2)0.9938
P(X≥\geq≥ 67.5)
Mean=70, sd=6, sample size =36
Z=(X−μ)∗nSDZ=\frac {(X-\mu) *\sqrt n} {SD}Z=SD(X−μ)∗n
Z=(67.5−70)∗366Z=\frac{(67.5-70)*\sqrt{36}}{6}Z=6(67.5−70)∗36
=-2.5
Φ(2.5)\Phi(2.5)Φ(2.5)
The value is 0.9938 from the z-tables. The answer is (2) 0.9938.
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