/*! 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} Q 50. Normal curve Estimate the mean a... [FREE SOLUTION] | 91影视

91影视

Normal curve Estimate the mean and standard deviation of the Normal density curve below.

Short Answer

Expert verified

Mean of the Normal density curve,

28

The standard deviation of the Normal density curve,

5

Step by step solution

01

Given information

02

Calculation

According to 689599.7rule:

In a normal distribution, 68percent of the data lies within 1standard deviation of the mean.

A normal distribution has 95percent of its data within two standard deviations of the mean.

A normal distribution has 99.7%of its data inside 1standard deviation of the mean.

Then

The general Normal density graph is represented as:

The mean is in the middle of the Normal density curve, near the peak of the distribution.

Note that

The peak of the given distribution appears to lie at 28

Thus,

The mean can be estimated as 28

=28

Now,

The values one standard deviation from the mean are generally at the inflection points of the distribution (where the curve appears roughly as a straight line and the curvature of the curve gets changed).

Note that

The inflection points appear to be at 23and 33both of which are 5 away from the average of 28

Thus,

The standard deviation can be estimated as 5

5

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!

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

Carbon dioxide emissions The following figure is a Normal probability plot of the emissions of carbon dioxide CO2 per person in 48 countries.15 Use the graph to determine if this distribution of CO2 emissions is approximately Normal.

Travel time (1.2) The dot-plot displays data on students鈥 responses to the question 鈥淗ow long does it usually take you to travel to school?鈥 Describe the distribution.

George鈥檚 average bowling score is180; he bowls in a league where the average for all bowlers is 150 and the standard deviation is 20Bill鈥檚 average bowling score is 190; he bowls in a league where the average is 160and the standard deviation is 15. Who ranks higher in his own league, George or Bill?

a. Bill, because his 190is higher than George鈥檚 180

b. Bill, because his standardized score is higher than George鈥檚.

c. Bill and George have the same rank in their leagues because both are 30pins above the mean.

d. George, because his standardized score is higher than Bill鈥檚.

e. George, because the standard deviation of bowling scores is higher in his league.

Jorge鈥檚 score on Exam 1in his statistics class was at the 64thpercentile of the scores for all students. His score falls

a. between the minimum and the first quartile.

b. between the first quartile and the median.

c. between the median and the third quartile.

d. between the third quartile and the maximum.

e. at the mean score for all students.

.Flight times An airline flies the same route at the same time each day. The flight time varies according to a Normal distribution with unknown mean and standard deviation. On 15%of days, the flight takes more than an hour. On localid="1649758011998" 3%of days, the flight lasts localid="1649758017692" 75minutes or more. Use this information to determine the mean and standard deviation of the flight time distribution

See all solutions

Recommended explanations on Math Textbooks

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.