Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the value of the third quartile q3. round your answer to the nearest tenth.

we need to find α=Q3 such that P(z≤α)=0.75
From the standard table, we get
P(Z≤0.675)=0.75
thus
α=0.675
⇒[H-63.6]/2.5=0.675
solving for H we get
H-63.6=1.6875
hence
H=65.2875
The 3rd quartile is 65.2875