Answer: (94.08, 101.92)
Step-by-step explanation:
The confidence interval for unknown population mean[tex](\mu)[/tex] is given by :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]\sigma[/tex] = Population standard deviation
z* = Critical z-value.
Given : A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.
[tex]\sigma= 12[/tex]
[tex]\overline{x}=98[/tex]
n= 36
Confidence interval = 95%
We know that the critical value for 95% Confidence interval : z*=1.96
Then, the 95% confidence interval for the mean of x  will be :-
[tex]98\pm (1.96)\dfrac{12}{\sqrt{36}}[/tex]
[tex]=98\pm (1.96)\dfrac{12}{6}[/tex]
[tex]=98\pm (1.96)(2)[/tex]
[tex]=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)[/tex]
Hence, the 95% confidence interval for the mean of x is (94.08, 101.92) .
Answer: Â 95% confidence interval would be (94.08,101.92).
Step-by-step explanation:
Since we have given that
n = 36
standard deviation = 12
sample mean = 98
At 95% confidence, z = 1.96
So, interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=98\pm 1.96\dfrac{12}{\sqrt{36}}\\\\=98\pm 3.92\\\\=(98-3.92,98+3.92)\\\\=(94.08,101.92)[/tex]
Hence, 95% confidence interval would be (94.08,101.92).
