Question

It has been reported that last year 25% of drivers have fallen asleep at the wheel. Source: National Sleep Foundation If 100 drivers are selected at random, find the following probabilities: [4a] Find the probability that at most 27 drivers will say that they have fallen asleep while driving. [4b] Use [4a] to find the probability that exactly 27 drivers will say that they have fallen asleep while driving. [4c] Find the probability that less than 24 drivers will say that they have fallen asleep while driving. [4d] Use [4a, b, c] to find the probability that between 24 and 27 drivers will say that they have fallen asleep while driving.

Answer 1

here mean of distribution=μ=np=	25.00
and standard deviation σ=sqrt(np(1-p))=	4.33
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

a)

probability =P(X<27.5)=(Z<(27.5-25)/4.33)=P(Z<0.58)=0.7190

b)

probability =P(26.5<X<27.5)=P((26.5-25)/4.33)<Z<(27.5-25)/4.33)=P(0.35<Z<0.58)=0.719-0.6368=0.0822

c)

probability =P(X<23.5)=(Z<(23.5-25)/4.33)=P(Z<-0.35)=0.3632

d)

probability =P(23.5<X<27.5)=P((23.5-25)/4.33)<Z<(27.5-25)/4.33)=P(-0.35<Z<0.58)=0.719-0.3632=0.3558