Question

1. Among students at a particular university, 90% have visited the main dining hall at least once. What is the probability that of 100 randomly selected students, between 90-93 of them will have visited the main dining hall at least once?

1. Answer this question using the binomial distribution.
2. Approximate this answer using the normal distribution without the continuity correction.
3. Approximate this answer using the normal distribution and the continuity correction. (Take note of how close your approximation is.)

Answer 1

using Normal approximation to Binomial

Mean = n * P = ( 100 * 0.9 ) = 90
Variance = n * P * Q = ( 100 * 0.9 * 0.1 ) = 9
Standard deviation = = 3

P ( 90 <= X <= 93 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 90 - 0.5 < X < 93 + 0.5 ) = P ( 89.5 < X < 93.5 )

P ( 89.5 < X < 93.5 )
Standardizing the value

Z = ( 89.5 - 90 ) / 3
Z = -0.17
Z = ( 93.5 - 90 ) / 3
Z = 1.17
P ( -0.17 < Z < 1.17 )
P ( 89.5 < X < 93.5 ) = P ( Z < 1.17 ) - P ( Z < -0.17 )
P ( 89.5 < X < 93.5 ) = 0.8783 - 0.4338
P ( 89.5 < X < 93.5 ) = 0.4445