Aloan company knows that 5% of its loans will be delinquent. of the company’s 400 loan accounts, what is the probability

a) that exactly 25 accounts will be delinquent?
b) that fewer than 30 accounts will be delinquent?
c) more than 24 accounts will be delinquent?

Step-by-step explanation:

Given that a loan company knows that 5% of its loans will be delinquent.

since each loan is independent of the other p , probability for any random loan to be delinquent is constant 0.05

X no of delinquent loan accounts is binomial with n =400 and p = 0.05

Since n is very large and also np = 20 and nq >5 we can approximate to normal

Mean = np = 20 :  Variance = npq = 19

Std dev = 4.36

X is N(1, 4.36)

With continuity correcton we calculate the prob

a) that exactly 25 accounts will be delinquent?

=

b) that fewer than 30 accounts will be delinquent?

=P(X<29.5)

= 0.9854

c) more than 24 accounts will be delinquent

=P(X>24,5)

=0.1509

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .