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StClowers
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A survey found that women's heights are normally distributed with mean 62.8 inches and standard deviation of 2.5 inches. This survey also found that men's heights are normally distributed with a mean 67.7 and a standard deviation 2.8.
A. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in and maximum of 6 ft 2 in. Find the percentage of women meeting height requirements. __%
B. Find the percentage of men meeting the height requirements. __%
C. If height requirements are changed to exclude only the tallest 5% of men & the shortest 5%, what are the height requirements?
The new height requirements are at least __in and at most__in.
 7 months ago
 7 months ago
StClowers Group Title
A survey found that women's heights are normally distributed with mean 62.8 inches and standard deviation of 2.5 inches. This survey also found that men's heights are normally distributed with a mean 67.7 and a standard deviation 2.8. A. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in and maximum of 6 ft 2 in. Find the percentage of women meeting height requirements. __% B. Find the percentage of men meeting the height requirements. __% C. If height requirements are changed to exclude only the tallest 5% of men & the shortest 5%, what are the height requirements? The new height requirements are at least __in and at most__in.
 7 months ago
 7 months ago

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ybarrap Group TitleBest ResponseYou've already chosen the best response.0
You might want to create a zscore for each: $$ z=\cfrac{x\mu}{\sigma} $$ x = 4ft 9in in and 6ft 2 in Then find the pvalue associated with each and subtract one from the other to find the percentage of the populations that are between. An alternative approach would be to use the standard normal distribution to compute $$ P(\text{4ft 9in} \le X \ \le \text{6ft 2in}) $$ You will need the error function to estimate these: $$ \Phi(\cfrac{\text{6ft 2in}\mu}{\sigma})\Phi(\cfrac{\text{4ft 9in}\mu}{\sigma}) $$ You'll need to convert to inches of course. http://en.wikipedia.org/wiki/Standard_normal_distribution#Cumulative_distribution Here is a calculator you can use: http://www.wolframalpha.com/input/?i=standard+normal+distribution+cdf+calculator
 7 months ago

StClowers Group TitleBest ResponseYou've already chosen the best response.0
so would that be 7462.8/2.5 minus 5762.8/2.5 4.48  2.32 = 2.16 Im stuck now...
 7 months ago
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