Question

The mean number of students who went to examinations is 450 students with a standard deviation of 75 students. The distribution of the number of students is normal.

1. What number of students who correspond to the last 15%.
2. Fin the probability that less than 430 students want on campus examinations.
3. Calculate the probability that more than 400 students wanted on campus examinations
4. Determine the probability that between 452 and 463 students wanted on campus examinations

Answer 1

i)

X ~ N ( µ = 450 , σ = 75 )
P ( X < x ) = 15% = 0.15
To find the value of x
Looking for the probability 0.15 in standard normal table to calculate Z score = -1.0364
Z = ( X - µ ) / σ
-1.0364 = ( X - 450 ) / 75
X = 372.27
P ( X < 372.27 ) = 0.15

ii)

X ~ N ( µ = 450 , σ = 75 )
P ( X < 430 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 430 - 450 ) / 75
Z = -0.2667
P ( ( X - µ ) / σ ) < ( 430 - 450 ) / 75 )
P ( X < 430 ) = P ( Z < -0.2667 )
P ( X < 430 ) = 0.3949

iii)

X ~ N ( µ = 450 , σ = 75 )
P ( X > 400 ) = 1 - P ( X < 400 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 400 - 450 ) / 75
Z = -0.6667
P ( ( X - µ ) / σ ) > ( 400 - 450 ) / 75 )
P ( Z > -0.6667 )
P ( X > 400 ) = 1 - P ( Z < -0.6667 )
P ( X > 400 ) = 1 - 0.2525
P ( X > 400 ) = 0.7475

iv)

X ~ N ( µ = 450 , σ = 75 )
P ( 452 < X < 463 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 452 - 450 ) / 75
Z = 0.0267
Z = ( 463 - 450 ) / 75
Z = 0.1733
P ( 0.03 < Z < 0.17 )
P ( 452 < X < 463 ) = P ( Z < 0.17 ) - P ( Z < 0.03 )
P ( 452 < X < 463 ) = 0.5688 - 0.5107
P ( 452 < X < 463 ) = 0.0581