Question

QUESTION PART A: A manufacturer knows that their items have a lengths that are skewed right, with a mean of 12.6 inches, and standard deviation of 2.6 inches.

If 42 items are chosen at random, what is the probability that their mean length is greater than 12.5 inches?

(Round answer to four decimal places)

QUESTION PART B: A particular fruit's weights are normally distributed, with a mean of 679 grams and a standard deviation of 19 grams.

If you pick 13 fruits at random, then 5% of the time, their mean weight will be greater than how many grams?

Give your answer to the nearest gram.

Answer 1

Answer A

We know that the paretn distribution is not normal (skewed right).

Mean of parent distribution= 12.6 inches

Standard deviation of parent distribution= 2.6 inches

Sample size, n= 42

Since the sample size is sufficiently large, from the Central Limit Theorem, we know that the sampling distribution of the sample means will be approximately normal.

Thus, P(sample mean > 12.5)= 1- P(sample mean < 12.5)

= 1- P(Z < (12.5-12.6) / 2.6/sqrt(42)

= 1- P(Z < -0.1/0.4011887)

= 1- P(Z < -0.2492593)

= 1- P(Z < -0.25)

= 0.59871 ~ 0.5987

Answer B

We know that the paretn distribution is normal (not skewed).

Mean of parent distribution= 679 grams

Standard deviation of parent distribution= 19 grams

We know that the parent distribution is normal, hence the sampe size will not impact the shape of the sampling distribution.

Thus, we know that

0.95= (X-679) / 19/sqrt(13)

0.95*5.269652= X-679

X= 679+5.006169

~ 684 grams