Question

1. The technology underlying hip replacements has changed as these operations become more popular. Still, for many patients the increased durability has been counterbalanced by an increased incidence of squeaking. Suppose that the probability of a hip squeaking is 0.4. A random sample of 35 people will be taken. Let X = # of subjects whose hips developed squeaking.

(a) Determine the exact and approximate distributions we could use for X.

(b) Estimate P(X ≤ 10). Compare your answer with and without the continuity correction. (You should always use the continuity correction.)

Answer 1

a)

here exact distribution is binomial with parameter n=35 and p=0.4

approximate distributions is normal with parameter =np=35*0.4 =14 and σ=sqrt(np(1-p))=2.8983

b)

for normal distribution z score =(X-μ)/σ

without continuity correction:

probability =

P(X<10)

=

P(Z<-1.38)=

0.0838

with continuity correction:

probability =

P(X<10.5)

=

P(Z<-1.21)=

0.1131

with  continuity correction , probability is higher and closer to exact probability.