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Good readers and good spellers correctly read an average of [tex]93.8 \pm 2.9[/tex] (Mean ± Standard Deviation), whereas women scored an average of [tex]16.84 \pm 5.66[/tex] (Mean ± Standard Deviation) on this scale. Assuming these data are normally distributed, what percentage of participants correctly read at least 90 words in the good readers and good spellers group?

Answer :

Final answer:

Approximately 90.5% of the good readers and good spellers group read at least 90 words. This is calculated using the normal distribution, with a provided mean of 93.8 and standard deviation of 2.9, and determining the cumulative distribution function for a score of 90.

Explanation:

To determine what percentage of participants correctly read at least 90 words in the good readers and good spellers group, we need to use the concept of the normal distribution. We were provided with a mean (M) of 93.8 and a standard deviation (SD) of 2.9 for the group. To find the percentage who read at least 90 words, we need to calculate the z-score for a score of 90 and then consult a standard normal distribution table or use a calculator that provides the cumulative distribution function for the normal distribution.

The z-score is calculated by the formula:

z = (X - M) / SD

Where:

X = the score of interest (90 in this case),

M = the mean of the distribution (93.8),

SD = the standard deviation of the distribution (2.9).

Plugging in the numbers:

z = (90 - 93.8) / 2.9 = -1.31

A z-score of -1.31 corresponds to the left tail of the normal distribution. By looking at standard normal distribution tables or using a calculator, we find that approximately 90.5% of the scores fall above a z-score of -1.31. This means that approximately 90.5% of participants in the good readers and good spellers group read at least 90 words.

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