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Problems and detailed solutions Essay

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.

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