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Chapter 6: Problem 38

Find the probabilities for each, using the standard normal distribution. $$ P(1.12

Short Answer

Expert verified
The probability \( P(1.12 < z < 1.43) \) is approximately 0.0550.

Step by step solution

01

Understand the Problem

We have to find the probability that the standard normal variable \( z \) falls between 1.12 and 1.43. This is represented as \( P(1.12 < z < 1.43) \).
02

Use the Z-Table

A Z-table provides the area (or probability) to the left of a given Z-score. Find the probabilities for \( z < 1.43 \) and for \( z < 1.12 \) using the Z-table.
03

Calculate the Cumulative Probability for z < 1.43

From the Z-table, locate the area to the left of \( z = 1.43 \). Assuming the Z-table gives a value of approximately 0.9236, this represents the cumulative probability \( P(z < 1.43) \).
04

Calculate the Cumulative Probability for z < 1.12

From the Z-table, find the area to the left of \( z = 1.12 \). Assuming the Z-table gives a value of approximately 0.8686, this is the cumulative probability \( P(z < 1.12) \).
05

Find the Desired Probability

Use the cumulative probabilities calculated in Steps 3 and 4. The probability that \( z \) is between 1.12 and 1.43 is the difference \( P(1.12 < z < 1.43) = P(z < 1.43) - P(z < 1.12) = 0.9236 - 0.8686 = 0.0550 \).

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Key Concepts

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

Understanding the Z-score
When diving into probability and statistics, the Z-score is a fundamental concept. The Z-score represents how many standard deviations a data point (or a value on a distribution) is away from the mean. In simple terms, it tells us how unusual or typical a value is in a data set.
  • A positive Z-score signifies that the value is above the mean.
  • A negative Z-score indicates it is below the mean.
  • A Z-score of 0 means the value is exactly at the mean.
To compute a Z-score for a value, use the formula:\[Z = \frac{(X - \mu)}{\sigma}\]where \( X \) is the value, \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation. This transformation allows us to use the standard normal distribution for probability calculations.
Navigating the Z-table
The Z-table is an incredibly handy tool when working with Z-scores. It helps in determining the probability that a standard normal variable (denoted as \( z \)) is less than or equal to a given Z-score.Here's how you can utilize a Z-table:
  • Locate the row corresponding to the first one or two digits of the Z-score.
  • Find the column that matches the second decimal place of the Z-score.
  • The intersection provides the cumulative probability from the Z-score to the left.
For example, if you wish to find the probability for \( z < 1.43 \), check the table where 1.4 intersects with 0.03.This probability signifies the cumulative probability up to that Z-score.
Decoding Cumulative Probability
Cumulative probability is all about accumulation. It tells us the probability that a random variable is less than or equal to a particular value.To understand this:
  • If you know the cumulative probability is 0.9236 for \( z = 1.43 \), it means there's a 92.36% chance that a randomly selected value is less than 1.43.
  • Similarly, 0.8686 for \( z = 1.12 \) indicates an 86.86% chance to be less than 1.12.
Cumulative probabilities are key in understanding the area under the curve, which translates to the probability of events within a range on the standard normal distribution.
Conducting Probability Calculations
Probability calculations allow us to determine the likelihood of outcomes within a given range. In the context of a standard normal distribution, these calculations are often concerned with finding the probability between two Z-scores.To calculate the probability that a Z-score falls between two values:
  • Find the cumulative probability for each Z-score individually using the Z-table.
  • Subtract the smaller cumulative probability from the larger cumulative probability.
Using our example, the calculation for \( P(1.12 < z < 1.43) \) is represented as:\[P(1.12 < z < 1.43) = P(z < 1.43) - P(z < 1.12)\]This subtraction results in 0.9236 - 0.8686, concluding with a probability of 0.0550. This shows that there is a 5.50% chance that a randomly selected value from the standard normal distribution falls between these two Z-scores.

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Most popular questions from this chapter

What is the standard deviation of the sample means called? What is the formula for this standard deviation?

To qualify for a police academy, applicants are given a test of physical fitness. The scores are normally distributed with a mean of 64 and a standard deviation of \(9 .\) If only the top \(20 \%\) of the applicants are selected, find the cutoff score.

Assume that the sample is taken from a large population and the correction factor can be ignored. Ages of Proofreaders At a large publishing company, the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. Assume the variable is normally distributed. a. If a proofreader from the company is randomly selected, find the probability that his or her age will be between 36 and 37.5 years. b. If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders in the sample will be between 36 and 37.5 years.

Assume that the sample is taken from a large population and the correction factor can be ignored. Medicare Hospital Insurance The average yearly Medicare Hospital Insurance benefit per person was \(\$ 4064\) in a recent year. If the benefits are normally distributed with a standard deviation of \(\$ 460,\) find the probability that the mean benefit for a random sample of 20 patients is a. Less than \(\$ 3800\) b. More than \(\$ 4100\)

Assume that the sample is taken from a large population and the correction factor can be ignored. Per Capita Income of Delaware Residents In a recent year, Delaware had the highest per capita annual income with \(\$ 51,803 .\) If \(\sigma=\$ 4850,\) what is the probability that a random sample of 34 state residents had a mean income greater than \(\$ 50,000 ?\) Less than \(\$ 48,000 ?\)
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