Question

The First National Mortgage Company has noted that 9% of its customers pay their mortgage payments after the due date. Upload a file detailing all of the work needed to answer the questions. Assume the data is normally distributed. a. b. Given a random sample of 150 customers, calculate the mean of the sample. (1 point) Given a random sample of 150 customers, calculate the standard deviation of the sample. (1 point) What is the probability that in a random sample of 150 customers less than 20 will be late on their payments? (2 points) What is the probability that in a random sample of 150 customers at least 10 will be late on their payments? (2 points) c. d.

Answer 1

P( customers pay their mortgage payments after the due dat) = 0.09 = p

a) Sample of 150,n=150

, Mean = np = 150 * 0.09 = 13.5

b) Standard deviation =

Vnp(1 - p)

= $\sqrt{150*0.09(1-0.09)}$

= 3.5049

c) Let X be the number of customers who will be late on their payments

X~ Binomial ( 150, 0.09)

Now, since the sample is large we will use normal approximation

X~ Normal ( 13.5, 3.5049)

P( X < 20) = P( X < 19.5 ) : Using the continuity correction

= P( $\frac{X-\mu}{\sigma}$ < $\frac{19.5-13.5}{3.5049}$ )

= P( z < 1.71)

= 0.9564

d) P( X >= 10) = P( X > 9.5) ; Using the continuity correction

= P(  $\frac{X-\mu}{\sigma}$ > $\frac{9.5-13.5}{3.5049}$ )

= P( z > -1.14)

= P( z < 1.14)

= 0.8729