The lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and astandard deviation of 13 days. A distribution of values is normal with a mean of 262 and a standard deviationof 13.What percentage of pregnancies last fewer than 223 days?P(X< 223 days) =Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained usingexact z-scores or z-scores rounded to 3 decimal places are accepted.

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Answered: The lengths of pregnancies in a small…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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In Math 150 class the points of the final exam are normally distributedWith a mean of 72 and standard deviation of 10. The top 15% will receive an A gradeFind the lowest score in the final exam that would qualify a student for an A grade.
A distribution of values is normal with a mean of 120 and a standard deviation of 4.Find the interval containing the middle-most 68% of scores: Enter your answer using interval notation. Example: (2.1,5.6)
A distribution of values is normal with a mean of 197.6 and a standard deviation of 24.6.Find P40, which is the score separating the bottom 40% from the top 60 P40 =
The SAT scores of students are normally distributed with a mean of 950 and a standard deviation of 200. Q19 What percentage of students score between 1100 and 1200? Your answer should be in % and have two digits after the period. So, if your answer is 50.723%, write 50.72. Your Answer:
A distribution of values is normal with a mean of 185.8 and a standard deviation of 24.3.Find P8, which is the score separating the bottom 8% from the top 92%.P8 =
IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the minimum IQ required to be able to correctly claim to be in the "top 5%" of IQ scores. Round to the nearest whole number.
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 28 and the age of the winner for best actress was 44. For all best actors, the mean age is 42.3 years and the standard deviation is 8.4 years. For all best actresses, the mean age is 33.3 years and the standard deviation is 10.7 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z = and the z score for the actress is z=, the had the more extreme age. (Round to two decimal places.)
A distribution of values is normal with a mean of 38.3 and a standard deviation of 84.2.Find P98, which is the score separating the bottom 98% from the top 2%.P98 =
A distribution of values is normal with a mean of 116.2 and a standard deviation of 60.1. Find P43, which is the score separating the bottom 43% from the top 57%. P43 = Enter your answer as a number accurate to 4 decimal places.
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 27 and the age of the winner for best actress were 47. For all best actors, the mean age is 42.7 years and the standard deviation is 7.7 years. For all best actresses, the mean age is 32.9 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z= and the z score for the actress is z=,the▼actress actor had the more extreme age.
2000 freshmen at state university took a biology test. the scores were normally distributed with a mean of 70 and a standard deviation of 5. Use z scores. (1) the percent of students that scored less than 77
z Scores LeBron James, one of the most successful basketball players of all time, has a height of 6 feet 8 inches, or 203 cm. Based on statistics from Data Set 1 “Body Data” in Appendix B, his height converts to the z score of 4.07. How many standard deviations is his height above the mean?

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