Answer: c. [tex]70\pm (1.96)\dfrac{4.5}{\sqrt{50}}[/tex]
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= sample size
[tex]\overline{x}[/tex] = sample mean.
z*= Critical value.
[tex]\sigma[/tex]= population standard deviation.
As per given , we have
n= 50
[tex]\sigma=4.5[/tex]
[tex]\overline{x}=70[/tex]
Also, we know that the critical value for 95% confidence interval : z*= 1.96
Then, the 95% confidence interval (CI) to estimate the true average score of golfers on this particular course will be :
[tex]70\pm (1.96)\dfrac{4.5}{\sqrt{50}}[/tex] (substitute all the value in the above formula.)
Hence, the correct option is c. [tex]70\pm (1.96)\dfrac{4.5}{\sqrt{50}}[/tex]
