The number of words in the active vocabulary of children of a particular age is normally distributedwith a mean of 3000 and a standard deviation of 500.
(a)Using the normal curve approximation rule (50%‐34%‐14%), how many words would a child ofthis age have to know to be in the top 2%, the top 16%, and the top 98%?
(b)Illustrate your answers with sketches.
(c)Explain your answers to a person who is familiar with mean, standard deviation, and Z scores,but not with normal curve percentages
(a) To find the number of words a child would need to know to be in the top 2%, top 16%, and top 98%, we can use Z-scores and the normal curve approximation rule.
For the top 2%:
Using the rule, we know that 2% of the data falls beyond 2 standard deviations above the mean. So we need to find the Z-score that corresponds to the top 2%. Using a Z-score table or a calculator, we find that the Z-score for the top 2% is approximately 2.05. We can then convert this Z-score to the number of words by using the formula:
Number of words = Z-score * standard deviation + mean
Number of words = 2.05 * 500 + 3000
Number of words = 4050
So, a child would need to know around 4050 words to be in the top 2%.
For the top 16%:
Similarly, we can find the Z-score that corresponds to the top 16%, which is approximately 1. If we use the same formula as above, we get:
Number of words = 1 * 500 + 3000
Number of words = 3500
Therefore, a child would need to know around 3500 words to be in the top 16%.
For the top 98%:
The Z-score corresponding to the top 98% is approximately 2.33. Using the formula:
Number of words = 2.33 * 500 + 3000
Number of words = 4165
So, a child would need to know around 4165 words to be in the top 98%.
(b) Below are sketches illustrating the distribution of the number of words in children's vocabularies:
[Sketches here]
(c) To explain these findings to someone familiar with mean, standard deviation, and Z-scores, you can say that the mean number of words is 3000 and the standard deviation is 500. This means that the majority of children (68%) fall within one standard deviation of the mean, which is between 2500 and 3500 words, according to the normal curve. To be in the top 2% of children, a child would need to know 4050 words, which is 2 standard deviations above the mean. Similarly, to be in the top 16%, a child would need to know 3500 words, which is 1 standard deviation above the mean. Finally, to be in the top 98%, a child would need to know 4165 words, which is 2.33 standard deviations above the mean.