Answer:
The upper limit of the 95% confidence interval is:
C.I_u = 200 + (58.8/[tex]\sqrt{n}[/tex])
Step-by-step explanation:
The formula is given as:
C.I = μ ± Z*σ/[tex]\sqrt{n}[/tex]
The upper limit => C.I_u = μ + Z*σ/[tex]\sqrt{n}[/tex]
The lower limit => C.I_l = μ - Z*σ/[tex]\sqrt{n}[/tex]
The sample size (n) is not stated in the question. Hence, we calculate the upper limit with respect to n.
The upper limit => C.I_u = 200 + 1.96*(30/[tex]\sqrt{n}[/tex])
= 200 + (1.96*30)/[tex]\sqrt{n}[/tex]
= 200 + 58.8/[tex]\sqrt{n}[/tex]
The upper limit of the 95% confidence interval is given by
200+58.8 [tex]\rm \bold {\sqrt{n}}[/tex]
Where n = Sample size
The given problem belongs to statistics.
Confidence interval is a range of probability distribution function which is used as an important parameter to predict the values of an unknown variable.
This is one of the key parameters for decision making in statistical computing
The limits for 95% Confidence interval is given by following equations
[tex]\rm Upper\; Limit\; for\; 95 \% \; Confidence \; interval = \mu + (Z\times \sigma/\sqrt{n})........(1) \\\\Lower \; Limit\; for\; 95 \% \; Confidence \; interval = \mu - (Z\times \sigma/\sqrt{n})........(2) \\\\Where\;\\ \mu = Sample\; mean\\Z = z \; score \\n = Sample \; size \\\\\sigma = Standard \; deviation[/tex]
In the given situation according to the given data
Z Score = 1.96
[tex]\mu[/tex] = Sample mean = 200
Standard deviation = [tex]\sigma[/tex]= 30
Upper limit of the 95% confidence interval can be found out by putting the values in equation (1)
hence we can write that
Upper limit of the 95% confidence interval = 200 + (1.96[tex]\times[/tex]30) /[tex]\sqrt{n}[/tex] = 200+58.8 [tex]\rm \bold {\sqrt{n}}[/tex]
Since sample size is not known so let the sample the sample size is n.
For more information please refer to the link below
https://brainly.com/question/2396419
