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Find the indicated probabilities. $$ P(0 \leq Z \leq 1.5) $$

Short Answer

Expert verified
The probability that the standard normal variable Z falls in the range \( 0 \leq Z \leq 1.5 \) is 0.4332.

Step by step solution

01

Understand the standard normal distribution

The standard normal distribution is a continuous probability distribution with mean 0 and standard deviation 1. It is also known as the z-distribution or z-score. We will use a standard normal distribution table to look up the probabilities of \( Z \leq z \) for any value z.
02

Identify the given value and range

In this case, we're asked to find the probability that a random variable \( Z \) from a standard normal distribution falls in the range \( 0 \leq Z \leq 1.5 \).
03

Use the standard normal distribution table

We will use a standard normal distribution table to find \( P(Z \leq 1.5) \) and \( P(Z \leq 0) \). The rows represent the whole number and tenths place, and the columns represent the hundredths place of the Z value. 1. Look up the probability for \( Z \leq 1.5 \) in the table. In the row for 1.5, we find the value 0.9332. 2. Look up the probability for \( Z \leq 0 \) in the table. In the row for 0.0, we find the value 0.5.
04

Calculate the required probability

Now that we have found \( P(Z \leq 1.5) \) and \( P(Z \leq 0) \), we will subtract the two values to find the probability that a standard normal variable Z falls in the given range (\( 0 \leq Z \leq 1.5 \)): \[ P(0 \leq Z \leq 1.5) = P(Z \leq 1.5) - P(Z \leq 0) \] \[ P(0 \leq Z \leq 1.5) = 0.9332 - 0.5 = 0.4332 \] Therefore, the probability that the standard normal variable Z falls in the range \( 0 \leq Z \leq 1.5 \) is 0.4332.

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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 fundamental concept in statistics, representing a special case of the normal distribution. It's defined as a normal distribution with a mean (average) of 0 and a standard deviation of 1. This distribution is termed 'standard' because variables measured across different scales can be easily converted to the standard normal distribution using a z-score.

This allows for a direct comparison of different datasets and simplifies the process of finding probabilities associated with specific intervals of a normal distribution. Since the mean is 0, values on the standard normal distribution can be thought of as deviations from the mean measured in units of the standard deviation.

An important property of the standard normal distribution is its symmetry around the mean. Half of the values will fall below the mean and half above, making the median and the mean equal. The tails of the distribution are where extreme values or outliers would lie, and the probability of a value occurring farther from the mean decreases exponentially.
Z-Score
A z-score, also known as a standard score, is a numerical measurement that describes a value's relationship to the mean of a group of values. It is a way of counting the number of standard deviations a particular score is above or below the mean. The formula for calculating a z-score in a standard normal distribution is given by:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

If a z-score is 0, it indicates that the data point's score is identical to the mean score. A positive z-score signifies that the data point is above the mean, and a negative z-score signifies that it is below the mean. Z-scores are invaluable for determining how unusual a particular result is within the context of a given data set and are widely used for statistical analysis in a variety of fields, including psychology, education, and finance.
Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. It is a statistical tool that helps to understand the likelihood of various results.

For continuous variables, such as those associated with a normal distribution, the probability of a single, precise value occurring is technically zero. Instead, probabilities are calculated over ranges of values, leading to the concept of the probability density function (PDF) for continuous distributions.

The area under the PDF for a given interval represents the probability that the value of the random variable falls within that interval. With the standard normal distribution, the total area under the curve adds up to 1, signifying that the probability of some value occurring within the distribution is certain. One key application of probability distributions is in making predictions about a population from sample data, which is a cornerstone in the field of inferential statistics.
Normal Distribution Table
The normal distribution table, often called a z-table, is an essential tool for statisticians and students alike – it provides the probabilities of certain z-scores in a standard normal distribution. Since calculating these probabilities by hand is complex and time-consuming, the table simplifies it significantly by offering pre-calculated values.

The table is organized with z-scores along the left-hand column and top row, and corresponding probabilities in the body. To find the probability for a z-score, one simply locates the z-score on the leftmost column, follows the appropriate row to the column that matches the z-score's decimal point, and reads the probability.

The table is typically one-sided, representing the probability of a value occurring at or below a given z-score. To find the probability between two z-scores, one can subtract the probability associated with the lower z-score from the probability associated with the higher z-score, thus utilizing the cumulative property of probability distributions. This concept is directly applied when solving problems like the exercise outlined, where one can find the probability lying between two points on the standard normal curve.

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Most popular questions from this chapter

I \(\vee\) Supermarkets A survey of supermarkets in the United States yielded the following relative frequency table, where \(X\) is the number of checkout lanes at a randomly chosen supermarket: \(^{49}\) \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \(\boldsymbol{x}\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \(\boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x})\) & \(.01\) & \(.04\) & \(.04\) & \(.08\) & \(.10\) & \(.15\) & \(.25\) & \(.20\) & \(.08\) & \(.05\) \\ \hline \end{tabular} a. Compute the mean, variance, and standard deviation (accurate to one decimal place). b. As financial planning manager at Express Lane Mart, you wish to install a number of checkout lanes that is in the range of at least \(75 \%\) of all supermarkets. What is this range according to Chebyshev's inequality? What is the least number of checkout lanes you should install so as to fall within this range?

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