Question

A study on aggressive driving found that 50 % of drivers say that they have yelled at another driver. A random sample of 180 drivers in Halifax is selected and are asked whether they have yelled at other drivers.

a. What is the probability that at least 85 drivers will say ”yes”, they have yelled at another driver? [Hint: use normal approximation]

b. What is the probability that at most 86 drivers will say ”yes”, they have yelled at another driver? [Hint: use normal approximation]

c. Use the normal approximation to calculate the probability that there will be exactly 86 sampled drivers who will say yes. [Hint: use normal approximation]

Answer 1

n=

180

p=

0.5000

here mean of distribution=μ=np=		90.00
and standard deviation σ=sqrt(np(1-p))=		6.71
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

a)

probability =P(X>84.5)=P(Z>(84.5-90)/6.708)=P(Z>-0.82)=1-P(Z<-0.82)=1-0.2061=0.7939

b)

probability =P(X<86.5)=(Z<(86.5-90)/6.708)=P(Z<-0.52)=0.3015

c)

probability =P(85.5<X<86.5)=P((85.5-90)/6.708)<Z<(86.5-90)/6.708)=P(-0.67<Z<-0.52)=0.3015-0.2514=0.0501