Challenges and thorough solutions essay

Statistics

Introduction

As defined by distinct scholars and mathematicians, stats refers to technology of expanding conclusions and learning from info, calculating and making informed decisions about a phenomena and its particular behavior by making use of data from calculated presumptions such as mean, mode, regular deviation, variance, and possibility among many others (Pestman & Alberink, 2008). The subsequent calculations can adopt many of the statistics presumptions to estimate results and make conclusions about the behaviour of the data.

Question six. 1

Given a standardised normal division mean zero and normal 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 can be between 1 ) 57 and 1 . 84

Z is no more than or higher than 1 . 84

The solution

Z-score = (data point “mean) / regular deviation.

(1. 57 ” 0) /s 1 sama dengan 1 . 57

Reading through the Z-scores desk, = 0. 9418

Possibility that Z is higher than 1 . 57 = 94. 18%

(0 -1. 84)/ 1=- 1 . 84

In the Z- ratings table =0. 0329

Consequently Probability that Z is less than 1 . 84 = 3. 29%

c)(94. 18 ” 3. 29) %

The chances that Unces is between 1 . 57 and 1 ) 84 is usually = 90. 89 %

d) (0. 94. 18 +0. 0329) / a couple of

The possibility that Z is less than 1 . 57 or perhaps greater than 1 ) 84 sama dengan 0. 4874

Probability sama dengan 68. seventy nine %

Problem 6. several

In 2011 the per capita consumption of coffee in United States was reported to become 4. 16 kg and 9. 56 pounds (data extracted from www.ico.org). Assume that the every capita consumption of coffee in Usa is approximately normally distributed having a mean of 9. 152 pounds and a standard change of 4. 16

Precisely what is the probability that an individual in Usa consumed much more than 10 pounds of caffeine in 2011?

What is the probability that someone in u United States used between three or more and five pounds of coffee this summer?

What is the probability that someone in United States used less than a few pounds of coffee this summer?

The Solution

(9. 152 ” 10) /3 = -0. 2837

The Z- ratings will be =. 3897

Hence the probability that someone in United States used more than 10 pounds of coffee this summer = 35. 97 %

i. (3 -4. 16) / three or more = zero. 3867

= 0. 3483

Probability = 34. 83%

ii. (5 ” 4. 16) / 3

sama dengan 0. 2800

Reading through the Z- results table sama dengan 0. 6103

Therefore the likelihood that a person in us consumed among 3 and 5 pounds 2011 will probably be = sixty one. 03 %

C) (5 ” some. 16)/ 3 = zero. 2800

sama dengan 0. 6103

Probability that the individual in united states consumed less than five pounds this year = sixty one. 03 %

Reference

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

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