Use z scores to compare the given values.Based on sample data, newborn males have weights with a mean of 3263.8 g and a standarddeviation of 841.2 g. Newborn females have weights with a mean of 3066.6 g and a standarddeviation of 570.2 g. Who has the weight that is more extreme relative to the group from whichthey came: a male who weighs 1500 g or a female who weighs 1500 g?Since the z score for the male is z =weight that is more extreme.(Round to two decimal places.)and the z score for the female is z =thehas thefemalemale

Based on sample data, newborn males have weights with a mean of = 3263.8 g and a standard deviation…

Answered: Use z scores to compare the given…

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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Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3282.4 g and a standard deviation of 704.5 g. Newborn females have weights with a mean of 3091.3 g and a standard deviation of 704.9 g. Who has the weight that is more extreme relative to the group from which they came: a male who weighs 1500 g or a female who weighs 1500 g? ..... Since the z score for the male is z= and the z score for the female is z = the has the weight that is more extreme. (Round to two decimal places.)
A person records the speeds of various cars as they drive past their house. After 100 readings, the speeds of the cars are normally distributed with a mean of 23.84 mph and a standard deviation of 3.56 mph. a. If a car's speed has a z-score of -2, what speed was that car going? b. Calculate the z-score for a car that is driving at 28 mph. c. One car is observed as driving at 30 mph, and another at 15 mph. Which of the two is more unusual?
i.
3. A certain breed of cat has weights which are normally distributed with a mean of 6.2 pounds, and a standard deviation of .8 pounds. What would be the z-score of a cat which weighed 7.4 pounds? 1.50 1.20 -1.50 -1.20
Calculate z scores from two samples The tallest living man at one time had a height of 252 cm. The shortest living man at that time had a height of 30.3 cm. Heights of men at that time had a mean of 170.25 cm and a standard deviation of 8.47 cm. Answer Z score for the tallest man Z-score for the shortest men
The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 119 and standard deviation of 21 .Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher.a. Around what percentage of adults in the USA have stage 2 high blood pressure?b. If you sampled 4500 people, how many would you expect to have BP> 160? Give your answer to the nearest person. peoplec. Stage 1 high BP is specified as systolic BP between 140 and 160. What percentage of adults in the US qualify for stage 1?
A particular group of men have heights with a mean of 181 cm and a standard deviation of 8 cm. Richard had a height of 193 cm. Convert Richard's height to a z score. The z score is (Type an integer or decimal rounded to two decimal places as needed.)
A grocer finds that the cantaloupes in his store have weights with mean 0.7 kg and standard deviation 53 grams. If this is true, then 99.7% of cantaloupes will have weights between how many grams?
The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.63 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.3 inches and a standard deviation of 2.54 inches. a. If a man is 6 feet 3 inches tall, what is his z-score (to 4 decimal places)? b. If a woman is 5 feet 11 inches tall, what is her z-score (to 4 decimal places)? Z = c. Who is relatively talle? O The 5 foot 11 inch American woman O The 6 foot 3 inch American man Submit Question
3. Use the following information about the heights measured in feet of a sample of 40 Ponderosa Pine trees and 40 Eastern White Pine trees. Ponderosa Pine Trees: Mean of 77.5 ft with a standard deviation of 11.55 ft Eastern White Pine Trees: Mean of 67.25 ft with a standard deviation of 10.33 ft Is it more unusual for a Ponderosa Pine tree to have a height of 71 ft or an Eastern White Pine tree to have a height of 71 ft relative to the data of each species of tree? Use z-scores to explain your solution.
Exam 1 scores have a mean of 500and a standard deviation of 110, while exam 2 scores have a mean of 20 and a standard deviation of 3. Assuming both types of scores have distributions that are unimodal and symmetric, which is more unusual: an exam 1 score of 790 or an exam 2 score of 28? A. The exam 2 score is more unusual. B. The exam 1 score is more unusual. C. They are the same. D. It cannot be determined which exam score is more unusual
The University of Montana admission standards require students to have an ACT score of at least 22. We know that Montana ACT scores are normally distributed with a mean of 20.1 and a standard deviation of 4. A: Students in Oregon have a mean ACT score of 21.1 with a standard deviation of 2.8. If we chose one student each at random from Oregon and from Montana, which student would be more likely to meet the ACT requirements of the University of Montana? Explain your answer using the appropriate statistics and a correctly labeled and shaded normal distribution. B: We know that the Montana ACT scores are normally distributed with a mean of 20.1 and a standard deviation of 4. Assume that the top 2% of individuals in the state are eligible for a state-funded scholarship. What score does an individual need on the ACT to be eligible? Explain your answer using the appropriate statistics and a correctly labeled and shaded normal distribution

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