Question

"Suppose that past records indicate that the probability is 0.04 that a new car will need repair in the first 90 days of use. What is the probability that, for the next 500 cars sold, exactly 16 cars need a repair?"

-Please show Excel commands

Answer 1

n=

500

p=

0.0400

here mean of distribution=μ=np=		20.00
and standard deviation σ=sqrt(np(1-p))=		4.3818

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

therefore from normal approximation of binomial distribution and continuity correction:

probability that, for the next 500 cars sold, exactly 16 cars need a repair

=P(15.5<X<16.5) =norm.dist(16.5,20,4.3818,true)-norm.dist(15.5,20,4.3818,true)

=0.0600

(Excel command is given in bold letters, please reply if any clarification is required for above)