Question

Suppose Canadian home-owners owe an average of $188,000 on their mortgages. Assume that mortgage debt is normally distributed in Canada with a standard deviation of $100,000.

Standard Normal Distribution Table

a. Albertans are reported to owe $242,000 in mortgage debt, much higher than the Canadian average. What is the probability of randomly selecting a Canadian with mortgage debt that exceeds $242,000?

Round to four decimal places if necessary

b. What is the probability of randomly selecting a Canadian with mortgage debt below $99,000?

Round to four decimal places if necessary

c. Determine the minimum mortgage debt owing by the 20% of Canadians with the largest mortgages.

Round to the nearest dollar

Answer 1

Solution:

Given in the question

Average of canadian home owners owe on their mortgeges ()= $188000

Standard deviation () = $100000

Assume that mortgege debt is normally distributed

Solution(a)

We need to find probability of randomly selected a Canadian with mortgege debt that exceeds $242000 or P(X>242000) = ?

Here we will use normal distribution table so we will calculate Z-Score which can be calculated as

Z-score = (X-)/ = (242000-188000)/100000 = 0.54

From Normal distribution table, we found probability

P(X>242000) = 1- P(X<=242000) = 1- 0.7054 = 0.2946

So there is 29.46% probability that a randomly selected a Canadian with mortgege debt that exceeds $242000.

Solution(b)

We need to calculate that a randomly selected a Canadian with mortgege debt below $99000 or P(X<$99000)

Z = (99000-188000)/100000 = -0.89

P-value can be found from Z table

P(X<$99000) = 0.1867

So there is 18.67% probability that a randomly selected Canadian with mortgege debt below $99000.

Solution(c)

Here given in the question

P-value = 0.80

Z-score can be found from Z table

Z-score = 0.8416

So minimum mortgege debt can be calculated as

X = + zScore * = 188000 +0.8416*100000 = $272160

So minimum mortgege debt = $272160