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Chapter 6: Problem 32

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(z \geq 0)$$

Short Answer

Expert verified
The probability is 0.5, as the area to the right of 0 is half of the standard normal curve.

Step by step solution

01

Understand the Standard Normal Distribution

The standard normal distribution is a probability distribution with a mean of 0 and a standard deviation of 1. It is symmetric around the mean.
02

Define the Probability

We need to find the probability that the variable \( z \) is greater than or equal to 0. This is represented as \( P(z \geq 0) \).
03

Symmetry Property

Due to the symmetry of the standard normal distribution, the probability of \( z \geq 0 \) is equal to the area from 0 to infinity. The entire curve has an area of 1, so the area to the right of 0 is half of it.
04

Calculate the Probability

Since the curve is symmetric, \( P(z \geq 0) = \frac{1}{2} \). This is because from \(-\infty\) to 0, the area is \(\frac{1}{2}\), and from 0 to \(\infty\) is also \(\frac{1}{2}\).
05

Conclusion

Thus, the area under the standard normal curve that represents \( P(z \geq 0) \) is \(0.5\), meaning that half of the data lies above 0.

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Key Concepts

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

Understanding Probability in Normal Distribution
Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. In the context of the standard normal distribution, probability is the area under the curve for a specific interval. When dealing with the question of finding the probability that a variable is greater than or equal to 0, such as in the standard normal distribution, we focus on the area of the curve from 0 to infinity.
  • The total area under the standard normal curve is 1, representing a 100% probability.
  • When calculating probability, we look for the region of the curve that corresponds to the event we're evaluating, such as events happening above or below a certain point (e.g., 0 in this case).
The probability concerning the curve is critical because it provides a visual and quantitative measure of how likely an event is relative to the entire distribution. By understanding this, you can solve various probability questions in statistics with confidence.
Exploring Standard Deviation
Standard deviation is a crucial measure of dispersion that tells us how spread out the numbers in a data set are. In a standard normal distribution, the standard deviation is always 1. This consistent measure helps in comparing different data sets and understanding variability.
  • A smaller standard deviation indicates that data points are closer to the mean.
  • A larger standard deviation means greater spread in data points, implying more variability.
In the context of the standard normal distribution, the standard deviation defines the width of the curve, making it a pivotal component for data analysis and interpretation.
Unpacking the Mean of Distribution
The mean is a central value in statistics, often referred to as the "average" of a data set. For a standard normal distribution, the mean is 0, and this center point plays a vital role. All measures of central tendency (like mean, median, mode) are located at the peak of this bell-shaped curve.
  • In a normal distribution, the mean divides the curve into two equal halves.
  • This point of symmetry is where the highest probability density occurs, meaning data values here are most frequent.
The mean being at 0 makes it an easy reference point for calculating probabilities or understanding how data points distribute around it, hence simplifying many statistical analyses.
The Symmetry of Normal Distribution
When discussing the normal distribution, symmetry is a defining characteristic. This curve is perfectly symmetrical around the mean; thus, normal distribution is often described as bell-shaped.
  • Each side of the distribution is a mirror image of the other.
  • This symmetry implies that equal areas lie on both sides of the mean.
This property of symmetry means that the probability of a value falling to the left of the mean is the same as it falling to the right. For instance, in the standard normal distribution, the probability of a value being greater than or equal to the mean (0) is exactly 0.5. This understanding of symmetry helps simplify the calculation of probabilities and makes the normal distribution a powerful tool in statistics.

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Most popular questions from this chapter

(a) If we have a distribution of \(x\) values that is more or less mound-shaped and somewhat symmetric, what is the sample size needed to claim that the distribution of sample means \(\bar{x}\) from random samples of that size is approximately normal? (b) If the original distribution of \(x\) values is known to be normal, do we need to make any restriction about sample size in order to claim that the distribution of sample means \(\bar{x}\) taken from random samples of a given size is normal?

Let \(x\) be a random variable that represents checkout time (time spent in the actual checkout process in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the \(x\) distribution is about \(\mu=2.7\) minutes, with standard deviation \(\sigma=0.6\) minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of \(x\) values is more or less symmetric and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let \(x_{i}(\text { for } i=1,2,3, \ldots, 30)\) represent the checkout time for each customer. For example, \(x_{1}\) is the checkout time for the first customer, \(x_{2}\) is the checkout time for the second customer, and so forth. Each \(x_{i}\) has mean \(\mu=2.7\) minutes and standard deviation \(\sigma=0.6\) minute. Let \(w=x_{1}+x_{2}+\cdots+x_{30}\). Explain why the problem is asking us to compute the probability that \(w\) is less than \(90 .\) (b) Use a little algebra and explain why \(w<90\) is mathematically equivalent to \(w / 30<3 .\) since \(w\) is the total of the \(30 x\) values, then \(w / 30=\bar{x} .\) Therefore, the statement \(\bar{x}<3\) is equivalent to the statement \(w<90 .\) From this we conclude that the probabilities \(P(\bar{x}<3)\) and \(P(w<90)\) are equal. (c) What does the central limit theorem say about the probability distribution of \(\bar{x} ?\) Is it approximately normal? What are the mean and standard deviation of the \(\bar{x}\) distribution? (d) Use the result of part (c) to compute \(P(\bar{x}<3) .\) What does this result tell you about \(P(w<90) ?\)

Let \(\alpha\) and \(\beta\) be any two constants such that \(\alpha<\beta .\) Suppose we choose a point \(x\) at random in the interval from \(\alpha\) to \(\beta .\) In this context the phrase at random is taken to mean that the point \(x\) is as likely to be chosen from one particular part of the interval as any other part. Consider the rectangle. The base of the rectangle has length \(\beta-\alpha\) and the height of the rectangle is \(1 /(\beta-\alpha),\) so the area of the rectangle is \(1 .\) As such, this rectangle's top can be thought of as part of a probability density curve. since we specify that \(x\) must lie between \(\alpha\) and \(\beta,\) the probability of a point occurring outside the interval \([\alpha, \beta]\) is, by definition, \(0 .\) From a geometric point of view, \(x\) chosen at random from \(\alpha\) to \(\beta\) means we are equally likely to land anywhere in the interval from \(\alpha\) to \(\beta .\) For this reason, the top of the (rectangle's) density curve is flat or uniform. Now suppose that \(a\) and \(b\) are numbers such that \(\alpha \leq a

Find the \(z\) value described and sketch the area described.Find the \(z\) value such that \(98 \%\) of the standard normal curve lies between \(-z\) and \(z\).

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. To the right of \(z=1.52\)
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