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The average number of words in a romance novel is 64,182 and the standard deviation is 17,154 A. Find the proportion of all novels that are between 50,000 and 60,000 words. square disappointed B. The 90^th percentile for novels is [Select] square words. C. On a shelf with 350 novels, how many would be estimated to have more than 70,000 words? square

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The average number of words in a romance novel is 64,182 and the standard deviation is 17,154 A. Find the proportion of all novels that are between 50,000 and 60,000 words. square disappointed B. The 90^th percentile for novels is [Select] square words. C. On a shelf with 350 novels, how many would be estimated to have more than 70,000 words? square

The average number of words in a romance novel is 64,182 and the standard deviation is 17,154 A. Find the proportion of all novels that are between 50,000 and 60,000 words. square disappointed B. The 90^th percentile for novels is [Select] square words. C. On a shelf with 350 novels, how many would be estimated to have more than 70,000 words? square

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ImogenVeteran · Tutor for 10 years

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<p> <br />A. 0.1607 <br />B. 88,770 words<br />C. 44<b></b> novels</p>

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<p> This question involves understanding the usage and application of mathematical statistics, mainly knowledge of means, standard deviations, percentiles, and ideas relating to normal distribution. <br /><br />A. Given that the mean (average number of words) is 64,182 and the standard deviation is 17,154, z-scores are to be calculated next to find the proportion of novels with a word count between 50,000 and 60,000. The formula for a z score ("value - mean/standard deviation") gives values corresponding to 50,000 and 60,000 under the normal curve.<br /><br />B. The 90th percentile is equivalent to the Ogive percentage. We use that percentage value in the standard normal cumulative distribution function for the z-variable. Then we revert the z-variable for the cumulative(Q) back into x which represents original percentile in term of the word's amount (x=σz+µ).<br /><br />C. With a mean of 64,182 and a standard deviation of 17,154, developing a z-score with a value of 70,000 can return a probability. Multiplying this probability by 350 gives the estimated number of novels likely to contain more than 70,000 words. </p>
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