Question

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?

Answer 1

Answer:

Probability that the mean outstanding bill is over $200 is 0.0526.

Step-by-step explanation:

We are given that the average outstanding bill for delinquent customer accounts for a national department store chain is $187.50 with standard deviation $54.50.

Also, a random sample of 50 delinquent accounts is selected.

Firstly, Let $\bar X$ = mean outstanding bill

The z score probability distribution for is given by;

          Z = $\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = average outstanding bill = $187.50

            $\sigma$ = standard deviation = $54.50

            n = sample of delinquent accounts = 50

Probability that the mean outstanding bill is over $200 is given by = P($\bar X$ > $200)

  P($\bar X$ > 200) = P( $\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{ 200-187.50}{\frac{54.50}{\sqrt{50} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

                                                         = 1 - 0.94738 = 0.0526

Therefore, probability that the mean outstanding bill is over $200 is 0.0526.

Answer 2

Answer: 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