Monkeyseat

03-05-2008, 11:47 AM

Hi,

Question:

5) A food processor produces large batches of jars of pickles. In each batch, the gross weight of a jar is known to be normally distributed with standard deviation 7.5g. (The gross weight is the weight of the jar plus the weight of the pickles). The gross weights, in grams, of a random sample from a particular batch were:

514, 485, 501, 486, 502, 496, 509, 491, 497,

501, 506, 486, 498, 490, 484, 494, 501, 506,

490, 487, 507, 496, 505, 498, 499

a) Calculate a 90% confidence interval for the mean gross weight of this batch.

---

The weight of an empty jar is known to be exactly 40g.

b)

i) What is the standard deviation of the weight of the pickles in a batch of jars?

ii) Assuming that the mean gross weight is at the upper limit of the confidence interval calculated in (a), calculate limits within which 99% of the weights of the pickles would lie.

Working:

a) I figured the mean out as 497.16g.

Mean +- z value * (standard deviation/sqr. n)

497.16 +- 1.6449 * (7.5/sqr.25)

Therefore: 494.69g to 499.63g

b)

i) It would be the same, 7.5g.

ii) Okay this is the bit I'm stuck on.

So I used the upper limit calculated in (a):

499.63 +- 2.5758 * 7.5

This gave 480.31g to 518.95g. The book says the answer is 440.3g - 478.9g... Where have I gone wrong?

Just (bii) I'm wondering about, the other bits I checked and were correct. Thanks.

EDIT:

I just thought, is question (bii) asking for the weights of the pickles excluding the jar. So would you take 40g off? If so, take it off 499.63 or the final answer, or is either correct? Cheers. :D

Question:

5) A food processor produces large batches of jars of pickles. In each batch, the gross weight of a jar is known to be normally distributed with standard deviation 7.5g. (The gross weight is the weight of the jar plus the weight of the pickles). The gross weights, in grams, of a random sample from a particular batch were:

514, 485, 501, 486, 502, 496, 509, 491, 497,

501, 506, 486, 498, 490, 484, 494, 501, 506,

490, 487, 507, 496, 505, 498, 499

a) Calculate a 90% confidence interval for the mean gross weight of this batch.

---

The weight of an empty jar is known to be exactly 40g.

b)

i) What is the standard deviation of the weight of the pickles in a batch of jars?

ii) Assuming that the mean gross weight is at the upper limit of the confidence interval calculated in (a), calculate limits within which 99% of the weights of the pickles would lie.

Working:

a) I figured the mean out as 497.16g.

Mean +- z value * (standard deviation/sqr. n)

497.16 +- 1.6449 * (7.5/sqr.25)

Therefore: 494.69g to 499.63g

b)

i) It would be the same, 7.5g.

ii) Okay this is the bit I'm stuck on.

So I used the upper limit calculated in (a):

499.63 +- 2.5758 * 7.5

This gave 480.31g to 518.95g. The book says the answer is 440.3g - 478.9g... Where have I gone wrong?

Just (bii) I'm wondering about, the other bits I checked and were correct. Thanks.

EDIT:

I just thought, is question (bii) asking for the weights of the pickles excluding the jar. So would you take 40g off? If so, take it off 499.63 or the final answer, or is either correct? Cheers. :D