Normal Distribution: Credit card balances, consisting of mean = $2870, standard deviation = $900.

Question

Normal Distribution question:

Credit card balances, consisting of:

Mean = $2870, standard deviation = $900.

25 credit card holders are chosen at random.

What is the probability that the mean balance on the credit card is under $2500 ?

Answer 1

Let us pull out the critical elements

Population: Normally distributed with mean of 2870

population standard deviation = 900

standard deviation of the sample mean (standard error) is 900 / √25. or 180

(note: as population sigma is known, we do not have to apply any corrections)

Q) P(x_bar < $2500) = ?

Since μ and σ are known (population mean and standard deviation) we can use a simple Z test.

Z_test = (2500-2870 )/ 180 or -370/180 = -2.055

As this is very close to -2 one can use the rule of thumb for probability within ± 2 sigma being 95.44% to get a close estimate. If the area between -2 and 2 sigma = .9544 then the area outside = .0456. Half of this (the amount under the curve LESS THAN -2) would be .0228

Using a calculator or computer for an exact answer we find (using a TI-83/86):

nmcdf(-9E6,2500,2870,180) = .0199