Finding contract specification for towel length with normal distribution

[cyt91]

Homework Statement

A sales contract between a manufacturer and a buyer requires 20 towels to be subjected to a water absorption test. If no more than 2 towels fails the test, the batch is accepted. If 2 or 3 towels fail the test, an additional 20 towels are tested. The batch is then accepted if 3 or less (out of 40) fail the test. Otherwise, the batch is rejected.

(a) If a batch of towels contains 5 % which would be rejected by the test,what is the probability that the batch is accepted.

(b) The lengths of 20 towels are measured and if the mean length is less than a value a specified in the contract, the batch is rejected. What should the value of a be to give a probability of 0.99 of accepting a batch with mean length of 106 mm and standard deviation of 6 mm.

I solved (a). The probability that the batch is accepted is 0.8961.

But I can't solve (b).

Homework Equations

The Attempt at a Solution

Here is my attempt on (b) :

Let X:length of a towel.
Y:total length of 20 towels

Y~(2120,720)

P( mean of Y >a ) = 0.99

P ( Y >20a ) = 0.99

P( Z > [20a-2120]/sqrt(720) ) = -2.326

[20a-2120]/sqrt(720) = 0.01

a = 102.9 mm

But the answer provided is 104.3 mm.

Can anyone help me with (b)? Thanks!

 

[CompuChip]

Where do you get 720 from?
It looks like you multiplied the standard deviation of a single towel by the square of the batch size, rather than the square root?

 

[cyt91]

Where do you get 720 from?
It looks like you multiplied the standard deviation of a single towel by the square of the batch size, rather than the square root?

6 x 6 x 20 = 720

I'm not sure if my approach is correct.

 

[CompuChip]

But why is your standard deviation $6^2 \times 20$? Shouldn't it be $(\sqrt{6}) \times 20$?

 

[cyt91]

720 is the variance.
In the calculation of z score, I used sqrt(720) to get standard deviation.