The average credit card debt for college seniors is $22,199 with a standard deviation of $5300. What is the probability that a sample of 30 seniors owes a mean of more than $20,200? Round answer to 4 decimal places.

Step-by-step explanation:

The number of samples is large(greater than or equal to 30). According to the central limit theorem, as the sample size increases, the distribution tends towards normal. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 22199

σ = 5300

n = 30

the probability that a senior owes a mean of more than $20,200 is expressed as

P(x > 20200)

Where x is a random variable representing the average credit card debt for college seniors.

For n = 30,

z = (20200 - 22199)/(5300/√30) =

- 2.07

Looking at the normal distribution table, the probability corresponding to the z score is 0.0197

P(x > 20200) = 0.0197

because they could drop even lower..

oof. u are asking for t much