Question

The average monthly mortgage payment for all homeowners in a city is $2870. Suppose that the distribution of monthly mortgages paid by homeowners in this city follow an approximate normal distribution with a mean of $2870 and a standard deviation of $470. Find to 4 decimal places the probability that the monthly mortgage paid by a randomly selected homeowner from this city is

Answer 1

Answer:

The probability that the monthly mortgage is between 2300 and 3140 is: 0.23167

Step-by-step explanation:

Given

$\mu = 2870$ --- the average

$\sigma = 470$ --- the standard deviation

Required [Missing from the question]

Monthly mortgage is between 2300 and 3140

This is represented as:

$P(2300 < x < 3140)$

This is calculated as:

$P(a< x < b) = P(z_a < Z < z_b)$

$P(2300< x < 3140) = P(z_b < Z < z_a)$

Calculate the z scores

$x = 3140$

$z = \frac{3140 - 2850}{470}$

$z = \frac{290}{470}$

$z = 0.6170$

$x = 2300$

$z = \frac{3140 - 2300}{470}$

$z = \frac{840}{470}$

$z = 1.7872$

So, we have;

$P(2300< x < 3140) = P(1.1787 < Z < 0.6170)$

This is then calculated as:

$P(a < Z < b) = P(Z < a) - P(Z <b)$

$P(a < Z < b) = P(Z < 1.1782) - P(Z <0.6170)$

$P(2300< x < 3140) = P(Z < 1.1782) - P(Z <0.6170)$

Using the z table:

$P(Z<0.6170) =0.73138$

$P(Z<1.7872) =0.96305$

$P(2300< x < 3140) = 0.96305 - 0.73138$

$P(2300< x < 3140) = 0.23167$