### Statistics 200 Final

Question Details: #5703

**Statistics 200 Final**

Introduction to Statistics STAT 200 Final Exam

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- Posted by: ExAQUSN
- Subjects: Math Environmental Law
- Attachments:
*Stat_200_Final_Summer_2015_Upload.docx (309 KB)*

Solution Details: #5934

**STAT 200 Final Complete Solution**

Answers:
(a) True
(b)False, enough information is not given to find P(A and B)
(c) True
(d) False
(e) True
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**Statistics 200 Final solution correct answers**

Question 1:
True, if all the observations in a data set are identical then the mean would be the same value so the deviation from mean would be 0 for all values and thus the variance would be 0.
False, if A and B are independent events then only P(A and B) = P(A)*P(B) = 0.5*0.4 = 0.2.
True, the normal distribution is symmetric about the mean thus the mean, median and mode are always same.
True, the 95% confidence interval uses a larger Z multiplier thus is wider than the 90% confidence interval.
False, as this is a two tailed test so P-value = P(|X|> 2) = 2*P(X>2) = 0.06. Which is larger than the significance level and thus the conclusion.
Question 2:
The complete table is given below,
Study Time (in hours) Frequency Relative Frequency
0.0-4.9 5 5/100 = 0.05
5.0-9.9 13 13/100 = 0.13
10.0-14.9 0.22*100 = 22 0.22
15.0-19.9 42 42/100 = 0.42
20.0-24.9 100-(5+13+22+42) = 18 18/100 = 0.18
Total 100 100/100 = 1
Question 3:
Required percentage = P(X ≥ 15)
= Relative frequency of class 15.0-19.9 + Relative frequency of class20.0-24.9
= 0.42+0.18 = 0.60 = 60%
Question 4:
There are 100 observations in total out of which the 1st 50 lies on or below the class interval 15.0-19.9. Thus the class interval in which median lie is 15.0-19.9.
Question 5:
Considering 15.0-19.9 as the median class we can see the class right of this class has higher frequency and smaller in number (on right we have only one class and on left we have 3 classes) as compare to the classes in left. Thus the distribution is left skewed. It can be seen more clearly from the histogram which is given below.
Question 6:
Note that each roll of a dice has 6 outcomes as 1,2,3,4,5 and 6. Now rolling two times means the same number would appear on the 2nd roll as well. As the rolls are independent hence for each of the outcomes in 1st roll there are 6 outcomes in 2nd roll. As we are rolling a 6 faced dice twice thus the number of elements in the sample space is 6*6 = 36. The sample space would look like, {(1,1); (1,2); (1,3);…(6,5),(6,6)}.
Question 7:
If the 1st roll is a multiple of 3 then...

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