Question

ndustry standards suggest that 14 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 10 Nissans yesterday. (Round your mean answer to 2 decimal places and the other answers to 4 decimal places.)

  

Compute the mean and standard deviation of this probability distribution.

Determine the probability that exactly two of these vehicles require warranty service.

What is the probability exactly one of these vehicles requires warranty service?

What is the probability that none of these vehicles requires warranty service?

Answer 1

Solution:

problem on binomal distribution

Here n=10

p=0.14

q=1-p=1-0.14=0.86

mean=np=10*0.14=1.4

standard deviation=sqrt(npq)=sqrt(10*0.14*0.86)=1.0973

MEAN=1.4

STD.DEVIATION=1.0973

Solution-b

Determine the probability that exactly two of these vehicles require warranty service.

P(X=2)

=10c2*0.14^2*0.86^10-2

= 0.2639102

0.2639

Solution-c

What is the probability exactly one of these vehicles requires warranty service?

P(X=1)

=10c1*0.14^1*0.86^10-1

=0.3602584

0.3603

Solution-d:

What is the probability that none of these vehicles requires warranty service?

P(X=0)

=10c0*0.14^0*0.86^10-0

=0.2213

ANSWER;

0.2213