Question

The amount of cereal dispensed into "16-ounce" boxes of
*Captain Crisp* cereal was normally distributed with mean
16.18 ounces and standard deviation 0.25 ounces.

**a.** What proportion of boxes are "underfilled"?
That is, what is the probability that the amount dispensed into a
box is less than 16 ounces? Give your answer to 4 decimal
places.

**b.** Find the probability that exactly 3 out of 9
randomly and independently selected boxes of cereal contain less
than 16 ounces. Give your answer to **4 significant
figures**.

**c.** Find the probability that the sample mean
amount of cereal for a random sample of 9 boxes is less than 16
ounces. Give your answer to 4 decimal places.

**d.** Suppose that the machine can be adjusted to
change the mean while the standard deviation remains at 0.25
ounces. What must the mean be so that only 20% of all the boxes are
"underfilled"? Give your answer to 3 decimal places. (Mean in
ounces)

Showing work is much appreciated.

Answer #1

Let X is a random variable shows the amount of cereal. Here X has normal distribution with parameters as follows:

(a)

The z-score for X = 16 is

The probability that the amount dispensed into a box is less than 16 ounces is

P(X < 16) = P(z < -0.72) = 0.2358

b)

Let Y is a random variable shows the number of boxes of cereal contain less than 16 ounces out of 9. Here Y has binomial distribution with parameters n=9 and p=0.2358.

The probability that exactly 3 out of 9 randomly and independently selected boxes of cereal contain less than 16 ounces is

c)

The z-score for is

The probability that the sample mean amount of cereal for a random sample of 9 boxes is less than 16 ounces is

d)

Here we need z-score that has 0.20 area to its left. The z-score for -0.84 has 0.20 area to its left. The requried X is

Answer: 16.21 ounces

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