Answer:
It is given that

for
we have

using the standard normal distribution table we have area above
equals

b)
Similarly for scores less than 1150 we have

using the standard normal distribution table we have area below
equals

c)
The score that would put the applicant in top 10% shall correspond score whose value gives 90% area of the normal distribution graph
For area of 90% we have Z=1.281
Thus we have

d)
The score that would put the applicant in bottom 25% shall correspond score whose value gives 25% area of the normal distribution graph
For area of 25% we have Z=-0.674
Thus we have
