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 .
Z can be between 1 ) 57 and 1 . 84
Z is no more than or higher than 1 . 84
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?
(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 %
Pestman, W. R., & Alberink, I. B. (2008). Concerns and detailed solutions. Berlin: De Gruyter.
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