Answer to Question #185813 in Statistics and Probability for Betty

Question

The home loan department of BRAC Bank Limited sanction a significant number of loans per month. In this month the amount of money requested on home loan applications at a Bank follow a normal distribution with a mean of Tk. 73 lacs and a standard deviation of Tk. 22 lacs. A loan application is received this morning. Find the probability that:

a. The amount requested is Tk. 75 lacs or more?

b. The amount requested is Tk. 45 lacs or less?

c. The amount requested is between Tk. 55 lacs and Tk. 90 lacs?

d. The amount requested is exactly Tk. 65 lacs?

e. The amount requested that would be in the lowest 8%?

f. Also, what is the the median and mode amount?

Answer 1

Solution:

All amount of money given in the question is in lacs. So, we just remove lacs from each amount and use the remaining number.

Given, $\mu=73, \sigma=22$

$X\sim N(\mu,\sigma)$

(a): $P(X\ge75)=1-P(X<75)$

$=1-P(z<\dfrac{75-73}{22})=1-P(z<0.09)=1-0.53586=0.46414$

(b): $P(X\le45)=P(z\le \dfrac{45-73}{22})=P(z\le-1.27)=0.10204$

(c): $P(55\le X\le90)=P(\dfrac{55-73}{22}\le z\le \dfrac{90-73}{22})$

$=P(-0.82\le z\le0.78)=P(z\le0.78)-P(z\le-0.82)$

$=P(z\le0.78)-[1-P(z\le0.82)]=0.78230-[1-0.79389]=0.57619$

(d): $P(X=65)=P(z= \dfrac{65-73}{22})=P(z=-0.36)=0.35942$

(e): $P(X=x)=P(z= \dfrac{x-73}{22})=0.08$

$\Rightarrow \dfrac{x-73}{22}=-1.4$

$\Rightarrow x=42.4$

which is 42 lacs 40 thousands.

(f): Mean = median = mode for normal distribution, so median = mode = 73 lacs