MA205 Chapter 6 Driving Distances Travel Plans & Movie Length Questions

User Generated

mnvqafu

Mathematics

MA205

Description

I want someone to answer this question and write a sentence for each one explain the answer

Unformatted Attachment Preview

Scanned with CamScanner
Purchase answer to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Explanation & Answer

Find the solution....

1. The point estimate of the population mean µ is
∑ 𝑥 286
𝑥̅ =
=
= 9.533
𝑛
30
The point estimate of the population mean µ of driving distances (in miles) to work for
people is 9.533 miles
Margin of error for a 95% confidence interval is
𝐸 = 𝑧0.95

𝜎

√𝑛
Here, 𝑧0.95 is the one-side critical value of the standard normal distribution at 95 %
confidence level. From the standard normal tables, the one –side critical value of normal
distribution at 95% confidence level is found to be 1.96.
Substituting 𝜎 = 8, 𝑧0.95 = 1.96 and n = 30, in the margin of error
8
𝐸 = 1.96
= 2.863
√30
The margin of error for a 95% confidence interval for the driving distances (in miles) to
work for people is 2.863 miles.
The confidence interval is ̅̅̅
(𝑥 ± 𝐸) which is (6.67, 12.396)
2. In this question the objective is to calculate the value of n
Let n represent the number of people need for survey
From the given information,
Margin of error E = 2
Population standard deviation 𝜎 = 8
C = 0.99 so

1−𝐶
2

= 0.005

The critical value of Z is �...


Anonymous
Really helpful material, saved me a great deal of time.

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4

Related Tags