Answer: upper bound = 89.36 and lower bound = 94.64
Step-by-step explanation:
Confidence interval for mean : [tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex], where [tex]\overline{x}[/tex] = sample mean , n= sample size , z* = two-tailed critical value, [tex]\sigma[/tex]= population standard deviation
As per given , we have
[tex]\overline{x}[/tex] = 92
n = 55
[tex]\sigma[/tex]= 10
Critical value for 95% confidence interval = 1.96
Required confidence interval : [tex]92\pm (1.96)\dfrac{10}{\sqrt{55}}[/tex]
[tex]92\pm (1.96)(1.3484)\approx92\pm2.64\\\\= (92-2.64,\ 92+2.64)\\\\=(89.36,\ 94.64)[/tex]
Hence, for 95% confidence interval for the mean IQ of primary school students:
upper bound = 89.36 and lower bound = 94.64