Question

Suppose a local health clinic sees 40 cases. The exact binomial probability that 15 or more cases are resistant to any antibiotic is 0.193 .

(a) What is the Normal approximation to ?(?≥15) ? (Enter your answer to four decimal places.)

P(X≥15)= _____

It can be observed that the Normal approximation using continuity correction gives a much closer value to the true binomial probability than the Normal approximation.

(b) What is the Normal approximation using the continuity correction? (Enter your answer to four decimal places.)

?(?≥14.5)=_____

Answer 1

NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 40 * 0.193 = 7.72
standard deviation ( √npq )= √40*0.193*0.807 = 2.496
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
the Normal approximation to ?(?≥15)
P(X < 15) = (15-7.72)/2.496
= 7.28/2.496= 2.9167
= P ( Z <2.9167) From Standard Normal Table
= 0.9982
P(X > = 15) = (1 - P(X < 15))
= 1 - 0.9982 = 0.0018
with binomial value approximate
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 40 * 0.193
= 7.72
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 40 * 0.193 * 0.807
= 6.23
III.
standard deviation = sqrt( variance ) = sqrt(6.23)
=2.496

P( X < 15) = P(X=14) + P(X=13) + P(X=12) + P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5)
= ( 40 14 ) * 0.193^14 * ( 1- 0.193 ) ^26 + ( 40 13 ) * 0.193^13 * ( 1- 0.193 ) ^27 + ( 40 12 ) * 0.193^12 * ( 1- 0.193 ) ^28 + ( 40 11 ) * 0.193^11 * ( 1- 0.193 ) ^29 + ( 40 10 ) * 0.193^10 * ( 1- 0.193 ) ^30 + ( 40 9 ) * 0.193^9 * ( 1- 0.193 ) ^31 + ( 40 8 ) * 0.193^8 * ( 1- 0.193 ) ^32 + ( 40 7 ) * 0.193^7 * ( 1- 0.193 ) ^33 + ( 40 6 ) * 0.193^6 * ( 1- 0.193 ) ^34 + ( 40 5 ) * 0.193^5 * ( 1- 0.193 ) ^35
= 0.9944
P( X > = 15 ) = 1 - P( X < 15) = 0.0056

b.
Continuity Correction. A continuity correction may be applied
when a continuous distribution is used to approximate a discrete distribution.
It is used in order to improve the approximation.
Probably the most straight forward example involves approximating the binomial distribution
with the normal distribution.

?(?≥14.5)
P(X < 14.5) = (14.5-7.72)/2.496
= 6.78/2.496= 2.7163
= P ( Z <2.7163) From Standard Normal Table
= 0.9967
P(X > = 14.5) = (1 - P(X < 14.5))
= 1 - 0.9967 = 0.0033