Question

the score of shs students in their first quarter in statistic and probability brainly were analyzed and found to have a mean of 54.5and standard deviation of 7.5. find a 90% confidence interval

Answer 1

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(54.5-1.6449\times\dfrac{7.5}{\sqrt{n}}, 54.5+1.6449\times\dfrac{7.5}{\sqrt{n}})$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $54.5-1.6449\times\dfrac{7.5}{\sqrt{n}}<\mu< 54.5+1.6449\times\dfrac{7.5}{\sqrt{n}},$

which indicates that we are 90% confident that the true population mean $\mu$

is contained by the interval 

$$(54.5-1.6449\times\dfrac{7.5}{\sqrt{n}}, 54.5+1.6449\times\dfrac{7.5}{\sqrt{n}}).$$