Answer :
Final answer:
To be in the top 84% of a normal distribution, a child's vocabulary needs to be about one standard deviation above the mean. In this case, with a mean of 3000 and a standard deviation of 500 words, this amounts to a minimum of 3500 words.
Explanation:
To find out the minimum number of words a child must know to be in the top 84%, we need to know the z-score (standard deviation from the mean) corresponding to that percentile. For a normally distributed dataset, the 84th percentile corresponds to a z-score of approximately 1. This means that to be in the top 84% of a normal distribution, a child's vocabulary needs to be about one standard deviation above the mean.
In this case, we are given a mean of 3000 and a variance of 250000. With variance being the square of the standard deviation, the standard deviation is the square root of the variance, i.e., sqrt(250000) = 500 words. These 500 words represent one standard deviation.
Therefore, one standard deviation above the mean (3000 + 500) would be 3500 words. So a child needs to know a minimum of 3500 words to be in the top 84%.
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