We use cookies to give you the best experience possible. By continuing we’ll assume you’re on board with our cookie policy

Check Writers' Offers

What's Your Topic?

Hire a Professional Writer Now

The input space is limited by 250 symbols

What's Your Deadline?

Choose 3 Hours or More.
Back
2/4 steps

How Many Pages?

Back
3/4 steps

Sign Up and Get Writers' Offers

"You must agree to out terms of services and privacy policy"
Back
Get Offer

Problems and detailed solutions

Paper type: Essay
Pages: 2 (459 words)
Downloads: 12
Views: 342

Statistics

Introduction

       As defined by different scholars and mathematicians, statistics refers to the science of developing conclusions and learning from data, calculating and making informed decisions about a phenomena and its behavior through the use of data from calculated assumptions such as mean, mode, standard deviation, variance, and probability among many others (Pestman & Alberink, 2008). The following calculations will adopt several of the statistics assumptions to calculate results and make conclusions about the behavior of the data.

Question 6.1

Given a standardized normal distribution mean 0 and standard deviation of 1 as in table E .

2. What is the probability that:

Z is greater than 1.57

Z is less than 1.

84

Z is between 1.57 and 1.84

Z is less than or greater than 1.84

The solution

Z-score = (data point –mean) / standard deviation.

(1.57 – 0) /s 1 = 1.57

Reading from the Z-scores table, = 0.9418

Probability that Z is greater than 1.57 = 94.18%

(0 -1.84)/ 1=- 1.84

From the Z- scores table =0.0329

Therefore Probability that Z is less than 1.84 = 3.29%

c)(94.18 – 3.29) %

The chances that Z is between 1.57 and 1.84 is = 90.89 %

d) (0.94.18 +0.0329) / 2

The probability that Z is less than 1.57 or greater than 1.84 = 0.4874

Probability = 68.79 %

Question 6.7

In 2011 the per capita consumption of coffee in United States was reported to be 4.16 kg and 9.56 pounds (data extracted from www.ico.org). Assume that the per capita consumption of coffee in United States is approximately normally distributed with a mean of 9.152 pounds and a standard deviation of 4.16

What is the probability that someone in United States consumed more than 10 pounds of coffee in 2011?

What is the probability that someone in u United States consumed between 3 and 5 pounds of coffee in 2011?

What is the probability that someone in United States consumed less than 5 pounds of coffee in 2011?

The Solution

(9.152 – 10) /3 = -0.2837

The Z- scores will be = .3897

Therefore the probability that someone in United States consumed more than 10 pounds of coffee in 2011 = 38.97 %

i. (3 -4.16) / 3 = 0.3867

= 0.3483

Probability = 34.83%

ii. (5 – 4.16) / 3

= 0.2800

Reading from the Z- scores table = 0.6103

Therefore the probability that a person in united states consumed between 3 and 5 pounds 2011 will be = 61.03 %

C) (5 – 4.16)/ 3 = 0.2800

= 0.6103

Probability that an individual in united states consumed less than 5 pounds in 2011 = 61.03 %

Reference

Pestman, W. R., & Alberink, I. B. (2008). Problems and detailed solutions. Berlin: De Gruyter.

Source document

Cite this page

Problems and detailed solutions. (2015, Nov 15). Retrieved from https://studymoose.com/statistics-essay

How to Avoid Plagiarism
  • Use multiple resourses when assembling your essay
  • Use Plagiarism Checker to double check your essay
  • Get help from professional writers when not sure you can do it yourself
  • Do not copy and paste free to download essays
Get plagiarism free essay

Not Finding What You Need?

Search for essay samples now

image

Your Answer is very helpful for Us
Thank you a lot!