/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Find the probability that \(z\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 65

Find the probability that \(z\) lies between \(z=-1.48\) and \(z=1.48\)

Short Answer

Expert verified
Answer: The probability that z lies between -1.48 and 1.48 is 0.8612 or 86.12%.

Step by step solution

01

Collect the necessary information and tools

To solve this problem, we need a Z-table which can be found online or in any standard statistics textbook. The Z-table provides the probabilities for different Z-scores in the standard normal distribution.
02

Read the Z-table for z = -1.48

Locate -1.48 in the Z-table which will typically have rows and columns to help you navigate. The row usually corresponds to the first two digits and one decimal place of the Z-score (-1.4 in this case), and the column corresponds to the second decimal place (0.08 in this case). The value in the cell where the row and column intersect is the probability. In this case, for z = -1.48, the probability is 0.0694.
03

Read the Z-table for z = 1.48

Similar to Step 2, locate the value corresponding to z = 1.48 in the Z-table. The probability for z = 1.48 is 0.9306.
04

Calculate the probability for z between -1.48 and 1.48

Subtract the probability of z = -1.48 from the probability of z = 1.48 to find the probability of z lying between the two values: \(P(-1.48 \leq z \leq 1.48) = P(z \leq 1.48) - P(z\leq -1.48) = 0.9306 - 0.0694 = 0.8612\). The probability that z lies between -1.48 and 1.48 is 0.8612 or 86.12%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

standard normal distribution
The standard normal distribution is an essential statistical concept. It's a bell-shaped curve, symmetric around the mean. The mean of this distribution is 0, and the standard deviation is 1.
This makes it a useful tool for examining random variables that have been standardized. In practice, the standard normal distribution is used to understand data behaviors and predict probabilities.
It serves as a fundamental basis in statistics, allowing for the comparison of different data sets with diverse properties.
z-scores
Z-scores are standardized scores used to understand where a data point stands in a distribution. They represent the number of standard deviations a data point is from the mean.
To calculate a z-score, use the formula:
\[ z = \frac{x - \mu}{\sigma} \]
where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Z-scores help identify how typical or atypical a particular measurement is compared to the others in the data set.
This method standardizes different data points, making them easily comparable.
probability calculation
Probability calculation involves finding the likelihood of a particular outcome. In the context of the standard normal distribution, this often involves z-scores.
When you know the z-scores, you can use a Z-table to find the probabilities associated with those scores.
In the given exercise, the probability calculation for z-scores between -1.48 and 1.48 involves checking the Z-table for each score. Then, subtracting the smaller probability from the larger gives the probability of z lying between the two values.
This process quantifies the chance of a data point falling within a specific range in the distribution.
Z-table interpretation
Interpreting a Z-table is vital for understanding probabilities in the standard normal distribution. A Z-table lists the probabilities that a standard normal random variable is less than or equal to a given z-score.
  • The rows reflect the first two digits and the first decimal of the z-score.
  • The columns capture the second decimal.
By combining these, you locate the probability that corresponds to a particular z-score.
For example, a z-score of -1.48 would have its probability found by intersecting the row for -1.4 and the column for 0.08 in the Z-table.
Understanding these readings is crucial to make accurate probability predictions using the Z-table.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercise \(6.28,\) we suggested that the IRS assigns auditing rates per state by randomly selecting 50 auditing percentages from a normal distribution with a mean equal to \(1.55 \%\) and a standard deviation of \(.45 \%\) a. What is the probability that a particular state would have more than \(2 \%\) of its tax returns audited? b. What is the expected value of \(x\), the number of states that will have more than \(2 \%\) of their income tax returns audited? c. Is it likely that as many as 15 of the 50 states will have more than \(2 \%\) of their income tax returns audited?

One method of arriving at economic forecasts is to use a consensus approach. A forecast is obtained from each of a large number of analysts, and the average of these individual forecasts is the consensus forecast. Suppose the individual 2013 January prime interest rate forecasts of economic analysts are approximately normally distributed with the mean equal to \(4.75 \%\) and a standard deviation equal to \(0.2 \% .\) If a single analyst is randomly selected from among this group, what is the probability that the analyst's forecast of the prime rate will take on these values? a. Exceed \(4.25 \%\) b. Be less than \(4.375 \%\)

Let \(x\) be a binomial random variable with \(n=36\) and \(p=.54 .\) Use the normal approximation to find: a. \(P(x \leq 25)\) b. \(P(15 \leq x \leq 20)\) c. \(P(x>30)\)

A stringer of tennis rackets has found that the actual string tension achieved for any individual racket will vary as much as 6 pounds per square inch from the desired tension set on the stringing machine. If the stringer wishes to string at a tension lower than that specified by a customer only \(5 \%\) of the time, how much above or below the customer's specified tension should the stringer set the stringing machine? (NOTE: Assume that the distribution of string tensions produced by the stringing machine is normally distributed, with a mean equal to the tension set on the machine and a standard deviation equal to 2 pounds per square inch.)

A soft drink machine can be regulated to discharge an average of \(\mu\) ounces per cup. If the ounces of fill are normally distributed, with standard deviation equal to .3 ounce, give the setting for \(\mu\) so that 8 -ounce cups will overflow only \(1 \%\) of the time.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.