Problems and detailed solutions
Problems and detailed solutions
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
Zscore = (data point –mean) / standard deviation.
(1.57 – 0) /s 1 = 1.57
Reading from the Zscores 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.
A+

Subject: Business & Economy,

University/College: University of Chicago

Type of paper: Thesis/Dissertation Chapter

Date: 15 November 2015

Words:

Pages:
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