This is a problem involving the normal distribution. We are asked to find the percentage of cereal packets that weigh more than the stated weight of 550 grams. Here's how to solve it step by step:

Step 1: Standardize the given value.

The problem states that the weights are normally distributed with a mean (μ) of 555 grams and a standard deviation (σ) of 5 grams.

We are looking for the probability that the weight of a packet is greater than 550 grams. To do this, we will first compute the z-score for 550 grams using the formula:

$z = \frac{x - \mu}{\sigma}$

Where:

• $x = 550$ (the weight we are comparing to),
• $\mu = 555$ (mean weight),
• $\sigma = 5$ (standard deviation).

$z = \frac{550 - 555}{5} = \frac{-5}{5} = -1$

Step 2: Find the probability corresponding to this z-score.

Using standard normal distribution tables or a calculator, we look up the cumulative probability for a z-score of -1. The table tells us that the cumulative probability up to z = -1 is approximately 0.1587.

This means that 15.87% of packets weigh less than 550 grams. Therefore, the percentage of packets that weigh more than 550 grams is:

$1 - 0.1587 = 0.8413 \text{ or } 84.13\%$

Conclusion:

The percentage of cereal packets that contain more than the stated weight of 550 grams is 84.13%.

The correct answer is:

b. 84.13%

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Do you want more details or have any questions about this problem?

Here are some related questions to explore:

1. What is the probability that a packet weighs less than 545 grams?
2. What weight corresponds to the 90th percentile of the distribution?
3. If the manufacturer wants 95% of packets to weigh at least 550 grams, what should the mean be?
4. What percentage of packets weigh between 550 and 560 grams?
5. How would the probability change if the standard deviation were 10 grams instead of 5 grams?

Tip: The z-score is a useful tool to compare individual data points to a normal distribution. It tells you how many standard deviations a value is from the mean.