Question

Based on past experience, a bank believes that 9.3 % of the people who receive loans will not make payments on time. The bank has recently approved 240 loans. What must be true to be able to approximate the sampling distribution with a normal model? Before proceeding, think about whether the conditions have been met. What are the mean and standard deviation of the sampling distribution of the proportion of people who will not make payments on time in samples of 240? mean, as a decimal =_______

standard deviation (accurate to 3 decimal places) =_______

What is the probability that over 10% of the clients in the group of 240 clients will not make timely payments? (accurate to 3 decimal places) (Remember: do not use rounded results in later calculations; rather use the value from prior-to-rounding.)________

Answer 1

It is given that sample size n = 240 and proportion = 9.3/100 = 0.093

We know that before finding the mean and standard deviation value, we have to check for the following two conditions

and

putting n = 240 and p = 0.093, we get

and

So, both the conditions are satisfied.

Using the formula for mean and standard deviation, we can write

Mean = n*p = 240*0.093 = 22.32

And

We have to find the probability that over 10% of the clients in the group of 240 clients will not make timely payment

so, 10% of 240 = (10/100)*240 = 0.1*240 = 24

Using the formula

setting the values, we get

this gives us

Using the identity

we can write it as

using z distribution, check 0.3 in the left most column and 0.07 in the top row, select the intersection cell. we get

rounding to 3 decimals, we get

Probability over 10% = 0.356