To calculate the probability that a randomly picked box contains less than 490g of cereal, we'll assume that the weights of the cereal boxes are normally distributed. We can use the properties of the normal distribution to find this probability.

Step 1: Identify the Given Information

• Mean weight ($\mu$): 500g
• Variance ($\sigma^2$): 20g
• Standard deviation ($\sigma$): $\sqrt{20} \approx 4.47g$
• The value we are interested in ($X$): 490g

Step 2: Standardize the Value

To find the probability, we first need to standardize the value of 490g to a z-score using the formula: $z = \frac{X - \mu}{\sigma}$ Substituting the values: $z = \frac{490 - 500}{4.47} \approx \frac{-10}{4.47} \approx -2.24$

Step 3: Find the Probability

Now, we need to find the probability corresponding to the z-score of -2.24. This is the probability that a standard normal variable is less than -2.24.

Using standard normal distribution tables or a calculator, we find: $P(Z < -2.24) \approx 0.0125$

Step 4: Conclusion

The probability that a randomly picked box contains less than 490g of cereal is approximately 0.0125 or 1.25%.

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If you have any further questions or need clarification, let me know! Here are some related questions you might be interested in:

1. What is the probability that a box contains more than 510g of cereal?
2. How would the probability change if the variance increased to 30g?
3. What is the probability that a box contains between 495g and 505g of cereal?
4. How do you find the z-score for a different weight, such as 495g?
5. What assumptions are made when using the normal distribution in this context?

Tip: Remember that the z-score tells you how many standard deviations away a value is from the mean. It helps in finding the probability for any value in a normal distribution.