Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Q.6.68 (page 390)

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 probability of spending more than two days in recovery?

a. 0.0580

b. 0.8447

c. 0.0553

d. 0.9420

Short Answer

Expert verified

The correct answer is option dthat is value ofP(X>2)is0.9420.

Step by step solution

01

Given information

Given in the question that

Mean=5.3

Standard deviation=2.1

02

Step: Solution

Here we need to find the probability of spending more than two days in recovery

X~N(5.3,2.1)

So, the probability can be calculated as

P(x>2)=1−P(x<2)

=1−Px−μσ<2−μσ

=1−PZ<2−5.32.1

=1−P(Z<−1.5714)

=1−0.0580From standardnormal table

=0.9420

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

The life of Sunshine CD players is normally distributed with a mean of 4.1years and a standard deviation of 1.3years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.

a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.

b. P(0<x<____________)=___________(Use zero for the minimum value of x.)

Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80cm girls in the reference population had a mean µ=10.2kg and standard deviation σ=0.8kg. Weights are normally distributed. X~N(10.2,0.8). Calculate the z-scores that correspond to the following weights and interpret them.

a. 11 kg

b. 7.9 kg

c. 12.2 kg

The scores on a college entrance exam have an approximate normal distribution with mean, µ=52points and a standard deviation, σ=11points.

a. About 68%of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.

b. About 95%of the yvalues lie between what two values? These values are ________________. The z-scores are ________________, respectively.

c. About 99.7% of the y values lie between what two values? These values are ________________. Thez-scores are ________________, respectively.

Jerome averages 16 points a game with a standard deviation of four points. X~N(16,4). Suppose Jerome scores ten points in a game. The z–score when x=10 is –1.5. This score tells you that x=10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).

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
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Calculus

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