Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Q.6.61 (page 389)

The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

What is the z-score for a patient who takes ten days to recover?

a. 1.5

b. 0.2

c. 2.2

d. 7.3

Short Answer

Expert verified

Option (c) is the correct answer because the value of the z-score of ten days is2.2.

Step by step solution

01

Given information

Given in the question that

Mean=5.3

Standard deviation=2.1

02

Solution

Here we need to calculate the z-score of 10days

z−score=10−μσ

=10−5.32.1

=4.72.1

≈2.2

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.

On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.

a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30 .

b. Find the 95th percentile, and express it in a sentence.

In 2005,1,475,623students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ=520and standard deviation σ=115

  1. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence.
  2. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score?
  3. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took?

According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66inches and a standard deviation of 2.5inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.

a.X~_____(_____,_____)

b. Find the probability that the person is between 65and 69inches. Include a sketch of the graph, and write a probability statement.

c. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically.

The heights of the 430National Basketball Association players were listed on team rosters at the start of the 2005–2006season. The heights of basketball players have an approximately normal distribution with mean, µ=79inches and a standard deviation, σ=3.89inches. For each of the following heights, calculate the z-score and interpret it using complete sentences

a. 77inches

b. 85inches

c. If an NBA player reported his height had a z-score of 3.5, would you believe him? Explain your answer

Suppose X ~ N(9, 3). What is the z-score of x = 9?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

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.

Sign-up for free