SOLUTION:
8. An accountant is studying the time it takes a company to make payments on accounts payable. The mean time it takes to pay off an invoice is 13.8 days with a standard deviat
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Question 971479:
8. An accountant is studying the time it takes a company to make payments on accounts payable. The mean time it takes to pay off an invoice is 13.8 days with a standard deviation of 4.3 days, and the times are normally distributed.
(a) What is the probability that an account is paid off in less than 16 days?
(b) Find the 95th percentile for the payoff time- that is, a time for which 95% of the
accounts are paid off in less than that time.
(c) The company has just received 29 invoices. Find the probability that the mean payoff
time for these invoices is more than 16 days.
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
mean=13.8 d SD=4.3 d
z< (16-13.8)/4.3 =(2.2/4.3)
z<0.51
p=0.695
95% from the left is z=1.645
z* sd= 1.645 *4.3=7.07 days
13.8+7.07=20.9 days
z>(16-13.8)/(4.3/sqrt(29))
z>2.2/.798 =2.755
p=0.003
One invoice can certainly be more than 16 days but the MEAN of 16 of them is far less likely.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
8. An accountant is studying the time it takes a company to make payments on accounts payable. The mean time it takes to pay off an invoice is 13.8 days with a standard deviation of 4.3 days, and the times are normally distributed.
(a) What is the probability that an account is paid off in less than 16 days?
z(16) = (16-13.8)/4.3 = 2.2/4.3 = 0.5116
P(x < 16) = P(z<0.5116) = normalcdf(-100,0.5116) = 0.6955
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(b) Find the 95th percentile for the payoff time- that is, a time for which 95% of the accounts are paid off in less than that time.
Find the z-value with a right-tail of 0.95
invnorm(0.95) = 1.645
Find the corresponding time value::
x = 1.645*4.3+13.8 = 20.87
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(c) The company has just received 29 invoices. Find the probability that the mean payoff time for these invoices is more than 16 days.
t(16) = (16-13.8)/[4.3/sqrt(29)] = 2.7552
P(x > 16) = P(t> 2.7552 when df = 28) = tcdf(2.7552,100,28) = 0.005
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Cheers,
Stan H.
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