subtracting the mean from the score and then dividing by the standard deviation. In simple terms, if you score above the mean, you have a positive Z-Score; below the mean, a negative Z-Score.

For instance, a Z-Score of 1.0 signifies that the value is one standard deviation above the mean. Similarly, a Z-Score of -1.0 indicates one standard deviation below the mean. This concept supplies a standard way to compare scores from different distributions and is especially useful because it enables us to calculate the probability of a score occurring within a normal distribution.

When you see problems involving frequencies within certain ranges, you'll often be working with Z-Scores to find these probabilities.
Standard Normal Distribution
The Standard Normal Distribution is a special case of the normal distribution that has a mean (\(\mu\)) of 0 and a standard deviation (\(\sigma\)) of 1. It is also referred to as the Z-Distribution. When you convert a normal distribution to a Z-Score, you are standardizing the distribution.

This means that no matter what the original mean or standard deviation was, the standard normal distribution allows us to compare scores across different scales. The shape of the standard normal distribution is the bell curve, which is symmetrical about the mean (zero in this case).

The total area under this curve equals 1, or a probability of 100%, and each segment of the curve corresponds to a proportion of that total probability. Knowing how to interpret the standard normal distribution is vital for finding probabilities related to specific Z-Scores.
Standard Normal Tables
Standard Normal Tables, often referred to as Z-Tables, are a statistical tool used to find the probability of a Z-Score occurring within the standard normal distribution. These tables are integral because they give us the area to the left (or sometimes the right) of any Z-Score on the curve, which corresponds to the cumulative probability up to that point.

To use these tables, you locate the Z-Score in question along the left column and top row of the table. The intersecting value inside the table gives you the cumulative area or probability. It's critical to note whether you're finding the area to the left (less than) or to the right (greater than) because sometimes you may need to subtract the table value from 1 to find the latter.

Understanding how to use these tables allows you to efficiently find the probabilities for different segments of the standard normal curve without the need for complicated integrations or calculations.
Probability Calculation
Probability Calculation in a standard normal distribution is about determining the likelihood of a Z-Score falling within a particular range. To calculate these probabilities, first, determine if the Z-Score is looking for a cumulative area to the left, right, or between two values.

If you have a single Z-Score, like in the problems \(P(z < 0.10)\) or \(P(z > -1.25)\), you'll use the standard normal table to find the cumulative probability up to that Z-Score. For Z-Scores that fall between two values, you'll find the cumulative probabilities for each Z-Score separately and then determine the difference between them.

It's also essential to remember the symmetry of the standard normal distribution when dealing with negative Z-Scores. This symmetry can simplify calculations, as you can often subtract the cumulative probability from 1 or find equivalencies between negative and positive Z-Scores. Mastering these calculations is fundamental in statistics, as it provides a foundation for more complex inferential statistics.

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Most popular questions from this chapter

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the random variable \(x\) as \(x=\) the number of people who actually show up for a sold-out flight on this plane

Suppose a playlist on an MP3 music player consisting of 100 songs includes 8 by a particular artist. Suppose that songs are played by selecting a song at random (with replacement) from the playlist. The random variable \(x\) represents the number of songs until a song by this artist is played. a. Explain why the probability distribution of \(x\) is not binomial. b. Find the following probabilities. (Hint: See Example 6.31) i. \(p(4)\) ii. \(P(x \leq 4)\) iii. \(P(x>4)\) iv. \(P(x \geq 4)\) c. Interpret each of the probabilities in Part (b) and explain the difference between them.

State whether each of the following random variables is discrete or continuous: a. The number of defective tires on a car b. The body temperature of a hospital patient c. The number of pages in a book d. The number of draws (with replacement) from a deck of cards until a heart is selected e. The lifetime of a light bulb

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).

Sophie is a dog who loves to play catch. Unfortunately, she isn't very good at this, and the probability that she catches a ball is only \(0.1 .\) Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

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