document.write( "Question 314444: 2. Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 °F and a standard deviation of 0.62 °F (based on data from the University of Maryland researchers).
\n" );
document.write( "a. If you have a body temperature of 99.00 °F, what is your percentile score?
\n" );
document.write( "b. Convert 99.00 °F to a standard score (or a z-score).
\n" );
document.write( "c. Is a body temperature of 99.00 °F unusual? Why or why not?
\n" );
document.write( "d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?
\n" );
document.write( "e. A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or why not? What should you conclude?
\n" );
document.write( "f. What body temperature is the 95th percentile?
\n" );
document.write( "g. What body temperature is the 5th percentile?
\n" );
document.write( "h. Bellevue Hospital in New York City uses 100.6 °F as the lowest temperature considered to indicate a fever. What percentage of normal and healthy adults would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6 °F is appropriate?\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #224974 by Fombitz(32388)![]() ![]() You can put this solution on YOUR website! a. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "91st percentile. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "b. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "c. Depends on what you consider normal, unusual, etc., you should make that determination and be able to defend your position. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "d. Set up a t test. \n" ); document.write( "The null hypothesis is that the means are the same. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Now the challenge is in working out the critical t value. \n" ); document.write( "t(0.05,49)=1.647 \n" ); document.write( "t(0.01,49)=2.405 \n" ); document.write( "Depending on the critical t value you choose, you could reject the null hypothesis, in which case, you would say the means are not the same, there is a statistically significant difference in the findings. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "e. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Yes, that would be odd because very few people would normally have that temperature. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "f. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "g. What body temperature is the 5th percentile? \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "h. \n" ); document.write( " \n" ); document.write( "Normally and healthy would be ones who have higher than 100.6 but are not sick. \n" ); document.write( " \n" ); document.write( "That's pretty low, so you would think it's pretty unlikely that someone coming in off the street would have a 100.6 temperature and be normal. The cutoff is probably appropriate. Again check the caveats in section c. \n" ); document.write( " |