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Chapter 7: Problem 4

Explain why \(P(Z<-1.30)=P(Z \leq-1.30)\)

Short Answer

Expert verified
For a continuous normal distribution, P(Z < -1.30) = P(Z 鈮� -1.30) because there is no difference between 'less than' and 'less than or equal to' in this context.

Step by step solution

01

Understand the Standard Normal Distribution

A standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is symmetric around its mean, resulting in equal probabilities for corresponding positive and negative values of Z.
02

Define the Z-Score

A Z-score represents the number of standard deviations a data point is from the mean. Therefore, a Z-score of -1.30 means the data point is 1.30 standard deviations below the mean.
03

Understand the Concept of Cumulative Probability

In statistics, cumulative probability refers to the probability that a random variable is less than or equal to a certain value. This is often denoted as P(Z 鈮� z).
04

Consider Negative Z-Scores in Cumulative Probability

Since cumulative probability includes the probability of the value itself and all values less than it, for any Z-score, P(Z < z) and P(Z 鈮� z) are effectively equivalent because of the continuous nature of the distribution.
05

Specific Case for Z = -1.30

Specifically, for Z = -1.30 in a continuous normal distribution, the probabilities are equal, so P(Z < -1.30) = P(Z 鈮� -1.30). This is because there is no difference between 'less than' and 'less than or equal to' in an infinitely divisible distribution.

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Key Concepts

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

Z-Score
The term 'Z-score' is essential when dealing with standard normal distribution. It measures how many standard deviations a particular data point is from the mean, altering our understanding of where this point lies relative to the entire data set. For instance, a Z-score of 1.30 means the point is 1.30 standard deviations above the mean. Conversely, a Z-score of -1.30 means the point is 1.30 standard deviations below the mean. Using Z-scores helps in comparing data points from different distributions by standardizing them around a mean of 0.
Cumulative Probability
Cumulative probability refers to the likelihood that a random variable will be less than or equal to a certain value, often denoted as \(P(Z \leq z)\). This probability sums up the chances of all outcomes up to and including the specified value. In continuous distributions, like the standard normal distribution, there's no practical distinction between 'less than' and 'less than or equal to' when calculating the cumulative probability. This is because the probability of the continuous random variable taking any exact value is zero. Therefore, \(P(Z < -1.30) = P(Z \leq -1.30)\).
Continuous Distribution
In statistics, a continuous distribution implies that the variable can take an infinite number of possible values. A standard normal distribution is an example of this, with a symmetrical bell-shaped curve centered around the mean, which is 0. Due to its continuous nature, there鈥檚 no significant difference between including or excluding a particular point while calculating cumulative probabilities. Probabilities are considered over intervals rather than individual points. For example, the probability that Z is exactly -1.30 is technically zero but considering probabilities over intervals makes logical calculations meaningful.

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Most popular questions from this chapter

Compute \(P(x)\) using the binomial probability formula. Then determine whether the normal distribution can be used as an approximation for the binomial distribution. If so, approximate \(P(x)\) and compare the result to the exact probability. $$n=85, p=0.8, X=70$$

Find the \(Z\) -scores that separate the middle \(70 \%\) of the distribution from the area in the tails of the standard normal distribution. Find the \(Z\) -scores that separate the middle \(99 \%\) of the distribution from the area in the tails of the standard normal distribution.

Assume the random variable \(X\) is normally distributed with mean \(\mu=50\) and standard deviation \(\sigma=7 .\) Compute the following probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded. $$P(55 \leq X \leq 70)$$

Find the indicated \(Z\) -score. Be sure to draw a standard normal curve that depicts the solution. Find the \(Z\) -score such that the area under the standard normal curve to the left is 0.85

The following data represent the relative frequencies of live multiple- delivery births (three or more babies) in 2002 for women 15 to 44 years old. Suppose the ages of multiple-birth mothers are approximately normally distributed with \(\mu=31.77\) years and standard deviation \(\sigma=5.19\) years. (a) Compute the proportion of multiple-birth mothers in each class by finding the area under the normal curve. (b) Compare the proportion to the actual proportions. Are you convinced that the ages of multiple-birth mothers are approximately normally distributed?
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