Answer: The 95% confidence interval for the mean of x is (94.08, 101.92) .
Step-by-step explanation:
We are given that ,
A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.
i.e.
Also, it is given that , Sample mean having sample size : n= 36
For 95% confidence ,
Significance level :
By using the z-value table , the two-tailed critical value for 95% Confidence interval :
We know that the confidence interval for unknown population mean is given by :-
, where = Sample mean
= Population standard deviation
= Critical z-value.
Substitute all the given values, then the required confidence interval will be :
Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .
Always
In order for them to intersect they ave to be in the same plane.
Total = Principal * (1 + rate/n)^(n*years)
where "n" is the number of compounding periods per year
Total = 6,000 * (1 + .06/4)^(4*5)
Total = 6,000 * (1 + .015)^(20)
Total = 6,000 *
<span>
<span>
<span>
1.3468550066
</span>
</span>
</span>
Total =
<span>
<span>
<span>
8,081.13</span></span></span>
Source:
http://www.1728.org/compint.htm
Answer:
a = 5
Step-by-step explanation: