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

there is no points to select.

step-by-step explanation:

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629