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Chapter 6: Problem 37

Find the probabilities for each, using the standard normal distribution. $$ P(1.56

Short Answer

Expert verified
The probability \(P(1.56 < z < 2.13)\) is 0.0428.

Step by step solution

01

Understand the Standard Normal Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. We use the standard normal distribution table (z-table) to find the probabilities associated with standard normal variables.
02

Locate Probabilities in the Z-table

To solve for the probability that lies between two z-scores, we need to find the probability associated with each z-score from the z-table. First, locate the probability for \(z = 2.13\), which is approximately 0.9834.
03

Repeat for the Second Z-score

Next, we locate the probability associated with the z-score \(z = 1.56\), which is approximately 0.9406 from the z-table.
04

Calculate the Probability Between Two Z-scores

To find \(P(1.56 < z < 2.13)\), subtract the probability associated with the lower z-score from the probability associated with the higher z-score. So, \(P(1.56 < z < 2.13) = 0.9834 - 0.9406 = 0.0428\).

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Key Concepts

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

Understanding the Z-table
The z-table, also known as the standard normal distribution table, is a tool used to find the probability of a z-score in a standard normal distribution.
It lists the cumulative probability of a random variable falling to the left of a given z-score, which represents the number of standard deviations a data point is from the mean.

When reading a z-table:
  • The numbers to the left and top indicate the z-score values.
  • The numbers in the body of the table represent cumulative probabilities.
These probabilities help determine
  • How likely it is for a random variable to fall within a certain range.
  • The area under the standard normal curve to the left of the z-score.
What are Z-scores?
A z-score tells us how far and in what direction a data point is from the mean in terms of standard deviations.
In a standard normal distribution, the mean is 0, and the standard deviation is 1.
Calculating z-scores lets us compare different data points from different normal distributions as if they are part of the same dataset.

The formula for calculating a z-score is:\[ z = \frac{(X - \mu)}{\sigma} \]where
  • \(X\) is the value we are calculating the z-score for,
  • \(\mu\) is the mean of the distribution,
  • and \(\sigma\) is the standard deviation.
This standardization allows for easy comparison and use in probability calculations.
Probability Calculations with Z-scores
Calculating probabilities with z-scores involves determining the area under the curve of a standard normal distribution.
This process can tell us the likelihood of a z-score falling within a certain range.
To calculate the probability between two z-scores:
  • Identify the z-scores of interest.
  • Use the z-table to find the cumulative probability for each z-score.
  • Subtract the smaller probability from the larger one for the range.
In the exercise example, this process was used to calculate the probability between z-scores 1.56 and 2.13, resulting in a probability of 0.0428.
The Role of Normal Distribution
A normal distribution is a continuous probability distribution often called a "bell curve" due to its shape.
The 'standard' normal distribution specifically has a mean of 0 and a standard deviation of 1.
This special case allows us to use simplified tables like the z-table to perform probability calculations.

Features of a normal distribution include:
  • Symmetrical about the mean.
  • The mean, median, and mode are all equal.
  • Approximately 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three.
Understanding the normal distribution is crucial for interpreting data and performing statistical analyses with z-scores.

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Most popular questions from this chapter

Assume that the sample is taken from a large population and the correction factor can be ignored. Cell Phone Lifetimes A recent study of the lifetimes of cell phones found the average is 24.3 months. The standard deviation is 2.6 months. If a company provides its 33 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.8 months. Assume cell phone life is a normally distributed variable.

A mandatory competency test for high school sophomores has a normal distribution with a mean of 400 and a standard deviation of 100 . a. The top \(3 \%\) of students receive \(\$ 500 .\) What is the minimum score you would need to receive this award? b. The bottom \(1.5 \%\) of students must go to summer school. What is the minimum score you would need to stay out of this group

Find the area under the standard normal distribution curve. To the right of \(z=-1.92\)

To help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom \(5 \%\) of the students in the district, based on the scores on a reading achievement exam. If the average score for the students in the district is \(122.6,\) find the cutoff score that will make a student eligible for the program. The standard deviation is \(18 .\) Assume the variable is normally distributed.

Find the probabilities for each, using the standard normal distribution. $$ P(-1.23
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