Answer:
The mean and standard deviation of the number preferring the incumbent is mean = 330, standard deviation = 10.59.
Step-by-step explanation:
We are given that From previous polls, it is believed that 66% of likely voters prefer the incumbent.
A new poll of 500 likely voters will be conducted. In the new poll the proportion favoring the incumbent has not changed.
Let p = probability of voters preferring the incumbent = 66%
n = number of voters polled = 500
<u>So, the mean of the number preferring the incumbent is given by;</u>
Mean = 
= 
= 330 voters
<u>And, standard deviation of the number preferring the incumbent is given by;</u>
Variance = 
= 
= 112.2
So, Standard deviation = 
= 
= 10.59
It's 5. Just took this test and got this question right :)
Answer:
- 0.9503 ; r is not statistically significant ; 0.9031
Step-by-step explanation:
Given the following :
Age (X) :
37
41
57
65
73
Bone density (Y)
355
345
340
315
310
Using the pearson R value calculator :
The r value of the data % - 0.9503.
This value depicts a very strong negative correlation between age and density of bone.
Using the pearson R calculator to obtain the P- value, the P value obtained is .01332 and hence the r is not significant at P < 0.01.
The Coefficient of determination R^2 can be obtained by getting the square value of R
R^2 = - 0.9503^2
R^2 = 0.90307009
R^2 = 0.9031
What is the third quartile of this data set 20,21,24,25,28,29,35,36,37,43,44
SashulF [63]
Answer:
37
Step-by-step explanation:
The data set is in ascending order, thus the median is the middle value of the data set.
median = 29
The third quartile is the middle value of the data to the right of the median
= 37
Answer:
True
Step-by-step explanation:
