In: Statistics and Probability
A binomial probability distribution has p = 0.25 and n = 81.
A) What are the mean and standard deviation?
B) Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain.
C) What is the probability of exactly 28 successes?
D) What is the probability of 18 to 22 successes?
E)What is the probability of 24 or fewer successes?
Part a)
Mean = n * P = ( 81 * 0.25 ) = 20.25
Variance = n * P * Q = ( 81 * 0.25 * 0.75 ) = 15.1875
Standard deviation =
= 3.8971
Part b)
We can use Normal approximation to Binomial, in it satisfies the condition
np > 5 and nq > 5
np = 81 * 0.25 = 20.25 > 5
nq = 81 * ( 1 - 0.25 ) = 60.75 > 5
Part c)
P ( X = 28 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 28 - 0.5 < X < 28 +
0.5 ) = P ( 27.5 < X < 28.5 )
P ( 27.5 < X < 28.5 )
Standardizing the value
Z = ( 27.5 - 20.25 ) / 3.8971
Z = 1.86
Z = ( 28.5 - 20.25 ) / 3.8971
Z = 2.12
P ( 1.86 < Z < 2.12 )
P ( 27.5 < X < 28.5 ) = P ( Z < 2.12 ) - P ( Z < 1.86
)
P ( 27.5 < X < 28.5 ) = 0.9829 - 0.9686
P ( 27.5 < X < 28.5 ) = 0.0143
Part d)
P ( 18 <= X <= 22 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 18 - 0.5 < X < 22 +
0.5 ) = P ( 17.5 < X < 22.5 )
P ( 17.5 < X < 22.5 )
Standardizing the value
Z = ( 17.5 - 20.25 ) / 3.8971
Z = -0.71
Z = ( 22.5 - 20.25 ) / 3.8971
Z = 0.58
P ( -0.71 < Z < 0.58 )
P ( 17.5 < X < 22.5 ) = P ( Z < 0.58 ) - P ( Z < -0.71
)
P ( 17.5 < X < 22.5 ) = 0.7181 - 0.2402
P ( 17.5 < X < 22.5 ) = 0.4779
Part e)
P ( X <= 24 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 24 + 0.5 ) = P ( X < 24.5
)
P ( X < 24.5 )
Standardizing the value
Z = ( 24.5 - 20.25 ) / 3.8971
Z = 1.09
P ( X < 24.5 ) = P ( Z < 1.09 )
P ( X < 24.5 ) = 0.8621