Garantie de satisfaction à 100% Disponible immédiatement après paiement En ligne et en PDF Tu n'es attaché à rien 4.2 TrustPilot
logo-home
Examen

MATH 225N Week 5 Assignment: Central Limit Theorem for Means.

Note
-
Vendu
-
Pages
16
Grade
A+
Publié le
13-01-2021
Écrit en
2020/2021

MATH 225N Week 5 Assignment: Central Limit Theorem for Means. Question and well detailed explained answers. All correct. Happy studying

Établissement
Cours










Oups ! Impossible de charger votre document. Réessayez ou contactez le support.

École, étude et sujet

Établissement
Cours

Infos sur le Document

Publié le
13 janvier 2021
Nombre de pages
16
Écrit en
2020/2021
Type
Examen
Contient
Questions et réponses

Sujets

Aperçu du contenu

MATH 225N Week 5 Assignment: Central Limit Theorem for
Means.

1. Question

A family of statisticians is trying to decide if they can afford for their child to play youth
baseball. The cost of joining a team is normally distributed with a mean of $750 and a standard
deviation of $185 . If a sample of 40 teams is selected at random from the population, select the
expected mean and standard deviation of the sampling distribution below.

Correct answer:
σx¯=$29.25
μx¯=$750

The standard deviation of the sampling distribution

σx¯=σn−−√=$18540−−√=$29.25

When the distribution is normal the mean of the sampling distribution is equal to the mean of the
population μx¯=μ=$750 .


Question

A cupcake baker is planning a supplies order and needs to know how much flour he needs. He
knows that his recipes use an average of 100 grams of flour, normally distributed, with a
population standard deviation of 15 grams. If he is consulting a sample size of 30 recipes, select
the mean and standard deviation of the sampling distribution to help him order his supplies from
the options below.

σx¯=2.74 grams
μx¯=100 grams

The standard deviation of the sampling distribution is

σx¯=σn−−√=1530−−√=2.74 grams

,Likewise, when the distribution is normal the mean of the sampling distribution is equal to the
mean of the population μx¯=μ=100 grams.


Question

A head librarian for a large city is looking at the overdue fees per user system-wide to determine
if the library should extend its lending period. The average library user has $19.67 in fees, with a
standard deviation of $7.02 . The data is normally distributed and a sample of 72 library users is
selected at random from the population. Select the expected mean and standard deviation of the
sampling distribution from the options below.

Correct answer:
σx¯=$0.83
μx¯=$19.67

The standard deviation of the sampling distribution is

σx¯=σn−−√=$7.0272−−√=$0.83

When the distribution is normal, the mean of the sampling distribution is equal to the mean of
the population μx¯=μ=$19.67 .




2. Question

A well known social media company is looking to expand their online presence by creating
another platform. They know that they current average 2,500,000 users each day, with a standard
deviation of 625,000 users. If they randomly sample 50 days to analyze the use of their existing
technology, identify each of the following, rounding to the nearest whole number if necessary:

We are given population mean μ=2,500,000 and population standard deviation σ=625,000 , and
want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of
size n=50 .

By the Central Limit Theorem, the means of the two distributions are the same:

, μx¯=μ=2,500,000

To find the Standard Deviation of the sampling distribution, we divide the population standard
deviation by the square root of the sample size:

σx¯=σn−−√=625,00050−−√≈88,388




3. Question

A bank is reviewing its risk management policies with regards to mortgages. To minimize the
risk of lending, the bank wants to compare the typical mortgage owed by their clients against
other homebuyers. The average mortgage owed by Americans is $306,500 , with a standard
deviation of $24,500 . Suppose a random sample of 150 Americans is selected.

Identify each of the following, rounding your answers to the nearest cent when appropriate:

 1306500$306500$306500
 224500$24500$24500
 3150$150$150
 4306500$306500$306500
 $2000.42

We are given population mean μ=$306,500 and population standard deviation σ=$24,500 , and
want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of
size n=150 .

By the Central Limit Theorem, the means of the two distributions are the same:

μx¯=μ=$306,500

To find the Standard Deviation of the sampling distribution, we divide the population standard
deviation by the square root of the sample size:

σx¯=σn−−√=$24,500150−−−√=$2,000.42
€177,49
Accéder à l'intégralité du document:

Garantie de satisfaction à 100%
Disponible immédiatement après paiement
En ligne et en PDF
Tu n'es attaché à rien

Faites connaissance avec le vendeur

Seller avatar
Les scores de réputation sont basés sur le nombre de documents qu'un vendeur a vendus contre paiement ainsi que sur les avis qu'il a reçu pour ces documents. Il y a trois niveaux: Bronze, Argent et Or. Plus la réputation est bonne, plus vous pouvez faire confiance sur la qualité du travail des vendeurs.
bettergrades19 Chamberlain College Of Nursing
S'abonner Vous devez être connecté afin de suivre les étudiants ou les cours
Vendu
1206
Membre depuis
5 année
Nombre de followers
1097
Documents
354
Dernière vente
1 année de cela

4,1

121 revues

5
68
4
21
3
16
2
3
1
13

Récemment consulté par vous

Pourquoi les étudiants choisissent Stuvia

Créé par d'autres étudiants, vérifié par les avis

Une qualité sur laquelle compter : rédigé par des étudiants qui ont réussi et évalué par d'autres qui ont utilisé ce document.

Le document ne convient pas ? Choisis un autre document

Aucun souci ! Tu peux sélectionner directement un autre document qui correspond mieux à ce que tu cherches.

Paye comme tu veux, apprends aussitôt

Aucun abonnement, aucun engagement. Paye selon tes habitudes par carte de crédit et télécharge ton document PDF instantanément.

Student with book image

“Acheté, téléchargé et réussi. C'est aussi simple que ça.”

Alisha Student

Foire aux questions