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Do Homework - Jessica Dolan - Google Chrome Secure | https://www.mat nt/PlayerHomework.aspx?homeworkld=432476411&questionId;=22&flushed;=false&cld;=4521647&centerwi..; Q C ì secure l https://w..-O Q Math 152 Jessica Dolan | 12/6/17 11:14 PM Google Find the probability and interpret the res Homework: Review of Mods 1- 3 ave All mages Nows Shopping Score: 0 of 1 pt 10 of 35 (26 complete) Score: 65.95%, 23.08 of 35 pts About 17 results (1.10 seconds) and any subsequent words) was ignored beca. T 2 4 29 Question Help * SOLUTION: The population mean a www.algebra.com Probability and statistics Heights of men on a baseball team have a bell-shaped distribution with a mean of 167 cm and a standard deviation of 7 cm. Using the empirical rule A random sample of 3 specialists i drawn fron Probability-and-statiatios SOLUTION: The pops compliance specialists is about $62000. Wha is less than $60,000. Aaoume what is the approximate percentage of the men between the following values? a. 153 cm and 181 cm b. 160 cm and 174 cm PART 1. The Population Mean AnnL a. % of the men are between 153 cm and 181 cm. www.chegg com..... stetistics and probabili(Round to one decimal place as needed.) Rating: 1 00%-2 votes PART 1. The population mean annual salary fore A random sample of 39 specialists la drawn fron What is the probability thet the mean salary of th places aa Question: Find the probability and it www.chegg com·) statistics and probabilr Rating: 100%-1 vote Find the probability and interpret the results. If c population mean annual salary for environment sample of 33 specialists is drawn from chis popu the sample is less I have one other question if you are https:/www.justanswer.com Answers to Ho Jun 11, 2012 The population mean annual salar $63,000. A random sample of 42 specialists is d the mean salary of the sample is less than $59,5 salary of the Enter vour answer in the answer box and then click Check Answer. part remaining Clear All Check Answer O Type here to search 11:14 PM a ^ 4x 12/6/2017

Explanation / Answer

Mean = 167

Sd = 7

A) P(153 < X < 181) = P((153 - 167)/7 < Z < (181 - 167)/7)

= P(-2 < Z < 2)

= 95%

B) P(160 < X < 174) = P((160 - 167)/7 < Z < (174 - 167)/7)

= P(-1 < Z < 1)

= 68%

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