The average outstanding bill for delinquent customer accounts for a national department store chain is $187.50 with standard deviation $54.50. In a simple random sample of 50 delinquent accounts, what is the probability that the mean outstanding bill is over $200?

P(x > 200) = 0.053

Step-by-step explanation:

Assuming a normal distribution for the outstanding bill for delinquent customer accounts for the national department store chain, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

x = outstanding bill for delinquent customer accounts.

µ = mean

σ = standard deviation

n = number of samples

From the information given,

µ = $187.50

σ = $54.50

n = 50

The probability that the mean outstanding bill is over $200 is expressed as

P(x > 200) = 1 - P(x ≤ 200)

For x = 200,

z = (200 - 187.5)/54.5/√50 = 1.62

Looking at the normal distribution table, the probability corresponding to the z score is 0.947

P(x > 200) = 1 - 0.947 = 0.053

f(x)=2/3x-8, which give g(x) the greatest value.

step-by-step explanation: