Answer: 10.703%
Step-by-step explanation:
Let minimum height of the tallest 25% of young women be M.
Let Q be the random variable which denotes the height of young women.
Therefore, Q – N(64,2.70)
Now, P(Q˂M) = 1-0.25
i.e. P[(Q-64)/2.7 ˂ (M-64)/2.7] = 0.75
I.e. ф-1 [(M-64)/2.7] = 0.75 i.e. (M-64)/2.7 = ф-1 (0.75) = 0.675 i.e. M = 65.8198 inches
Let R be the random variable denoting the height of young men
Therefore, R – N (69.3, 2.8)
i.e. (R-69.3)/2.8 – N(0,1)
therefore the probability required = P(R ˂65.8198) = P[(R-69.3)/2.8 ˂ (65.8198 – 69.3)/2.8]
this gives P[(R-69.3)/2.8 ˂] = ф (-1.2429) = 0.107033
From this, the percentage of young men shorter than the shortest amongst the tallest 25% of young women is 10.703%
