Records maintained by the office of budget in a particular s

Records maintained by the office of budget in a particular state indicate that the amount of time elapsed between the submission of travel vouchers and the final reimbursement of funds has approximately a normal mean of 39 days and a population standard deviation of 12 days. A sample of 16 vouchers was pulled at random.

What is the standard error of the sampling distribution?

What is the probability that the sample mean will be less than 31 days for reimbursement?

What is the probability that the sample mean will indicate between 35 and 45 days for reimbursement?

What is the probability that the sample mean will indicate more than 46 days for reimbursement?

Solution

Records maintained by the office of budget in a particular state indicate that the amount of time elapsed between the submission of travel vouchers and the final reimbursement of funds has approximately a normal distribution.

Let X be a random variable reimbursement of funds.

Given that , X follows Normal distribution with mean (µ) 39 days and population standard deviation () of 12 days.

Also n=16.

a) What is the standard error of the sampling distribution?

We know that the standard error = / sqrt(n) (sqrt(n) in the square root of n)

Standard error (se) = 12 / sqrt(16) = 3

b) What is the probability that the sample mean will be less than 31 days for reimbursement?

In this problem we have to calculate probability that sample mean will be less than 31 days.

Means here we are deal with sample mean which is denoted by Xbar.

And we know that if X follows normal distribution with mean µ and standard deviation then Xbar also follows normal distribution with mean µ and standard deviation /sqrt(n).

Here we have to find P(Xbar <31) = P( (Xbar - µ) / se < ( 31 - µ ) /se ) (from a) )

= P ( Z < (31 - 39) / 3)

= P ( Z < -2.667)

This probability we can find by using EXCEL and command for that is NORMSDIST( z = -2.67 )

= 0.00379

c) What is the probability that the sample mean will indicate between 35 and 45 days for reimbursement?

Here we have to calculate P(35 < Xbar < 45)

P(35 < Xbar < 45) = P ( (35 - µ) / se < (Xbar - µ ) / se < (45 -µ ) / se )   

= P( (35 - 39) / 3 < Z < (45 - 39) / 3 )

= P ( -1.33 < Z < 2)

= P( Z < 2 ) - P( Z < -1.33)

= 0.9772 - 0.0918

= 0.8854

d) What is the probability that the sample mean will indicate more than 46 days for reimbursement?

That is we have to find the P( Xbar > 46 )

P( Xbar > 46 ) = P( (Xbar - µ) / se > ( 46 - µ ) / se )

= P( Z > ( 46 - 39) / 3)

= P( Z > 2.33 )

= 1 - P(Z 2.33)

= 1 - 0.9901

= 0.0099