To find the probability that a randomly selected newborn baby boy weighs less than 4362 grams, given that the weights follow a normal distribution with a mean and variance, we'll use the Z-score formula and standard normal distribution.

Given dataTo find the probability that a randomly selected newborn baby boy weighs less than 4362 grams, given that the weights follow a normal distribution with a mean and variance, we'll use the Z-score formula and standard normal distribution.

Given

• Mean weight ($\mu$) = 3284 grams
• Variance ($\sigma^2$) = 454,276
• Standard deviation ($\sigma$) = $\sqrt{454,276} = 673.99$ grams (approx.)
• Weight to compare = 4362 grams

Step 1: Calculate the Z-score

The Z-score formula is: $Z = \frac{X - \mu}{\sigma}$ Where:

• $X$ is the value we want to compare (4362 grams)
• $\mu$ is the mean
• $\sigma$ is the standard deviation

Substitute the values: $Z = \frac{4362 - 3284}{673.99} = \frac{1078}{673.99} \approx 1.598$

Step 2: Find the probability

Now we need to find the probability corresponding to $Z = 1.598$ using standard normal distribution tables or a calculator.

The cumulative probability for $Z = 1.598$ is approximately 0.9449.

Conclusion:

The probability that a randomly selected newborn baby boy weighs less than 4362 grams is approximately 0.9449.

Would you like more details or have any questions?

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Here are 5 related questions to explore:

1. What is the probability that a newborn weighs more than 4362 grams?
2. How would the probability change if the mean weight were 3300 grams?
3. What is the probability of a baby weighing between 3000 and 4000 grams?
4. If the standard deviation were larger, how would that affect the probability?
5. How would you calculate the probability if the distribution wasn't normal?

Tip: For normal distributions, Z-scores help compare any value to the mean in terms of standard deviations.