Respuesta :
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 [tex]\bar X[/tex] = mean outstanding bill
The z score probability distribution for is given by;
     Z = [tex]\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = average outstanding bill = $187.50
       [tex]\sigma[/tex] = standard deviation = $54.50
      n = sample of delinquent accounts = 50
Probability that the mean outstanding bill is over $200 is given by = P([tex]\bar X[/tex] > $200)
  P([tex]\bar X[/tex] > 200) = P( [tex]\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{ 200-187.50}{\frac{54.50}{\sqrt{50} } }[/tex] ) = P(Z > 1.62) = 1 - P(Z [tex]\leq[/tex] 1.62)
                             = 1 - 0.94738 = 0.0526
Therefore, probability that the mean outstanding bill is over $200 is 0.0526.
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
