Bob wants to know the average GPA of FC students. He surveys 150 random students, and their average GPA is 2.3. Bob reports that the average GPA of FC students is 2.3.
Mean,μ=x‾=2.3Mean, \mu=\overline{x}=2.3Mean,μ=x=2.3
σ=0.8\sigma =0.8σ=0.8
n=150n=150n=150
For 90% Confidence; z−score=1.645z-score=1.645z−score=1.645
CI=x‾ ±z(σn)CI=\overline{x}\:\:\pm z\left(\frac{\sigma }{\sqrt{n}}\right)CI=x±z(nσ)
CI=2.3±1.645(0.8150)CI=2.3\pm 1.645\left(\frac{0.8}{\sqrt{150}}\right)CI=2.3±1.645(1500.8)
CI=2.3±0.11CI=2.3\pm 0.11CI=2.3±0.11
CI=(2.19, 2.41)CI=\left(2.19,\:2.41\right)CI=(2.19,2.41)
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